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Notes solving polynomials using synthetic division

L
Lori Rapp

This document discusses using synthetic division to evaluate polynomials at specific values and factor polynomials. It provides examples of using synthetic division to: 1) Evaluate polynomials like f(x) = x^2 - x + 5 at specific values such as f(-2) 2) Factor polynomials when one factor is known, such as factoring x^3 - 3x^2 - 13x + 15 after determining (x + 3) is a factor 3) Find all zeros of a polynomial by setting each factor equal to 0 after factoring Step-by-step instructions and additional examples are provided to illustrate the process.

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Solving Polynomials Using Synthetic Division
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial.
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value.
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual.
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual. −2 1 −1 5 −2 6 1 −3 11
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual. • The remainder is the value. −2 1 −1 5 −2 6 1 −3 11
Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual. • The remainder is the value. −2 1 −1 5 • So −2 6 f ( −2 ) = 11 1 −3 11
Try it: Simplify using Synthetic Substitution Find g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2
Try it: Simplify using Synthetic Substitution Find g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2 5 −1 1 −7 8 −5 −20 −135 −1 −4 −27 −127
Try it: Simplify using Synthetic Substitution Find g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2 5 −1 1 −7 8 −5 −20 −135 −1 −4 −27 −127 Therefore, g ( 5 ) = −127.
Try it: Simplify using Synthetic Substitution Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2
Try it: Simplify using Synthetic Substitution Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2 −3 1 −2 −3 5 −6 −3 15 −36 93 1 −5 12 −31 87
Try it: Simplify using Synthetic Substitution Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2 −3 1 −2 −3 5 −6 −3 15 −36 93 1 −5 12 −31 87 Therefore, f ( −3) = 87.
Try it: Simplify using Synthetic Substitution Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2 Don’t be intimidated by the 2b. Use the same process as before.
Try it: Simplify using Synthetic Substitution Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2 Don’t be intimidated by the 2b. Use the same process as before. 2b 1 3 −2 1 2b 4b 2 + 6b 8b 3 + 12b 2 − 4b 2 3 2 1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1
Try it: Simplify using Synthetic Substitution Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2 Don’t be intimidated by the 2b. Use the same process as before. 2b 1 3 −2 1 2b 4b 2 + 6b 8b 3 + 12b 2 − 4b 2 3 2 1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1 Therefore, h ( 2b ) = 8b 3 + 12b 2 − 4b + 1.
Find Factors given a Polynomial & one factor • Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods.
Find Factors given a Polynomial & one factor • Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods. • It’s a great way to ‘break down’ the big, nasty looking polynomials.
Find Factors given a Polynomial & one factor • Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods. • It’s a great way to ‘break down’ the big, nasty looking polynomials. • Care to factor this polynomial? 3 2 x − 3x − 13x + 15
Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors.
Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors. • First step is to divide by the factor you know.
Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors. • First step is to divide by the factor you know. −3 1 −3 −13 15 −3 18 −15 1 −6 5 0
Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors. • First step is to divide by the factor you know. −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Remainder is zero so this indicates that (x + 3) is indeed a zero of the polynomial.
Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division.
Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division. 2 x − 6x + 5
Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division. 2 x − 6x + 5 • Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic.
Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division. 2 x − 6x + 5 • Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic. ( x − 5 ) ( x − 1)
Continued... 3 2 • So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1)
Continued... 3 2 • So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1) • When writing the factorization, be sure to include the first factor you were given because without it, you do not have the original polynomial!
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0.
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve.
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5 x =1
Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5 x =1 The zeros are {-3, 1, 5}.
Your turn... Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2
Your turn... Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2 3 1 4 −15 −18 3 21 18 1 7 6 0 Remainder is 0, therefor x - 3 is a factor.
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor.
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 Write the new polynomial.
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like.
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like. ( x + 6 ) ( x + 1)
Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like. ( x + 6 ) ( x + 1) x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve.
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6 x = −1
And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6 x = −1 The zeros are {-6, -1, 3}.
Practice some more... • Follow this link to practice using synthetic division to find the zeros of the polynomials.
Rational Roots Theorem • Watch this video to learn more about the Rational Roots Theorem and how to use it to find any possible rational roots of a polynomial. • Aquí está una versión española del vídeo racional del teorema de las raíces.
Factoring when you don’t know a factor
Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor.
Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor. • The video you just watched showed you how to determine if a polynomial has a rational root.
Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor. • The video you just watched showed you how to determine if a polynomial has a rational root. • Now we will use this test to find all factors of a polynomial.
Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor. • The video you just watched showed you how to determine if a polynomial has a rational root. • Now we will use this test to find all factors of a polynomial. • Watch this video to see how the Rational Roots Theorem is applied to find the solutions of a polynomial. • Aquí está una versión española para aplicar el teorema racional de las raíces para encontrar las soluciones de un polinomio.
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient.
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term.
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the factors of the leading coefficient. No need to repeat numbers and always simplify.
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the ±1, ±2, ±5, ±10 factors of the leading coefficient. No need to repeat numbers and always simplify.
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the ±1, ±2, ±5, ±10 factors of the leading coefficient. No need to repeat numbers and always simplify. Lots from which to choose but lets start with an easy one.
Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the ±1, ±2, ±5, ±10 factors of the leading coefficient. No need to repeat numbers and always simplify. f ( x ) = x 3 + 6x 2 + 3x − 10 Lots from which to choose but lets start with an easy one. 3 2 f (1) = (1) + 6 (1) + 3(1) − 10 = 0
Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial.
Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0
Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0 Write the new polynomial.
Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0 Write the new polynomial. 2 x + 7x + 10 = 0
Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0 Write the new polynomial. 2 x + 7x + 10 = 0 Look at the trinomial. Can it be factored?
Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor.
Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0
Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve.
Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve. ( x + 5) = 0 x = −5
Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve. ( x + 5) = 0 ( x + 2) = 0 x = −5 x = −2
Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve. ( x + 5) = 0 ( x + 2) = 0 x = −5 x = −2 Solutions: {−5, −2,1}
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient.
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term.
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28 Possible roots are leading coefficient factors over constant term factors without repeats.
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28 Possible roots are leading coefficient factors over constant term factors without repeats. ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5
Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28 Possible roots are leading coefficient factors over constant term factors without repeats. ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Wow, that’s a lot! Not to worry. Start with the smallest positive integer and look for a pattern to help find a root.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6 No good again because but notice our remainder is getting larger. This indicates we are moving away from 0 so we are going in the wrong direction. Try a negative number.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6 No good again because but notice our remainder is getting larger. This indicates we are moving away from 0 so we are going in the wrong direction. Try a negative number. 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6 No good again because but notice our remainder is getting larger. This indicates we are moving away from 0 so we are going in the wrong direction. Try a negative number. 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Now it’s way too small. Remember we are try to get 0.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1. Yes, that means we now get to try a fraction!
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1. Yes, that means we now get to try a fraction! 3 2  1  1  1  1 f   = 5   − 29   + 55   − 28 = −243 25  5  5  5  5
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1. Yes, that means we now get to try a fraction! 3 2  1  1  1  1 f   = 5   − 29   + 55   − 28 = −243 25  5  5  5  5 No good but it’s negative. This means the root is between 1/5 and 1.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0 Write the resulting polynomial.
Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0 Write the resulting polynomial. 5x 2 − 25x + 35 = 0
Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first!
Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0
Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0 Because this is an equation, divide by 5 to simplify.
Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0 Because this is an equation, divide by 5 to simplify. 2 x − 5x + 7 = 0
Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0 Because this is an equation, divide by 5 to simplify. 2 x − 5x + 7 = 0 The trinomial isn’t easily factored so use the quadratic formula or complete the square to find the solution.
Continued... x 2 − 5x + 7 = 0 The quadratic formula will be the better option because the middle term is odd and will result in a fraction when completing the square.
Continued... x 2 − 5x + 7 = 0 The quadratic formula will be the better option because the middle term is odd and will result in a fraction when completing the square. a = 1; b = −5; c = 7 2 2 −b ± b − 4ac 5 ± ( −5 ) − 4 ⋅1⋅ 7 5 ± 25 − 28 x= = = 2a 2 ⋅1 2 5 ± −3 5 ± i 3 x= = 2 2
Continued... 5x 3 − 29x 2 + 55x − 28 = 0 Our original polynomial was degree 3, which indicates 3 solutions. We found 3 solutions although only 1 solutions was real.
Continued... 5x 3 − 29x 2 + 55x − 28 = 0 Our original polynomial was degree 3, which indicates 3 solutions. We found 3 solutions although only 1 solutions was real. The solutions to the equation are...
Continued... 5x 3 − 29x 2 + 55x − 28 = 0 Our original polynomial was degree 3, which indicates 3 solutions. We found 3 solutions although only 1 solutions was real. The solutions to the equation are...  4 5 + i 3 5 − i 3    5, ,    2 2  
The end.

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Notes solving polynomials using synthetic division

  • 1. Solving Polynomials Using Synthetic Division
  • 2. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial.
  • 3. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value.
  • 4. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2
  • 5. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual.
  • 6. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual. −2 1 −1 5 −2 6 1 −3 11
  • 7. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual. • The remainder is the value. −2 1 −1 5 −2 6 1 −3 11
  • 8. Simplify Polynomials using Synthetic Substitution • Recall Synthetic Division can be used to divide a polynomial by a binomial. • Synthetic Division can also be used to evaluate a polynomial at a specific value. • Such as given f ( x ) = x − x + 5 , find f ( −2 ) 2 • The x value goes in the box and the coefficients of the polynomial make up the first line. Proceed as usual. • The remainder is the value. −2 1 −1 5 • So −2 6 f ( −2 ) = 11 1 −3 11
  • 9. Try it: Simplify using Synthetic Substitution Find g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2
  • 10. Try it: Simplify using Synthetic Substitution Find g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2 5 −1 1 −7 8 −5 −20 −135 −1 −4 −27 −127
  • 11. Try it: Simplify using Synthetic Substitution Find g ( 5 ) for g ( x ) = −x + x − 7x + 8 3 2 5 −1 1 −7 8 −5 −20 −135 −1 −4 −27 −127 Therefore, g ( 5 ) = −127.
  • 12. Try it: Simplify using Synthetic Substitution Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2
  • 13. Try it: Simplify using Synthetic Substitution Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2 −3 1 −2 −3 5 −6 −3 15 −36 93 1 −5 12 −31 87
  • 14. Try it: Simplify using Synthetic Substitution Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6 4 3 2 −3 1 −2 −3 5 −6 −3 15 −36 93 1 −5 12 −31 87 Therefore, f ( −3) = 87.
  • 15. Try it: Simplify using Synthetic Substitution Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2 Don’t be intimidated by the 2b. Use the same process as before.
  • 16. Try it: Simplify using Synthetic Substitution Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2 Don’t be intimidated by the 2b. Use the same process as before. 2b 1 3 −2 1 2b 4b 2 + 6b 8b 3 + 12b 2 − 4b 2 3 2 1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1
  • 17. Try it: Simplify using Synthetic Substitution Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1 3 2 Don’t be intimidated by the 2b. Use the same process as before. 2b 1 3 −2 1 2b 4b 2 + 6b 8b 3 + 12b 2 − 4b 2 3 2 1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1 Therefore, h ( 2b ) = 8b 3 + 12b 2 − 4b + 1.
  • 18. Find Factors given a Polynomial & one factor • Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods.
  • 19. Find Factors given a Polynomial & one factor • Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods. • It’s a great way to ‘break down’ the big, nasty looking polynomials.
  • 20. Find Factors given a Polynomial & one factor • Synthetic Division can be useful in factoring polynomials that cannot be factored using traditional methods. • It’s a great way to ‘break down’ the big, nasty looking polynomials. • Care to factor this polynomial? 3 2 x − 3x − 13x + 15
  • 21. Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors.
  • 22. Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors. • First step is to divide by the factor you know.
  • 23. Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors. • First step is to divide by the factor you know. −3 1 −3 −13 15 −3 18 −15 1 −6 5 0
  • 24. Factoring when know 1 factor 3 2 x − 3x − 13x + 15 • Say we know one factor of this nasty polynomial is (x + 3) or the zero of the function is x = -3. We can use synthetic division to help us find the remaining factors. • First step is to divide by the factor you know. −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Remainder is zero so this indicates that (x + 3) is indeed a zero of the polynomial.
  • 25. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division.
  • 26. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division. 2 x − 6x + 5
  • 27. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division. 2 x − 6x + 5 • Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic.
  • 28. Continued... −3 1 −3 −13 15 −3 18 −15 1 −6 5 0 • Write the new polynomial from your synthetic division. 2 x − 6x + 5 • Look, we now have a nice quadratic. Use any method you’ve learned to factor the quadratic. ( x − 5 ) ( x − 1)
  • 29. Continued... 3 2 • So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1)
  • 30. Continued... 3 2 • So the factorization of x − 3x − 13x + 15 is ( x + 3) ( x − 5 ) ( x − 1) • When writing the factorization, be sure to include the first factor you were given because without it, you do not have the original polynomial!
  • 31. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0.
  • 32. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1)
  • 33. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve.
  • 34. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0
  • 35. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3
  • 36. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5
  • 37. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5 x =1
  • 38. Continued... 3 2 Now find the zeros of x − 3x − 13x + 15 = 0. Remember the factors are ... ( x + 3) ( x − 5 ) ( x − 1) Set each factor to 0 and solve. x+3= 0 x−5=0 x −1= 0 x = −3 x=5 x =1 The zeros are {-3, 1, 5}.
  • 39. Your turn... Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2
  • 40. Your turn... Show ( x − 3) is a factor of x + 4x − 15x − 18 3 2 3 1 4 −15 −18 3 21 18 1 7 6 0 Remainder is 0, therefor x - 3 is a factor.
  • 41. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor.
  • 42. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0
  • 43. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 Write the new polynomial.
  • 44. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6
  • 45. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like.
  • 46. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like. ( x + 6 ) ( x + 1)
  • 47. Now take the last problem a little further... 3 2 Factor x + 4x − 15x − 18 Remember you were given a factor. 3 1 4 −15 −18 3 21 18 1 7 6 0 2 Write the new polynomial. x + 7x + 6 It’s a quadratic. Factor using any method you like. ( x + 6 ) ( x + 1) x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2
  • 48. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18
  • 49. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2
  • 50. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve.
  • 51. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0
  • 52. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3
  • 53. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6
  • 54. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6 x = −1
  • 55. And a little further... 3 2 Find the zeros of x + 4x − 15x − 18 Remember ... x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1) 3 2 Set each factor to 0 and solve. x−3= 0 x+6=0 x +1= 0 x=3 x = −6 x = −1 The zeros are {-6, -1, 3}.
  • 56. Practice some more... • Follow this link to practice using synthetic division to find the zeros of the polynomials.
  • 57. Rational Roots Theorem • Watch this video to learn more about the Rational Roots Theorem and how to use it to find any possible rational roots of a polynomial. • Aquí está una versión española del vídeo racional del teorema de las raíces.
  • 58. Factoring when you don’t know a factor
  • 59. Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor.
  • 60. Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor. • The video you just watched showed you how to determine if a polynomial has a rational root.
  • 61. Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor. • The video you just watched showed you how to determine if a polynomial has a rational root. • Now we will use this test to find all factors of a polynomial.
  • 62. Factoring when you don’t know a factor • The previous problems weren’t too bad because you knew a factor. • The video you just watched showed you how to determine if a polynomial has a rational root. • Now we will use this test to find all factors of a polynomial. • Watch this video to see how the Rational Roots Theorem is applied to find the solutions of a polynomial. • Aquí está una versión española para aplicar el teorema racional de las raíces para encontrar las soluciones de un polinomio.
  • 63. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0
  • 64. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient.
  • 65. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1
  • 66. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term.
  • 67. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10
  • 68. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the factors of the leading coefficient. No need to repeat numbers and always simplify.
  • 69. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the ±1, ±2, ±5, ±10 factors of the leading coefficient. No need to repeat numbers and always simplify.
  • 70. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the ±1, ±2, ±5, ±10 factors of the leading coefficient. No need to repeat numbers and always simplify. Lots from which to choose but lets start with an easy one.
  • 71. Try one: Solve using the Rational Roots Theorem 3 2 x + 6x + 3x − 10 = 0 Find factors of leading coefficient. ±1 Find factors of constant term. ±1, ±2, ±5, ±10 Possible roots are made up of the constant term factors over the ±1, ±2, ±5, ±10 factors of the leading coefficient. No need to repeat numbers and always simplify. f ( x ) = x 3 + 6x 2 + 3x − 10 Lots from which to choose but lets start with an easy one. 3 2 f (1) = (1) + 6 (1) + 3(1) − 10 = 0
  • 72. Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial.
  • 73. Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0
  • 74. Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0 Write the new polynomial.
  • 75. Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0 Write the new polynomial. 2 x + 7x + 10 = 0
  • 76. Continued... Solution was 0 so we lucked out on our first try. Use Synthetic Division to simplify the polynomial. 1 1 6 3 −10 1 7 10 1 7 10 0 Write the new polynomial. 2 x + 7x + 10 = 0 Look at the trinomial. Can it be factored?
  • 77. Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor.
  • 78. Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0
  • 79. Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve.
  • 80. Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve. ( x + 5) = 0 x = −5
  • 81. Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve. ( x + 5) = 0 ( x + 2) = 0 x = −5 x = −2
  • 82. Continued... 2 x + 7x + 10 = 0 Yes! It’s a factorable quadratic so use any method you would like to factor. ( x + 5 )( x + 2 ) = 0 Now set each factor to 0 and solve. ( x + 5) = 0 ( x + 2) = 0 x = −5 x = −2 Solutions: {−5, −2,1}
  • 83. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0
  • 84. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient.
  • 85. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5
  • 86. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term.
  • 87. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28
  • 88. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28 Possible roots are leading coefficient factors over constant term factors without repeats.
  • 89. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28 Possible roots are leading coefficient factors over constant term factors without repeats. ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5
  • 90. Try another: Solve using the Rational Roots Theorem 3 2 5x − 29x + 55x − 28 = 0 Find factors of leading coefficient. ±1, ±5 Find factors of constant term. ±1, ±2, ±4, ±7, ±14, ±28 Possible roots are leading coefficient factors over constant term factors without repeats. ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Wow, that’s a lot! Not to worry. Start with the smallest positive integer and look for a pattern to help find a root.
  • 91. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1.
  • 92. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3
  • 93. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer.
  • 94. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6
  • 95. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6 No good again because but notice our remainder is getting larger. This indicates we are moving away from 0 so we are going in the wrong direction. Try a negative number.
  • 96. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6 No good again because but notice our remainder is getting larger. This indicates we are moving away from 0 so we are going in the wrong direction. Try a negative number. 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117
  • 97. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 Start with 1. 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 Didn’t work so try the next highest integer. 3 2 f ( 2 ) = 5 ( 2 ) − 29 ( 2 ) + 55 ( 2 ) − 28 = 6 No good again because but notice our remainder is getting larger. This indicates we are moving away from 0 so we are going in the wrong direction. Try a negative number. 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Now it’s way too small. Remember we are try to get 0.
  • 98. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1.
  • 99. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1. Yes, that means we now get to try a fraction!
  • 100. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1. Yes, that means we now get to try a fraction! 3 2  1  1  1  1 f   = 5   − 29   + 55   − 28 = −243 25  5  5  5  5
  • 101. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2 f (1) = 5 (1) − 29 (1) + 55 (1) − 28 = 3 3 2 f ( −1) = 5 ( −1) − 29 ( −1) + 55 ( −1) − 28 = −117 Because 0 falls between 3 and -117, we know the root must fall between -1 and 1. Yes, that means we now get to try a fraction! 3 2  1  1  1  1 f   = 5   − 29   + 55   − 28 = −243 25  5  5  5  5 No good but it’s negative. This means the root is between 1/5 and 1.
  • 102. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5
  • 103. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial.
  • 104. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0
  • 105. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0 Write the resulting polynomial.
  • 106. Continued... ±1, ±2, ±4, ±7, ±14, ±28 1 2 4 7 14 28 f ( x ) = 5x 3 − 29x 2 + 55x − 28 ± ,± ,± ,± ,± ,± 5 5 5 5 5 5 3 2  4  4  4  4 f   = 5   − 29   + 55   − 28 = 0  5  5  5  5 Found it! So now use Synthetic Division to simplify the polynomial. 4 5 5 −29 55 −28 4 −20 28 5 −25 35 0 Write the resulting polynomial. 5x 2 − 25x + 35 = 0
  • 107. Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first!
  • 108. Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0
  • 109. Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0 Because this is an equation, divide by 5 to simplify.
  • 110. Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0 Because this is an equation, divide by 5 to simplify. 2 x − 5x + 7 = 0
  • 111. Continued... 5x 2 − 25x + 35 = 0 Factor the trinomial using any method but always look for a GCF first! 5 ( x 2 − 5x + 7 ) = 0 Because this is an equation, divide by 5 to simplify. 2 x − 5x + 7 = 0 The trinomial isn’t easily factored so use the quadratic formula or complete the square to find the solution.
  • 112. Continued... x 2 − 5x + 7 = 0 The quadratic formula will be the better option because the middle term is odd and will result in a fraction when completing the square.
  • 113. Continued... x 2 − 5x + 7 = 0 The quadratic formula will be the better option because the middle term is odd and will result in a fraction when completing the square. a = 1; b = −5; c = 7 2 2 −b ± b − 4ac 5 ± ( −5 ) − 4 ⋅1⋅ 7 5 ± 25 − 28 x= = = 2a 2 ⋅1 2 5 ± −3 5 ± i 3 x= = 2 2
  • 114. Continued... 5x 3 − 29x 2 + 55x − 28 = 0 Our original polynomial was degree 3, which indicates 3 solutions. We found 3 solutions although only 1 solutions was real.
  • 115. Continued... 5x 3 − 29x 2 + 55x − 28 = 0 Our original polynomial was degree 3, which indicates 3 solutions. We found 3 solutions although only 1 solutions was real. The solutions to the equation are...
  • 116. Continued... 5x 3 − 29x 2 + 55x − 28 = 0 Our original polynomial was degree 3, which indicates 3 solutions. We found 3 solutions although only 1 solutions was real. The solutions to the equation are...  4 5 + i 3 5 − i 3    5, ,    2 2  

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