Factoring PPT.pptx FOR ALL TYPE OF STUDENT

Factoring
Polynomials &
Trinomials
To factorize polynomial and trinomial functions

Factoring
•Distributive Property:
• ab + ac = a(b+c)
•Difference of two squares:
• A2
− b2
= (a–b) * (a+b)
•Sum of two cubes:
• A3
+ b3
= (a + b) * (a2
− ab + b2
)
•Difference of two cubes:
• a3
− b3
= (a − b) * (a2
+ ab + b2
)
• To factorize polynomial and trinomial functions

Factoring Trinomials of the Form x2
+ bx + c

Factoring Trinomials
Recall by using method
(x + 2)(x + 4) = x2
+ 4x + 2x + 8
= x2
+ 6x + 8
To factor x2
+ bx + c into (x + one #)(x + another #), note that b is
the sum of the two numbers and c is the product of the two numbers.
So, we’ll be looking for 2 numbers whose product is c and whose sum
is b.
Note: there are fewer choices for the product, so that’s why we start
there first.

Factor the polynomial x2
+ 13x + 30.
Factoring Polynomials
Example
a=1
b=1
c=-9
=
X= = 2.54
x == -3.54

+ 13x + 30.
Since our two numbers must have a product of 30 and a
sum of 13, the two numbers must both be positive.
Positive factors of 9 Sum of Factors
1, 30 31
2, 15 17
3, 10 13
Note, there are other factors, but once we find a pair
that works, we do not have to continue searching.
So x2
+ 13x + 30 = (x + 3)(x + 10).
Example

– 11x + 24.
Example

– 11x + 24.
sum of -11, the two numbers must both be negative.
Negative factors of 24 Sum of Factors
– 1, – 24 – 25
– 2, – 12 – 14
– 3, – 8 – 11
So x2
– 11x + 24 = (x – 3)(x – 8).
Example

– 2x – 35.
Example

– 2x – 35.
Since our two numbers must have a product of – 35 and a sum
of – 2, the two numbers will have to have different signs.
Factors of – 35 Sum of Factors
– 1, 35 34
1, – 35 – 34
– 5, 7 2
5, – 7 – 2
So x2
– 2x – 35 = (x + 5)(x – 7).
Example

– 6x + 10.
Prime Polynomials
Example

– 6x + 10.
sum of – 6, the two numbers will have to both be negative.
Negative factors of 10 Sum of Factors
– 1, – 10 – 11
– 2, – 5 – 7
Since there is not a factor pair whose sum is – 6,
x2
– 6x +10 is not factorable and we call it a prime
polynomial.
Prime Polynomials
Example

Factoring Trinomials of the Form ax2
+ bx + c

Factoring Trinomials
(3x + 2)(x + 4) = 3x2
+ 12x + 2x + 8
= 3x2
+ 14x + 8
To factor ax2
+ bx + c into (#1·x + #2)(#3·x + #4),
note that a is the product of the two first coefficients,
c is the product of the two last coefficients and b is
the sum of the products of the outside coefficients
and inside coefficients.
Note that b is the sum of 2 products, not just 2
numbers, as in the last section.

Factor the polynomial 25x2
+ 20x + 4.
Possible factors of 25x2
are {x, 25x} or {5x, 5x}.
Possible factors of 4 are {1, 4} or {2, 2}.
We need to methodically try each pair of factors until we find a combination that works, or
exhaust all of our possible pairs of factors.
Example

We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 20x.
{x, 25x} {1, 4} (x + 1)(25x + 4) 4x + 25x 29x
(x + 4)(25x + 1) x + 100x 101x
{x, 25x} {2, 2} (x + 2)(25x + 2) 2x + 50x 52x
Factors of
25x2
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
Factors of
4
{5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x + 10x 20x
Example Continued

– 41x + 10.
Example

– 41x + 10.
Possible factors of 21x2
are {x, 21x} or {3x, 7x}.
Since the middle term is negative, possible factors of 10
must both be negative: {-1, -10} or {-2, -5}.
We need to methodically try each pair of factors until we
find a combination that works, or exhaust all of our
possible pairs of factors.
Example

We will be looking for a combination that gives the sum of
the products of the outside terms and the inside terms equal
to 41x.
Factors of
21x2
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
Factors of
10
{x, 21x}{1, 10}(x – 1)(21x – 10) –10x 21x – 31x
(x – 10)(21x – 1) –x 210x – 211x
{x, 21x} {2, 5} (x – 2)(21x – 5) –5x 42x – 47x
(x – 5)(21x – 2) –2x 105x – 107x
(3x – 5)(7x – 2) 6x + 35x = 41x
Example Continued

– 7x + 6.
Example

– 7x + 6.
The only possible factors for 3 are 1 and 3, so we know that, if factorable, the polynomial
will have to look like (3x )(x ) in factored form, so that the product of the first
two terms in the binomials will be 3x2
.
Since the middle term is negative, possible factors of 6 must both be negative: {1,  6} or
{ 2,  3}.
We need to methodically try each pair of factors until we find a combination that works,
or exhaust all of our possible pairs of factors.
Example

We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 7x.
{1, 6} (3x – 1)(x – 6) 18x x 19x
(3x – 6)(x – 1) Common factor so no need to test.
{2, 3} (3x – 2)(x – 3) 9x 2x 11x
(3x – 3)(x – 2) Common factor so no need to test.
Factors of
6
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
Example Continued

Now we have a problem, because we have
exhausted all possible choices for the factors but
have not found a pair where the sum of the
products of the outside terms and the inside
terms is –7.
So 3x2
– 7x + 6 is a prime polynomial and will
not factor.
Example Continued

y2
– 2xy2
– 60y2
.
Example

y2
– 2xy2
– 60y2
.
Remember that the larger the coefficient, the greater the
probability of having multiple pairs of factors to check. So
it is important that you attempt to factor out any common
factors first.
6x2
y2
– 2xy2
– 60y2
= 2y2
(3x2
– x – 30)
The only possible factors for 3 are 1 and 3, so we know that,
if we can factor the polynomial further, it will have to look
like 2y2
(3x )(x ) in factored form.
Example

Since the product of the last two terms of the binomials
will have to be –30, we know that they must be different
signs.
Possible factors of –30 are {–1, 30}, {1, –30}, {–2, 15},
{2, –15}, {–3, 10}, {3, –10}, {–5, 6} or {5, –6}.
We will be looking for a combination that gives the sum
of the products of the outside terms and the inside terms
equal to –x.
Example Continued

Factors of
-30
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
{-1, 30} (3x – 1)(x + 30) 90x -x 89x
(3x + 30)(x – 1) Common factor so no need to test.
{1, -30} (3x + 1)(x – 30) -90x x -89x
(3x – 30)(x + 1) Common factor so no need to test.
{-2, 15} (3x – 2)(x + 15) 45x -2x 43x
(3x + 15)(x – 2) Common factor so no need to test.
{2, -15} (3x + 2)(x – 15) -45x 2x -43x
(3x – 15)(x + 2) Common factor so no need to test.
Example Continued

Factors of
–30
Resulting
Binomials
Product of Outside
Terms
Product of Inside
Terms
Sum of
Products
{–3, 10} (3x – 3)(x + 10) Common factor so no need to test.
(3x + 10)(x – 3) –9x 10x x
{3, –10} (3x + 3)(x – 10) Common factor so no need to test.
(3x – 10)(x + 3) 9x –10x –x
Example Continued

Factoring Trinomials of the Form x2
+ bx + c by
Grouping

Factoring polynomials often involves additional
techniques after initially factoring out the GCF.
One technique is factoring by grouping.
Factor xy + y + 2x + 2 by grouping.
Notice that, although 1 is the GCF for all four
terms of the polynomial, the first 2 terms have a
GCF of y and the last 2 terms have a GCF of 2.
xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 =
y(x + 1) + 2(x + 1) = (x + 1)(y + 2)
Factoring by Grouping
Example

1) x3
+ 4x + x2
+ 4
2) 2) 2x3
– x2
– 10x + 5
Factor each of the following polynomials by grouping.
Example

1) x3
+ 4x + x2
+ 4 = x · x2
+ x · 4 + 1 · x2
+ 1 · 4 =
x(x2
+ 4) + 1(x2
+ 4) =
(x2
+ 4)(x + 1)
2) 2x3
– x2
– 10x + 5 = x2
· 2x – x2
· 1 – 5 · 2x – 5 · (– 1) =
x2
(2x – 1) – 5(2x – 1) =
(2x – 1)(x2
– 5)
Factor each of the following polynomials by grouping.
Example

Factor 2x – 9y + 18 – xy by grouping.
Example

Factor 2x – 9y + 18 – xy by grouping.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2x + 18 – 9y – xy = 2 · x + 2 · 9 – 9 · y – x · y =
2(x + 9) – y(9 + x) =
2(x + 9) – y(x + 9) = (make sure the factors are identical)
(x + 9)(2 – y)
Example

Factoring Perfect Square Trinomials and the Difference of
Two Squares

In the last chapter we learned a shortcut for squaring a
binomial
(a + b)2
= a2
+ 2ab + b2
(a – b)2
= a2
– 2ab + b2
So, if the first and last terms of our polynomial to be
factored are can be written as expressions squared, and the
middle term of our polynomial is twice the product of
those two expressions, then we can use these two previous
equations to easily factor the polynomial.
a2
+ 2ab + b2
=(a + b)2
a2
– 2ab + b2
= (a – b)2
Perfect Square Trinomials

Difference of Two Squares
Another shortcut for factoring a trinomial is when we
want to factor the difference of two squares.
a2
– b2
= (a + b)(a – b)
A binomial is the difference of two square if
1.both terms are squares and
2.the signs of the terms are different.
9x2
– 25y2
– c4
+ d4

Difference of Two Squares
Example
– 9.
The first term is a square and the last term, 9, can be
written as 32
. The signs of each term are different, so
we have the difference of two squares
Therefore x2
– 9 = (x – 3)(x + 3).
Note: You can use FOIL method to verify that the
factorization for the polynomial is accurate.

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