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Ms. G. Martin
To solve square of trinomial use the formula of
F2 + M2 + L2 + 2FM + 2FL + 2ML
Where,
F2 = square the first term
M2 = square middle term
L2 = square the last term
2FM = twice the product of the first and second term
2FL = twice the product of the first and last term
2ML = twice the product of the middle term and the last term
Example 1. Find the product of (a + b + c)2 .
Solution:
Using the formula of
F2 + M2 + L2 + 2FM + 2FL +
2ML
F2 = (a)2 = a2
M2 = (b)2 = b2
L2 = (c)2 = c2
2FM = 2(a . b) = 2ab
2FL = 2(a. c) = 2ac
2ML = 2(b . c) = 2bc
= a2 + b2 + c2 + 2ab + 2ac + 2bc
Example 2. Find the product of (x + y – 3)2 .
Solution:
Using the formula of
F2 + M2 + L2 + 2FM + 2FL +
2ML
F2 = (x)2 = x2
M2 = (y)2 = y2
L2 = (-3)2 = 9
2FM = 2(x. y) = 2xy
2FL = 2(x . -3) = -6x
2ML = 2(y . -3) = -6y
= x2 + y2 + 9 + 2xy – 6x – 6y
= x2 + 2xy + y2 – 6x – 6y + 9
Example 3. Find the product of (x – 2y – 1)2.
Solution:
Using the formula of
F2 + M2 + L2 + 2FM + 2FL +
2ML
F2 = (x)2 = x2
M2 = (-2y)2 = 4y2
L2 = (-1)2 = 1
2FM = 2(x. -2y) = -4xy
2FL = 2(x -1) = -2x
2ML = 2(-2y . -1) = 4y
= x2 + 4y2 + 1 – 4xy – 2x + 4y
= x2 -4xy + 4y2 – 2x + 4y + 1
Example 4. Solve (2a – b + b2)2.
Solution:
Using the formula of
F2 + M2 + L2 + 2FM + 2FL +
2ML
F2 = (2a)2 = 4a2
M2 = (-b)2 = b2
L2 = (b2)2 = b4
2FM = 2(2a . –b) = -4ab
2FL = 2(2a . b2) = 4ab2
2ML = 2(-b . b2) = -2b3
= 4a2 + b2 + b4 – 4ab + 4ab2 – 2b3
= 4ab2 +4a2– 4ab + b4 - 2b3 + b2
THANK YOU!
SMSDGRADE8-MT.
FUJI@ODL2020

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Square of trinomial

  • 2. To solve square of trinomial use the formula of F2 + M2 + L2 + 2FM + 2FL + 2ML Where, F2 = square the first term M2 = square middle term L2 = square the last term 2FM = twice the product of the first and second term 2FL = twice the product of the first and last term 2ML = twice the product of the middle term and the last term
  • 3. Example 1. Find the product of (a + b + c)2 . Solution: Using the formula of F2 + M2 + L2 + 2FM + 2FL + 2ML F2 = (a)2 = a2 M2 = (b)2 = b2 L2 = (c)2 = c2 2FM = 2(a . b) = 2ab 2FL = 2(a. c) = 2ac 2ML = 2(b . c) = 2bc = a2 + b2 + c2 + 2ab + 2ac + 2bc
  • 4. Example 2. Find the product of (x + y – 3)2 . Solution: Using the formula of F2 + M2 + L2 + 2FM + 2FL + 2ML F2 = (x)2 = x2 M2 = (y)2 = y2 L2 = (-3)2 = 9 2FM = 2(x. y) = 2xy 2FL = 2(x . -3) = -6x 2ML = 2(y . -3) = -6y = x2 + y2 + 9 + 2xy – 6x – 6y = x2 + 2xy + y2 – 6x – 6y + 9
  • 5. Example 3. Find the product of (x – 2y – 1)2. Solution: Using the formula of F2 + M2 + L2 + 2FM + 2FL + 2ML F2 = (x)2 = x2 M2 = (-2y)2 = 4y2 L2 = (-1)2 = 1 2FM = 2(x. -2y) = -4xy 2FL = 2(x -1) = -2x 2ML = 2(-2y . -1) = 4y = x2 + 4y2 + 1 – 4xy – 2x + 4y = x2 -4xy + 4y2 – 2x + 4y + 1
  • 6. Example 4. Solve (2a – b + b2)2. Solution: Using the formula of F2 + M2 + L2 + 2FM + 2FL + 2ML F2 = (2a)2 = 4a2 M2 = (-b)2 = b2 L2 = (b2)2 = b4 2FM = 2(2a . –b) = -4ab 2FL = 2(2a . b2) = 4ab2 2ML = 2(-b . b2) = -2b3 = 4a2 + b2 + b4 – 4ab + 4ab2 – 2b3 = 4ab2 +4a2– 4ab + b4 - 2b3 + b2