by. G. Martin
The square of a binomial is always a trinomial and the result is a
Perfect Square Trinomial.
To solve for the square of trinomial, use the formula of:
F2 + 2FL + L2.
where,
1st term ( F2 ) = Square the first term of the binomial.
2nd term (2FL ) = Twice the product of the first term and the second
term of the binomial.
3rd term (L2 ) = Square of the second term / last term of the binomial
Example 1. Find the product of (x + 2)2
Solution:
(x + 2)2 means (x + 2) (x + 2)
F2 (square the first term) = (x)2 = x . X = x2
2FL ( twice the product of the first and second / last term) = 2(x
. 2) = 2(2x) = 4x
L2 ( square the last term)= (2)2 = 2 . 2 = 4
= x2 + 4x + 4
Example 2. Find the product of (x -3)2
Solution:
(x – 3)2 means (x - 3) (x – 3)
F2 = (x)2 = x . x = x2
2FL = 2(x . -3) = 2(-3x) = -6x
L2 = (-3)2 = -3 . -3 = 9
= x2 – 6x + 9
Example 3. Find the product of (m + 4)2
Solution:
F2 = (m)2 = m . m = m2
2FL = 2(m . 4) = 2(4m) = 8m
L2 = (4)2 = 4 . 4 = 16
= m2 + 8m + 16
Example 4. Find the product of (2x + 5)2 .
Solution:
F2 = (2x)2 = 2x . 2x = 4x2
2FL = 2(2x . 5) = 2(10x) = 20x
L2 = (5)2 = 5. 5 = 25
= 4x2 + 20x + 25
Example 5. Find the product of (x3 – 2y)2
Solution:
F2 = (x3)2 = x3 . x3 = x6
2FL = 2(x3 . -2y) = 2(-2x3y) = -4x3y
L2 = (-2y)2 = (-2y . -2y) = 4y2
= x6 – 4x3y + 4y2
Example 6. Find the product of (3 – 6x)2
Soulution:
F2 = (3)2 = 3. 3 = 9
2FL = 2(3 . -6x) = 2(-18x) = -36x
L2 = (-6x)2 = -6x . -6x = 36x2
= 9 – 36x + 36x2
Arrange in standard form = 36x2 – 36x + 9
THANK YOU!

Square of a binomial

  • 1.
  • 2.
    The square ofa binomial is always a trinomial and the result is a Perfect Square Trinomial. To solve for the square of trinomial, use the formula of: F2 + 2FL + L2. where, 1st term ( F2 ) = Square the first term of the binomial. 2nd term (2FL ) = Twice the product of the first term and the second term of the binomial. 3rd term (L2 ) = Square of the second term / last term of the binomial
  • 3.
    Example 1. Findthe product of (x + 2)2 Solution: (x + 2)2 means (x + 2) (x + 2) F2 (square the first term) = (x)2 = x . X = x2 2FL ( twice the product of the first and second / last term) = 2(x . 2) = 2(2x) = 4x L2 ( square the last term)= (2)2 = 2 . 2 = 4 = x2 + 4x + 4
  • 4.
    Example 2. Findthe product of (x -3)2 Solution: (x – 3)2 means (x - 3) (x – 3) F2 = (x)2 = x . x = x2 2FL = 2(x . -3) = 2(-3x) = -6x L2 = (-3)2 = -3 . -3 = 9 = x2 – 6x + 9
  • 5.
    Example 3. Findthe product of (m + 4)2 Solution: F2 = (m)2 = m . m = m2 2FL = 2(m . 4) = 2(4m) = 8m L2 = (4)2 = 4 . 4 = 16 = m2 + 8m + 16
  • 6.
    Example 4. Findthe product of (2x + 5)2 . Solution: F2 = (2x)2 = 2x . 2x = 4x2 2FL = 2(2x . 5) = 2(10x) = 20x L2 = (5)2 = 5. 5 = 25 = 4x2 + 20x + 25
  • 7.
    Example 5. Findthe product of (x3 – 2y)2 Solution: F2 = (x3)2 = x3 . x3 = x6 2FL = 2(x3 . -2y) = 2(-2x3y) = -4x3y L2 = (-2y)2 = (-2y . -2y) = 4y2 = x6 – 4x3y + 4y2
  • 8.
    Example 6. Findthe product of (3 – 6x)2 Soulution: F2 = (3)2 = 3. 3 = 9 2FL = 2(3 . -6x) = 2(-18x) = -36x L2 = (-6x)2 = -6x . -6x = 36x2 = 9 – 36x + 36x2 Arrange in standard form = 36x2 – 36x + 9
  • 9.