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Perfect Square Trinomials
Perfect Square Trinomialsa2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a b)2
In order for a polynomial to be a perfect squaretrinomial, two conditions must be satisfied:
1. The first and last terms must be perfect squares.
2. The “middle term” must equal 2 or – 2 times the product of the expressions being squared in the first and last term.
Perfect Square Trinomial Form: Example:
(Form 1)a2 + 2ab + b2 = (a + b)2 x2 + 6x+ 9 = ____________ = (x + 3)2
a = _____ , b
= _____ , 2ab = _______
(Form 2)a2 – 2ab + b2 = (a b)2 x2 – 10x + 25 = ____________ = (x 5)2
a = _____ , b = _____ , 2ab = _______
Condition 1:First and last term are
perfect squares.
Condition 1:First and last term are
perfect squares.
Perfect Square Trinomial Form: Example:
(Form 1)a2 + 2ab + b2 = (a + b)2 x2 + 6x+ 9 = ____________ = (x + 3)2
a = _____ , b
= _____ , 2ab = _______
(Form 2)a2 – 2ab + b2 = (a b)2 x2 – 10x + 25 = ____________ = (x 5)2
a = _____ , b = _____ , 2ab = _______
Condition 2:
The “middle term” must equal 2 or – 2 times the product of the expressions
being squared in the first and last term.
Condition 2:
The “middle term” must equal 2 or – 2 times the product of the expressions
being squared in the first and last term.
Perfect Square Trinomial Form: Example:
(Form 1)a2 + 2ab + b2 = (a + b)2 x2 + 6x+ 9 = ____________ = (x + 3)2
a = _____ , b
= _____ , 2ab = _______
(Form 2)a2 – 2ab + b2 = (a b)2 x2 – 10x + 25 = ____________ = (x 5)2
a = _____ , b = _____ , 2ab = _______
Perfect Square Trinomials
Examples: Factor.
x2 + 8x + 16 = ____________ = (x + 4)2
1.) x2 + 8x + 16The first term, x2, and third term, 16, are perfect squares.
The middle term, 8x, is 2 times the product of x and 4.
Difference of Two SquaresDifference of Squares Form: Example:a2 – b2 = (a + b)(a b) x2 – 64 = ( ____ ____ ) ( ____ ____ )
Check using FOIL:
Why do we call these DIFFERENCES What happens to the two
middle of two squares? terms when we FOIL a
difference of two squares?
Example 1: Factor x2 – 4
Notice the terms are both perfect squares
x2 = (x)2 4 = (2)2
x2 – 4 = (x)2 – (2)2
a2 – b2
and we have a difference
= (x – 2)(x + 2)
difference of squares
= (a – b)(a + b)factors as
Difference of Two Squares
Example 2: Factor 9p2 – 16
Notice the terms are both perfect squares
9p2 = (3p)2 16 = (4)2
9a2 – 16 = (3p)2 – (4)2
a2 – b2
and we have a difference
= (3p – 4)(3p + 4)
difference of squares
= (a – b)(a + b)factors as
Difference of Two Squares
Example 3: Factor 2y6 – 50
Difference of Two Squares
GCF First 2(y6 – 25)
Now it’s a difference of
squares!
Example 3: Factor 2y6 – 50
Notice the terms are both perfect squares
y6 = (y3)2 25 = (5)2
2(y6 – 25) = 2 ((y3)2 – (5)2)
2(a2 – b2)
and we have a difference
= 2(y3 – 5)(y3 + 5)
difference of squares
= 2(a – b)(a + b)factors as
Difference of Two Squares
GCF First 2(y6 – 25)
Example 4: Factor 81 – x2y2
Notice the terms are both perfect squares
81 = (9)2 x2y2 = (xy)2
81 – x2y2 = (9)2 – (xy)2
a2 – b2
and we have a difference
= (9 – xy)(9 + xy)
difference of squares
= (a – b)(a + b)factors as
Difference of Two Squares