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Section 6.5 - Factoring Perfect Square Trinomials and
a Difference of Squares
Concept #1:
Factoring Perfect Square Trinomials
Recall the perfect square trinomials from the previous course:
1)
F2 + 2FL + L2 = (F + L)2
2)
F2 – 2FL + L2 = (F – L)2
If a trinomial is a perfect square trinomial, we can use these formulas to
factor the trinomial. In order for a trinomial to fit the pattern, the first term
has to be a perfect square, the last term has to be a perfect square, the
signs in front of the first and last terms has to be positive, and the middle
term has to match the appropriate pattern.
Factor the following:
Ex. 1
9m2 + 24m + 16
Solution:
G.C.F. = 1. The first and last terms are perfect squares, so F = 3m
and L = 4. The middle term is positive so we want to use pattern #1
to check the middle term:
2FL = 2(3m)(4) = 24m which is a
match. So,
9m2 + 24m + 16 =(3m + 4)2
Ex. 2
64y2 – 144yz + 81z2
Solution:
G.C.F. = 1. The first and last terms are perfect squares, so F = 8y
and L = 9z. The middle term is negative so we want to use pattern
#2 to check the middle term:
– 2FL = – 2(8y)(9z) = – 144yz
which is a match. So,
64y2 – 144yz + 81z2 = (8y – 9z)2
Ex. 3
9y2 + 30y – 25
Solution:
G.C.F. = 1. Since there is a subtraction sign in front of the last term,
this is not a perfect square trinomials. Thus, we cannot use our
patterns on this. If we tried to use trial and error or the AC method to
factor this problem, we would find that it is prime.
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Ex. 4
– 4x2 + 29x – 25
Solution:
G.C.F. = – 1. So, – 4x2 + 30x – 25 = – (4x2 – 29x + 25)
The first and last terms are perfect squares, so F = 2x and L = 5.
The middle term is negative so we want to use pattern #2 to check
the middle term:
– 2FL = – 2(2x)(5) = – 20x
which is does not match. So, we have to use trial and error or the AC
method to factor this. When you do so, you will get – (4x – 1)(x – 25).
Ex. 5
50x4z – 80x2yz + 32y2z
Solution:
G.C.F. = 2z. So, 50x4z – 80x2yz + 32y2z = 2z(25x4 – 40x2y + 16y2)
The first and last terms are perfect squares, so F = 5x2 and L = 4y.
The middle term is negative so we want to use pattern #2 to check
the middle term:
– 2FL = – 2(5x2)(4y) = – 40x2y
which is a match. So
50x4z – 80x2yz + 32y2z = 2z(5x2 – 4y)2
1 2
r
4
Ex. 6
1
rs
3
+
+
1 2
s
9
Solution:
G.C.F. = 1. The first and last terms are perfect squares, so F =
and L =
1
s.
3
The middle term is positive so we want to use pattern
#1 to check the middle term: 2FL = 2
a match. So,
1 2
r
4
Concept #2
+
1
r
2
1
rs
3
+
1 2
s
9
=
(
1
r
2
+
( 21 r)( 31 s) =
1
rs
3
which is
1 2
s
3
)
Factoring Difference of Squares Binomials.
The third special product from the last chapter was the difference of
squares:
3)
F2 – L2 = (F – L)(F + L)
If a binomial is a difference of square, we can use this formula to factor the
binomial. In order for a binomial to fit the pattern, the first term has to be a
perfect square, the last term has to be a perfect square, and the operation
has to be subtraction.
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Factor the following:
Ex. 7
Ex. 8
9x2 – 16y2
Solution:
The G.C.F. = 1. The first and last terms are perfect squares, so
F = 3x and L = 4y. The operation is subtraction so this does fit
pattern #3. Thus,
9x2 – 16y2 = (3x – 4y)(3x + 4y)
–
9 3
x
64
+
Solution:
16 2
xy
49
The G.C.F. = – x, so –
9 3
x
64
+
16 2
xy
49
=–x
and last terms are perfect squares, so F =
( 649 x
3
x
8
2
–
and
)
16 2
y . The first
49
4
L = y. The
7
operation is subtraction so this does fit pattern #3. Thus,
–x
Ex. 9
( 649 x
2
–
16 2
y
49
) = – x( 38 x –
4
y
7
)( 38 x +
4
y
7
)
36x2 + 49
Solution:
The G.C.F. = 1. Although the first and last terms are perfect
squares, the operation is addition so this does not fit pattern #3.
In fact, the sum of squares is prime in the real numbers.
Ex. 10
– x8 + 256
Solution:
The G.C.F. = – 1. So, – x8 + 256 = – (x8 – 256)
The first and last terms are perfect squares, so F = x4 and L = 16.
The operation is subtraction so this does fit pattern #3. Thus,
– (x8 – 256) = – (x4 – 16)(x4 + 16)
But, x4 – 16 is a difference of squares with F = x2 and L = 4. We can
break that down using pattern #3 (x4 + 16 is prime).
– (x4 – 16)(x4 + 16) = – (x2 – 4)(x2 + 4)(x4 + 16)
Yet, x2 – 4 is a difference of squares with F = x and L = 2. We can
break that down using pattern #3 (x2 + 4 is prime).
– (x2 – 4)(x2 + 4)(x4 + 16) = – (x – 2)(x + 2)(x2 + 4)(x4 + 16)
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