Transcript Factoring Perfect Square Trinomials and the Difference of

Solve
x  6x  9
2
Notice that if you take ½ of the middle number and
square it, you get the last number. 6 divided by 2
is 3, and 32 is 9. When this happens you have a
1
special product.
This problem factors into
( )6  3
2
2
3 9
 x  3 x  3 or  x  3
2
2ab +b²
b²
(a + b)² = (a + b)(a + b) = a² +ab + ab + b² = a² +2ab
(a - b)² = (a - b)(a - b) = a² -ab - ab + b² = a² -2ab +b²
For example:
(3x + 5)² = (3x)² + 2(3x)(5) + (5)² = 9x² + 30x +25
a = 3x b = 5
To factor this trinomial , the grouping
method can be used. However, if we
recognize that the trinomial is a perfect
square trinomial, we can use one of the
following patterns to reach a quick solution
1. Check if the first and third terms are both
t² + 10t +25
perfect squares with positive coefficients.
2. If this is the case, identify a and b, and
determine if the middle term equals 2ab
• The first term is a perfect square: t² = (t)²
• The third term is a perfect square: 25 = (5)²
• The middle term is twice the product of t and 5: 2(t)(5)
t² + 10t +25
Perfect square trinomial
t² + 4t +1
1. Check if the first and third terms are both
perfect squares with positive coefficients.
2. If this is the case, identify a and b, and
determine if the middle term equals 2ab
• The first term is a perfect square: t² = (t)²
• The third term is a perfect square: 1 = (1)²
• The middle term is not twice the product of t and 1: 2(t)(1)
t² + 4t + 1
Is not perfect square trinomial
Remember: A perfect square trinomial is one that can
be factored into two factors that match each other
(and hence can be written as the factor squared).
x  12 x  36   x  6  x  6    x  6 
2
This is a perfect square trinomial because it factors
into two factors that are the same and the middle
term is twice the product of x and 6. It can be written
as the factor squared.
Notice that the first and last terms are perfect squares.
The middle term comes from the outers and inners when
FOILing. Since they match, it ends up double the
product of the first and last term of the factor.
2
The GCF is 1.
The first and third terms are positive
The first term is a perfect square: 25y² = (5y)²
The third term is a perfect square: 4 = (2)²
The middle term is twice the product of 5y and 2: 20y = 2(5y)(2)
Factor as (5y - 2)²
y² - 25
= (y)² - (5)²
= (y + 5)(y – 5)
The binomial is a difference of squares.
Write in the form: a² - b², where a = y, b = 5.
Factor as (a + b)(a – b)
9 z  25
2
difference
When you see two terms, look for the difference of squares.
Is the first term something squared? Is the second term
something squared but with a minus sign (the difference)?
3z   5  3z  53z  5
2
2
rhyme for the day
The difference of squares factors into conjugate pairs!
A conjugate pair is a set of factors that look the same but
one has a + and one has a – between the terms.
Factor
Completely:

20y  45
2
5 4y  9
2


Two terms left----is it the difference of
squares?
5 2 y   3
2
Look for something in
common (there is a 5)
2

Yes---so factor into
conjugate pairs.
52 y  32 y  3
8 p  24 p q  18 pq
3
2
Look for something in common
There is a 2p in each term
2 p(4 p  12 pq  9 p )
2
2
2
Check by FOILing and then
distributing 2p through
Three terms left---try
trinomial factoring
"unFOILing"
2 p(2 p  3q)(2 p  3q)
2 p(4 p  12 pq  9q )
2
2
8 p  24 p q  18 pq
3
2
2
Factoring a Sum and Difference of Cubes
Sum of Cubes:
a³+ b³ = (a + b)(a² -ab +b²)
Difference of Cubes:
a³ - b³ = (a - b)(a² +ab +b²)
Square the first term of the binomial
x³ + 8 = (x)³ + (2) ³
x - -(x)(2)
= (x
(x++2)2) ( (x)²
(x)(2) + (2)²)
2
Product of terms in the binomial
x³ and 8 are perfect cubes
Square the last term of the binomial.
•The factored form is the product of a binomial and a trinomial.
•The first and third terms in the trinomial are the squares of the
terms within the binomial factor.
•Without regard to signs, the middle term in the trinomial is the
product of terms in the binomial factor.
27s  64
can be sum or
difference here
3
If it's not the difference of squares, see if it is sum
or difference of cubes. Is the first term something
cubed (to the third power)? Is the second term
something cubed?
The first factor comes from what was cubed.


22



3
s

4
12ss  16
3s   4  3 499ss 12
3
3
multiply
together
but
change
sign
square
the first
term
square
the last
term
You must just memorize the steps to factor cubes.
You should try multiplying them out again to assure
yourself that it works.
Let's try one more:
64v  125
3
What cubed gives the first term?
What cubed gives the second term?
4v   5  4v  5516v  20v  25

The first factor comes from what was cubed.
3
3
2
multiply
together
but
change
sign
square
the first
term
square
the last
term
Try to memorize the steps to get the second factor:
First term squared---multiply together & change sign--last term squared