Algebra 2 Notes SOL AII.1 Factoring Polynomials
Mrs. Grieser
Name: _______________________________________ Date: ________________ Block: _________
Factoring Review
Factor: rewrite a number or expression as a product of primes; e.g. 6 = 2 ∙ 3
In algebra, factor by rewriting a polynomial as a product of lower-degree polynomials
In the example above, (x + 1)(x – 2) is the factored form of x2 - x – 2 (multiply to verify!)
We will look at 5 different factoring “types” (many thanks to Mrs. Donohue from
TJHSST!)
TYPE I Factoring: “Factor Out” GCF Monomials
Find a common monomial in the polynomial (the GCF) and factor it out (perform reverse
distribution)
IMPORTANT!!! Always factor common factors out first (type I factoring); then factor the
polynomial that remains, if necessary, using other factoring methods.
IMPORTANT!!! Always verify your factored results by multiplying the factors to get the
original polynomial
Examples: Factor the following polynomials…
a) 6x2 + 12x
b) 12x + 42y
c) 4x4 + 24x3
GCF = ________
GCF = ______
GCF = ______
Factored:____________
Factored: __________
Factored: ___________
Verify by distributing:
Verify!
Verify!
b) 9m3 – 3m2
c) 3x5 + 36x6
You Try… Factor:
a) 4m - 2
Type II Factoring – Sums and Differences of Perfect Squares and Cubes
Remember to factor out any common factors first!
Two Terms that are the Difference of Perfect Squares: a2 – b2
o Difference of squares: a2 – b2 = (a + b)(a – b)
Examples: Factor the binomials below…
a) y2 – 16
b) 4x2 - 25
c) x4 – 16
d) x8 - 1
e) 9x6 – y8
Algebra 2 Notes SOL AII.1 Factoring Polynomials
Mrs. Grieser Page 2
Two Terms that are Sum or Difference of Perfect Cubes: a3 + b3 or a3 – b3
o Sum of cubes: a3 + b3 = (a + b)(a2 – ab + b2)
o Difference of cubes: a3 – b3 = (a – b) (a2 + ab + b2)
o To remember the signs… SOAP: “Same” “Opposite” “Always Positive”’
o The trinomial will not be factorable
Examples: Factor the binomials below…
a) x3 – 8
b) 27x3 + 1
c) x3y6 – 64
d) 16y3 + 54
You Try: Factor the binomials below (factor out the GCF first if necessary!)
a) 3x2 - 27
b) 4x2 - 16
c) 8x2 – 50
d) x3 + 8
e) 8x3 – 27y3
f) x15 + y21
g) 2x3 – 18x
h) 4x3 - 108
i) 16x3 + 2
Type III Factoring – Trinomials with Leading Coefficient 1 (form: x2 + bx + c)
Notice a special product:
If it is of the form a2 + 2ab + b2, then its factored form is (a + b)2.
If it is of the form a2 - 2ab + b2, then its factored form is (a - b)2.
A quick test is to make a binomial of a and c and square it, and see if you get original
polynomial, or use complete the square rules (c is (b/2)2).
Examples:
a) x2 + 6x + 9
b) x2 – 10x + 25
c) x2 + 4x + 4
For all other trinomials of the form x2 + bx + c, you must ask yourself the question:
“What do you multiply to get the last number (c), and add to get the middle
number (b)?”
Algebra 2 Notes SOL AII.1 Factoring Polynomials
Mrs. Grieser Page 3
Examples:
a) x2 + 5x + 6
b) x2 – 6x + 8
c) x2 - x – 2
d) x10 - 3x5 – 10
Factors are:
Factors are:
Factors are:
Factors are:
______________
______________
______________
______________
Verify!
Verify!
Verify!
Verify!
a) x2 + 4x + 3
b) x2 – 11x + 24
c) x2 + 6x - 16
d) x2 – 2x - 24
e) x2 + 23x – 24
f) x2 – 14x + 24
g) x2 + 6xy – 7y2
h) x4 + 4x2 – 32
You Try…Factor:
Type IV Factoring – Factor by Grouping (Four Terms)
If we are given a four term polynomial, we split the polynomial into two sets of two
terms, and factor those sets using type I factoring.
If we find a common polynomial, we use type I factoring again to factor it out.
Factoring a common polynomial: Factor x(x – 5) + 3(x - 5)
Notice there is a common polynomial of x – 5. Use type I factoring to factor it out; we are
left with x + 3. So the factored form is (x – 5)(x + 3).
Examples:
a) 5x2(x – 2) + 3(x – 2)
b) 7y(5 – y) – 3(y – 5)
c) 11x(x – 8) + 3(8 – x)
Use this skill to factor a four term polynomial:
o Factor the first two terms, then factor the second two terms.
o Factor out the common polynomial.
Algebra 2 Notes SOL AII.1 Factoring Polynomials
Mrs. Grieser Page 4
Examples:
a) n3 + 6n2 + 5n + 30
b) m3 + 7m2 – 2m – 14
c) 9x3 – 7x + 9x2 - 7
You try: Factor the expression
a) 3y2(y – 2) + 5(2 – y)
b) x3 + 3x2 + 5x + 15
c) x2 – y2 + 4x + 4y
d) x3 + x2 + x + 1
e) y2 + y + xy + x
f) x3 – 6 + 2x – 3x2
(HINT: Rearrange
terms in degree order!)
Type V Factoring – Factor ax2 + bx + c
We can factor polynomials of the form x2 + bx + c (type III factoring). What do we do to
factor polynomials of this form when the leading coefficient is not 1?
IMPORTANT: Always factor out a GCF first; you may find that you really have a type III.
Example: 2x2 – 2x – 4 = 2(x2 – x – 2) = 2(x – 2)(x + 1)
Method 1: Guess and Check
Factor 2x2 – 7x + 3
Draw sets of parentheses: (
)(
)
In this case, the first terms in each must be 2x and x (why?) and the signs must be
negative (why?):
(2x - )(x - )
The factors of 3 are 1 and 3; test by multiplying back to see what works
o (2x – 3)(x – 1)
2x2 – 5x + 3 NOPE!
o (2x – 1)(x – 3)
2x2 – 7x + 3 YES!!
Factors are (2x – 1)(x – 3)
Fairly easy to do when a and c are prime numbers; gets harder if they are not!
Algebra 2 Notes SOL AII.1 Factoring Polynomials
Mrs. Grieser Page 5
Method 2: Factor by Grouping Method
If you are not a good guesser, it can be hard sometimes to use the guess and check
method. Factoring by grouping (type IV) can help us:
Factor 15x2 + 13x + 2
METHOD
1) Factor out GCF if there is one
EXAMPLE
1) No common factors
2) Multiply a x c
2) a = 15; c = 2; ac = 15 x 2 = 30
3) What factors of ac add to b?
3) What factors of 30 add to 13? 10 and 3
4) Split up middle term by factors found above
4) Split up middle term: 15x2 + 10x + 3x + 2
5) Apply grouping method (type IV)
5) Group: 5x(3x + 2) + (3x + 2)
6) Factor out polynomial
6) Factor out polynomial: (3x+2)(5x+1)
7) VERIFY (do not skip this step)
7) VERIFY: (3x + 2)(5x + 1) =
15x2 + 13x + 2
NOTE: ALWAYS FACTOR OUT ANY GCF PRIOR TO USING METHOD ABOVE!!
Examples:
a) 6x2 – 11x - 10
b) 3x2 + 14x - 5
c) 4x2 + 26x - 14
You try: Factor the polynomials
a) 3x2 + 8x + 4
b) 4x2 – 9x + 5
c) 2x2 – 13x + 6
d) -4x2 + 12x + 7
e) 4x2 + 11x - 3
f) 12x2 – x - 6