Factoring Quadratic Polynomials
January 2012
Factoring Quadratic Polynomials
Objective: The student will learn to factor
quadratic polynomials.
Quadratic Polynomials
Polynomials of the form,
ax2+bx+c
a≠0
are quadratic polynomials.
They are also called second degree polynomials.
The degree of the monomial with the highest degree
is 2.
Factoring quadratic polynomials that
are not perfect squares
We have learned to recognize quadratic
polynomials that are perfect squares and to
factor them.
Example: x2 + 4x + 4 = (x + 2)2
However, what if the polynomial is not a perfect
square?
Today, we will learn how to factor quadratic
polynomials that are not perfect squares.
Quadratic Trinomial
2nd degree polynomial with three terms
a≠0
b≠0
c≠0
a: leading coefficient
ax2: leading term
b: linear coefficient
bx: linear term
c: constant coefficient
c:
constant term, constant
Factoring Quadratic Polynomials when
a=1
Multiply the following binomials: (x+A)∙(x+B)
Use FOIL:
Comparing with Standard Form:
F:
O:
I :
L:
x2
Ax
Bx
AB
Combining:
x2
x2 + bx + c = x2 + (A+B)x + AB
Since x can be any number,
b = A+B
c = AB
+ (A+B)x + AB
Example
Factoring Quadratic Polynomials
a = 1 b > 0 c>0
x2 + 5x + 6
16
Factors of 6
23
Sum of the factors
1+6=7
x2 + 5x + 6 = (x + 2)(x + 3)
2+3=5
Class Work
Factor the following quadratic polynomial on a
piece of paper to hand in:
𝑥 2 + 6𝑥 + 8
𝑥+2 𝑥+4
Example
Factoring Quadratic Polynomials
a=1
b>0
c<0
x2 + 2x  15
Factors of 15
1  15
1  15
3  5
3  5
Sum of the factors
1 + 15 = 14
3 + 5 = 2
1 + (15) = 14
3 + (5) = 2
x2 + 2x  15 = (x  3)(x + 5)
Class Work
Factor the following quadratic polynomial on a
piece of paper to hand in:
𝑥 2 + 𝑥 − 12
𝑥+4 𝑥−3
Example
Factoring Quadratic Polynomials
a=1
b<0
c<0
x2  2x  15
Factors of 15
1  15
1  15
3  5
3  5
Sum of the factors
1 + 15 = 14
3 + 5 = 2
1 + (15) = 14
3 + (5) = 2
x2  2x  15 = (x + 3)(x  5)
Class Work
Factor the following quadratic polynomial on a
piece of paper to hand in:
𝑥 2 − 7𝑥 − 12
𝑥−4 𝑥−3
Example
Factoring Quadratic Polynomials
a=1
b<0
c>0
x2  5x + 6
Negative factors of 6
1  (6)
2  (3)
Sum of the factors
1 + (6) = 7
2 + (3) = 5
x2  5x + 6 = (x  2)(x  3)
Class Work
Factor the following quadratic polynomial on a
piece of paper to hand in:
𝑥 2 − 9𝑥 + 18
𝑥−3 𝑥−6
Example
Factoring Quadratic Polynomials
a = 1
2x + 15  x2
Put in standard form
x2 + 2x + 15
Factor out negative sign
(x2  2x  15)
Factor as usual:
2x + 15  x2 = (x + 3)(x  5)
Class work:
Oral Exercises:
P 190: 1-8
Homework:
P 191: 1-22,
P 186: 49-52
Factoring Quadratic Polynomials,
Part 2
January 2012
Factoring Quadratic Polynomials when
a>1
• First, try to factor out any common
monomial terms:
2𝑥 2 + 10𝑥 + 12
2 is a common monomial factor.
Factoring it out gives, 2(𝑥2 + 5𝑥 + 6)
• Then factor the quadratic trinomial inside
the parentheses.
(𝑥 + 2)(𝑥 + 3)
• Finally, multiply the result by 2.
2(𝑥 + 2)(𝑥 + 3)
Factoring Quadratic Polynomials when
𝑎>1
Multiply the following binomials: (𝐴𝑥 + 𝐵) ∙ (𝐶𝑥 + 𝐷)
Use FOIL:
Comparing with Standard Form:
F:
O:
I :
L:
𝐴𝐶𝑥2
𝐴𝐷𝑥
𝐵𝐶𝑥
𝐵𝐷
Combining:
𝐴𝐶𝑥2 + (𝐴𝐷 + 𝐵𝐶)𝑥 + 𝐵𝐷
𝑎𝑥2 + 𝑏𝑥 + 𝑐
= 𝐴𝐶𝑥2 + (𝐴𝐷 + 𝐵𝐶)𝑥 + 𝐵𝐷
Since x can be any number,
𝑎 = 𝐴𝐶
𝑏 = 𝐴𝐷 + 𝐵𝐶
𝑐 = 𝐵𝐷
Example :
Factoring Quadratic Polynomials
𝑎 > 1 𝑏 > 0 𝑐>0
𝟒𝒙𝟐 + 𝟖𝒙 + 𝟑
Factors of 4
14
22
Factors of 3
13
31
Arrange the factors, cross-multiply, and add.
1 4
1 3
3+4=7
1 4
3 1
1+12=13
2 2
1 3
6+2=8
4𝑥2 + 8𝑥 + 3 = (2𝑥 + 1)(2𝑥 + 3)
Class Work
Factor the following quadratic polynomial on a
piece of paper to hand in:
2𝑥 2 + 7𝑥 + 3
Factors of 2:
1∙2
Factors of 3:
1∙3
3∙1
1 2
1 3
2+3
1 2
3 1
6+1
𝑥 + 3 2𝑥 + 1
Example:
Factoring Quadratic Polynomials
𝑎>1 𝑏>0 𝑐<0
𝟔𝒙𝟐 + 𝟕𝒙𝟏𝟎
Factors of 6
16
23
Factors of 10
1 10
10 1
2 5
5 2
(1)10
(10)1
(2)5
(5)2
Arrange the factors, cross-multiply, and add.
(Sixteen possible combinations. See the next slide.)
1
6
1 10
1 6
1 10
1 6
10 1
1 6
10 1
10+6=4 106=4 1+60=59 160=59
1 6
1 6
1 6
1 6
2 5
2 5
5 2
5 2
5 +12=7 512=7 2+30=28 230=28
We need go no further.
6𝑥2 + 7𝑥10 = 1𝑥 + 2 6𝑥 5
= (𝑥 + 2)(6𝑥5)
Class Work
Factor the following quadratic polynomial on a
piece of paper to hand in:
2𝑥 2 + 3𝑥 − 5 = 𝑥 − 1 2𝑥 + 5
Factors of 2:
1∙2
Factors of −5: 1 ∙ −5
−5 ∙ 1
1 2
−5 1
−10 + 1
1
2
5 −1
10 − 1
−1 ∙ 5
5 ∙ −1
2
1
1 −5
2−5
1 2
−1 5
−2 + 5
Prime Polynomials
• A polynomial that has more than one term
and cannot be expressed as the product of
polynomials of lower degree taken from a
given factor set is irreducible over that set.
• An irreducible polynomial with integral
coefficients is prime if the greatest common
factor of its coefficients is 1.
Prime Polynomials
(Examples)
The following polynomial is prime:
𝑥2 + 4𝑥 – 3
No two integers have product 3 and sum 4.
The following polynomial is not prime:
2𝑥2 + 8𝑥 – 6
2 is a common monomial factor.
Factored Completely
A polynomial is factored completely when it is
written as a product of factors, and each
factor is a monomial, a prime polynomial, or a
power of a prime polynomial.
15x3(x2 + 4x – 3)(3x – 4)2
Factored Completely
(Example)
• Factor the following polynomial completely:
3x6  48x2
= 3x2(x4  16)
= 3x2(x2 + 4) (x2  4)
= 3x2(x2 + 4) (x + 2)(x  2)
Factor out 3x2
Factor x4  16
Factor x2  4
• This polynomial is completely factored.
The GCF of Polynomials
• To find the GCF of two or more polynomials,
o Find the prime factors of all the
polynomials.
o Determine the lowest power of each prime
factor.
o The GCF is the product of the lowest
powers of each prime factor.
The GCF of Polynomials
(Example)
Find the GCF of the following polynomials.
First, find the prime factors.
𝑥23𝑥 + 2 = (𝑥1)(𝑥2)
𝑥24𝑥 + 4 = (𝑥2)2
The GCF is the product of the lowest powers of
the common prime factors:
GCF = (𝑥2)
The LCM of Polynomials
• To find the LCM of two or more polynomials,
o Find the prime factors of all the
polynomials.
o Determine the largest power of each prime
factor.
o The LCM is the product of the largest
powers of each prime factor.
The LCM of Polynomials
(Example)
Find the LCM of the following polynomials.
First, find the prime factors.
𝑥23𝑥 + 2 = (𝑥1)(𝑥2)
𝑥24𝑥 + 4 = (𝑥2)2
The LCM is the product of the highest powers
of all the prime factors :
LCM = (𝑥1)(𝑥2)2
Class work:
Oral Exercises:
P 190: 9-16
Homework
p 191: 23-30,
31-49 odd