factoring quadratic polynomials

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.

However, what if the polynomial is not a perfect square?

Today, we will learn how to factor quadratic polynomials that are not perfect squares.

Example: x2 + 4x + 4 = (x + 2)2

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: F : x2

O: Ax

I : Bx

L : AB

Combining:

x2 + (A+B)x + AB

Comparing with Standard Form:

x2 + bx + c = x2 + (A+B)x + AB

Since x can be any number,

b = A+B

c = AB

Example Factoring Quadratic Polynomials

a = 1 b > 0 c>0

x2 + 5x + 6

Sum of the factors

1 + 6 = 7 2 + 3 = 5

Factors of 6 1 6 2 3

x2 + 5x + 6 = (x + 2)(x + 3)

Class Work

Factor the following quadratic polynomial on a piece of paper to hand in:

𝑥2 + 6𝑥 + 8

𝑥 + 2 𝑥 + 4


a = 1 b > 0 c < 0

x2 + 2x 15

Sum of the factors

1 + 15 = 14 1 + (15) = 14

3 + 5 = 2 3 + (5) = 2

Factors of 15

1 15 1 15 3 5 3 5

x2 + 2x 15 = (x 3)(x + 5)

Class Work


𝑥2 + 𝑥 − 12

𝑥 + 4 𝑥 − 3


a = 1 b < 0 c < 0

x2 2x 15

Sum of the factors

1 + 15 = 14 1 + (15) = 14

3 + 5 = 2 3 + (5) = 2

Factors of 15

1 15 1 15 3 5 3 5

x2 2x 15 = (x + 3)(x 5)

Class Work


𝑥2 − 7𝑥 − 12

𝑥 − 4 𝑥 − 3


a = 1 b < 0 c > 0

x2 5x + 6

Sum of the factors

1 + (6) = 7 2 + (3) = 5

Negative factors of 6

1 (6) 2 (3)

x2 5x + 6 = (x 2)(x 3)

Class Work


𝑥2 − 9𝑥 + 18

𝑥 − 3 𝑥 − 6


a = 1

2x + 15 x2

Factor out negative sign (x2 2x 15)

Put in standard form 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

• Finally, multiply the result by 2.

2(𝑥 + 2)(𝑥 + 3)

• Then factor the quadratic trinomial inside

the parentheses. (𝑥 + 2)(𝑥 + 3)

• First, try to factor out any common

monomial terms: 2𝑥2 + 10𝑥 + 12

2 is a common monomial factor.

Factoring it out gives, 2(𝑥2 + 5𝑥 + 6)

Factoring Quadratic Polynomials when 𝑎 > 1

Multiply the following binomials: (𝐴𝑥 + 𝐵) ∙ (𝐶𝑥 + 𝐷)

Use FOIL: F : 𝐴𝐶𝑥2

O: 𝐴𝐷𝑥

I : 𝐵𝐶𝑥

L : 𝐵𝐷

Combining: 𝐴𝐶𝑥2 + (𝐴𝐷 + 𝐵𝐶)𝑥 + 𝐵𝐷

Comparing with Standard Form:

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 3

1 3 3 1

Arrange the factors, cross-multiply, and add.

1 4 1 4 2 2

1 3 3 1 1 3

3+4=7 1+12=13 6+2=8

Factors of 4 1 4 2 2

4𝑥2 + 8𝑥 + 3 = (2𝑥 + 1)(2𝑥 + 3)

Class Work


2𝑥2 + 7𝑥 + 3

Factors of 2: 1 ∙ 2

11

23

2 + 3

𝑥 + 3 2𝑥 + 1

Factors of 3: 1 ∙ 3 3 ∙ 1

13

21

6 + 1

Example: Factoring Quadratic Polynomials

𝑎 > 1 𝑏 > 0 𝑐 < 0 𝟔𝒙𝟐 + 𝟕𝒙𝟏𝟎

Factors of 10 1 10 (1)10

10 1 (10)1

2 5 (2)5

5 2 (5)2

Arrange the factors, cross-multiply, and add. (Sixteen possible combinations. See the next slide.)

Factors of 6

16 23

1 6 1 6 1 6 1 6

1 10 1 10 10 1 10 1

10+6=4 106=4 1+60=59 160=59

We need go no further.

6𝑥2 + 7𝑥10 = 1𝑥 + 2 6𝑥 5

= (𝑥 + 2)(6𝑥5)

1 6 1 6 1 6 1 6

2 5 2 5 5 2 5 2

5 +12=7 512=7 2+30=28 230=28

Class Work


2𝑥2 + 3𝑥 − 5

Factors of 2: 1 ∙ 2

11

2−5

2 − 5

Factors of −5: 1 ∙ −5 −1 ∙ 5

−5 ∙ 1 5 ∙ −1

1−1

2 5

−2 + 5

1−5

21

−10 + 1

15

2−1

10 − 1

= 𝑥 − 1 2𝑥 + 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)

2 is a common monomial factor.

The following polynomial is not prime:

2𝑥2 + 8𝑥 – 6

No two integers have product 3 and sum 4.

The following polynomial is prime:

𝑥2 + 4𝑥 – 3

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) Factor out 3x2

= 3x2(x2 + 4) (x2 4) Factor x4 16

= 3x2(x2 + 4) (x + 2)(x 2) Factor x2 4

• This polynomial is completely factored.

The GCF of Polynomials

• To find the GCF of two or more polynomials,

oFind the prime factors of all the

polynomials.

oDetermine the lowest power of each prime

factor.

oThe GCF is the product of the lowest

powers of each prime factor.

The GCF of Polynomials (Example)

𝑥23𝑥 + 2 = (𝑥1)(𝑥2)

𝑥24𝑥 + 4 = (𝑥2)2

The GCF is the product of the lowest powers of the common prime factors:

GCF = (𝑥2)

First, find the prime factors.

Find the GCF of the following polynomials.

The LCM of Polynomials

• To find the LCM of two or more polynomials,

oFind the prime factors of all the

polynomials.

oDetermine the largest power of each prime

factor.

oThe LCM is the product of the largest

powers of each prime factor.

The LCM of Polynomials (Example)

𝑥23𝑥 + 2 = (𝑥1)(𝑥2)

𝑥24𝑥 + 4 = (𝑥2)2

The LCM is the product of the highest powers of all the prime factors :

LCM = (𝑥1)(𝑥2)2

First, find the prime factors.

Find the LCM of the following polynomials.

Homework

p 191: 23-30,

31-49 odd

Class work: Oral Exercises: P 190: 9-16

factoring quadratic polynomials

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