Factoring PolynomialsFactoring Polynomials

Finding GCF

If you remember multiplication with real numbers, then you should remember the

following facts:

If 6 • 2 = 12, then we know• 12 is the product of 6 and 2• 6 and 2 are factors, or divisors, of 12.• the quotient of 12 divided by 6 is 2.• the quotient of 12 divided 2 is 6.So, to factor a number is to write it as the

product of two or more numbers, usually natural numbers. Factoring and division are closely related.

Review: a factor - is a number that is multiplied by another number to produce a product. A prime number - is any natural number greater than 1 whose only factors are

1 and itself. A composite number - is a number greater than 1 that has more than two

factors.The prime factorization - is the factorization of a natural

number that contains only prime numbers or powers of

prime numbers.

The figure below shows 24 square tiles arranged to form a rectangle. (4 tiles wide,

6 tiles long)Sketch other ways that the 24 tiles can be

arranged to form a rectangle.

ANSWER

If you notice each rectangle has an area of 24, which includes (4 x 6, 3 x 8, 2 x 12, 1 x

24).Each of the numbers involved in these

multiplications is a factor of 24. There are no other natural number pairs that have a

product of 24, so 1, 2, 3, 4, 6, 8, 12, and 24 are the only factors of 24.

Here are two examples

Write 3 different factorizations of 16. Use natural numbers.

16 divided by 1 2 3 4 5 6 7 8

Natural number 16

8 no

4 no

no

no

2

1 x 162 x 8 4 x 4 8 x 2

Finding Greatest Common FactorFinding Greatest Common Factor

GCF’s

Let’s make a list of factors of 36, written in order from least to greatest.

36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Make a list of factors for the number 54.54: 1, 2, 3, 6, 9, 18, 27, 54

Examine the two lists for factors that appear in both list.

36: 1, 2, 3, 6, 9, 12, 18, 3654: 1, 2, 3, 6, 9, 18, 27, 54

common factors: 1, 2, 3, 6, 9, 18GCF: 18

What if the numbers in our previous example were expressions 36c3 and 54c2?

36c3

2 ∙ 2 ∙ 3 ∙ 3 ∙ c ∙ c ∙ c

21 ∙ 32 ∙ c2

18 ∙ c2

54c2

2 ∙ 3 ∙ 3 ∙ 3 ∙ c ∙ c

21 ∙32 ∙c2

18 ∙ c2

GCF 18c2

Try again

Find the GCF:

a) 18d5 and 108d

b) 18d and 5

c) 3m3n3 and 9m2n2

d) 4mn3, 4m2n3, and 16m2n2

Factoring a monomial from a polynomial

Factoring a monomial from a polynomial

Using GCF

Factoring a Monomial from a

Polynomial Factoring a polynomial reverses the

multiplication process. To factor a monomial

from a polynomial, first find the greatest

common factor (GCF) of its terms.

Find the GCF of the terms of: 4x3 + 12x2 – 8xList the prime factors of each

term.4x3 = 2 · 2 · x · x x12x2 = 2 · 2 · 3 · x · x 8x = 2 · 2 · 2 · x

The GCF is 2 · 2 · x or 4x.

Find the GCF of the terms of each polynomial.

a) 5v5 + 10v3

b) 3t2 – 18

c) 4b3 – 2b2 – 6b

d) 2x4 + 10x2 – 6x

Use the GCF to factor each polynomial.

a) 8x2 – 12x

b) 5d3 + 10d

c) 6m3 – 12m2 – 24m

d) 4x3 – 8x2 + 12x

Try to factor mentally by scanning the coefficients of each term to find the GCF.

Next, scan for the least power of the variable.

Factoring Out a Monomial

Factor 3x3 – 12x2 + 15xStep 1 Find the GCF3x3 = 3 · x · x · x12x2 = 2 · 2 · 3 · x · x15x = 3 · 5 · x

The GCF is 3 · x or 3x

Step 2Factor out the GCF3x3 – 12x2 + 15x = 3x(x2) + 3x(-4x) +

3x(5) = 3x(x2 – 4x + 5)

To factor a polynomial completely, you must

factor until there are no common factors other than

1.

Factoring x2 + bx + cFactoring x2 + bx + c

Factoring x2 + bx + c when c is positive

Observe the two columns below, the multiplication of binomials on the left and

the products on the right.What do you notice about the two list?

1. (x + 5)(x + 6)

2. (x + 3)(x + 10)

3. (x + 2)(x + 15)

4. (x + 1)(x + 30)

[x2 + 11x + 30]

[x2 + 13x + 30]

[x2 + 17x + 30]

[x2 + 31x + 30]

The product of the constants = 30The sum of the constants = the coefficient

of the x-terms

1. (x + 5)(x + 6)

2. (x + 3)(x + 10)

3. (x + 2)(x + 15)

4. (x + 1)(x + 30)

[x2 + 11x + 30]

[x2 + 13x + 30]

[x2 + 17x + 30]

[x2 + 31x + 30]

TRY THIS

Remember the patterns you say earlier.

x2 + 17x + 30

Write the binomial multiplication that gives this product.

( )( )

What two constants multiplied together

gives you 30?

The sum of what two of those

constants give you 17?

In other words, to factor x2 + bx + c, look for the factor pairs (the two numbers) whose

product is c. Then choose the pair whose sum is b.

Factor x2 + 5x + 61. c = 6: write all (+ and -) the factor pairs of

6. 1 • 6 2 • 3 -1 • -6 -2 • -3

2. b = 5: choose the pair whose sum is 5.1 + 6 = 7 2 + 3 = 5 -1 - 6 = -7 -2 – 3 =

-5

3. Write the product using 2 and 3. Thus(x + 2)(x + 3) = x2 + 5x + 6

a2 + 9a + 20

TRY THIS

NEXT

Factor y2 – 10y + 24

1. c = 24 Write all (+ and -) factor pairs of 24.1 • 24 -1 • -24 4 • 6 -4 • -62 • 12 -2 • -12 3 • 8 -3 • -8

2. b = -10: choose the pair whose sum is -10.1 + 24 = 25 -1 +(-24) = -25 4 + 6 = 10 -4 + (-

6) = -10

3. Write the product using -4 and -6.(y – 4)(y – 6) = y2 -10y + 24

TRY THIS n2 -13n + 36

Practice and Problem Solving.

1. t2 + 7t + 10 =(t + 2)(t + □)

2. x2 – 8x + 7 =(x – 1)(x - □)

3. x2 + 9x + 18 =(x + 3)(x + □)

Factor each expression1. r2 +4r + 3

2. n2 – 3n + 2

3. k2 + 5k + 6

4. x2 – 2x + 1

5. y2 + 6y + 8

Factoring x2 + bx + cFactoring x2 + bx + c

Factoring x2 + bx + c, when c is negative

Find each product.What do you notice about the c term in

each?

(x – 5)(x + 6)(x – 3)(x + 10)(x – 2)(x + 15)(x – 1)( x + 30)

(x + 5)(x - 6)(x + 3)(x - 10)(x + 2)(x – 15)(x + 1)(x – 30)

x2 + x – 30x2 + 7x – 30x2 + 13x – 30x2 + 29x – 30

x2 – x – 30x2 – 7x – 30x2 – 13x – 30x2 – 29x - 30

What do you notice about

c in each expression?

What do you notice about c

and the coefficient of the x-term?

You can also factor x2 + bx – c

Factor x2 + x – 201. c = -20: write all (+ and -) factors of -20.

(-1) • 20 1 • (-20) (-2) • 10 2 • (-10) (-4) • 5 4 • (-5)

2. b = 1: find the pair whose sum is 1.(-1) • 20 1 • (-20) (-2) • 10 2 • (-10) (-4) • 5 4

• (-5)

3. Write the product using -4 and 5 = 1(x – 4)(x + 5) = x2 + x - 20

TRY THIS n2 + 3n - 40

Factor z2 – 4z – 12

1. c = -12: write all (+ and -)factors of -12.(-1) • 12 1 • (-12) (-2) • 6 2 • (-6) (-3) • 4 3

• (-4)

2. b = -4: find the factor pair of -4.(-1) • 12 1 • (-12) (-2) • 6 2 • (-6) (-3) • 4 3

• (-4)

3. write the product using 2 and -6.(x + 2)(x – 6) = x2 – 4z - 12

TRY THIS n2 – 3n - 40

Not every polynomial of the form x2 + bx + c

is factorable.

Factor x2 + 3x – 1

c = -1: the only factors of -1, are 1 and -1.

b = 3: because -1 + 1 ≠ 3,x2 + 3x – 1 cannot be factored

TRY THIS t2 + 5t - 8

Factor:

1. y2 + 10y – 11

2. x2 – x – 42

3. b2 – 17b – 38

4. s2 + 4s – 5

5. y2 + 2y – 63

Factoring ax2 + bx + c

Factoring ax2 + bx + c

Factoring ax2 + bx + c, when c is positive

Before we tackle this factoring let’s go back and

review the F O I L method of how the product of two binomials works.

(2x + 3)(5x + 4) =

10x2 + (8x +15x) + 12 =

10x2 + 23x + 12

First Terms

Outer Terms

Inner Terms

Last Terms

Notice what happens when the multiply the Outer Terms and Inner

Terms.

NEXT

10x2 + 23x + 12To factor it, think of 23x as 8x + 15x.

8x + 15x = 23x10x2 + 23x + 12 = 10x2 + 8x + 15x + 12

Where did we get 8x and 15x?Notice that multiplying (a) 10 and (c) 12 gives you 120, which is the product of the x2-coefficient (10)

and the constant term (12).

8 and 15 are factors of 1201•120 2•60 3•40 4•30 5•24 6•20 8•15 10•12

also 8x + 15x = 23This example suggest that, to factor a trinomial, you should look for factors of the product ac that have a

sum of b.

Let’s see if it works.

In the form ofax2 + bx + c

Consider the trinomial 6x2 + 23x + 7. To factor it, think of 23x as 2x + 21x.

Where did we get 2 and 21?

If we multiply 6 and 7 we get 42, which is the product of the x2-coefficient (6) and the constant

(7).

2 and 21 are factor of 421•42 2•21 3•14 6•7

and2x + 21x = 23x

Yes it does work!So we must find the product of ac that have the

sum b.

NEXT

Now we must rewrite the trinomial using the factors you found for b. (2 and 21)

6x2 + 2x + 21x + 7

Now we are going to find the GCF by grouping terms. (Remember the Associative Property)

(6x2 + 2x) + (21x + 7)

2x(3x + 1) + 7(3x + 1)

Now lets use the Distributive Property to write the two binomials.

(2x + 7)(3x + 1)

FACTOR FACTOR

What do you notice about the

terms in the parenthesis?

Now we have factored6x2 + 23x + 7 to(2x + 7)(3x + 1)

Factor 5x2 + 11x + 2

5x2 + 11x + 2 = 5x2 + 1x + 10x + 2 Rewrite bx: 11x = 1x + 10x.

= (5x2 + 1x)(10x + 2) Group terms, Associative Property.

= x(5x + 1) + 2(5x + 1) Factor GCF of each pair of terms.

= (x + 2)(5x + 1) Use Distributive Property to write factored terms.

Step 1:Find factors of ac that

have a sum b.

Factors of 10 1 x 10 2 x 5

Sums of factors

11 7

Since ac = 10 and b = 11, find the positive factors of 10 that have a sum 11.

Step2:To factor the trinomial,

use the factors you found (1 + 10) to

rewrite bx.

TRY ANOTHER6x2 + 13x + 5

What is the factored form of 6x2 + 13x + 5?

6x2 + 13x + 5 = 6x2 + 3x + 10x + 5 Rewrite bx: 13x = 3x + 10x.

(6x2 + 3x)(10x + 5) Group terms together to factor.

3x(2x + 1) + 5(2x + 1) Factor GCF of each pair of terms.

(3x + 5)(2x + 1) Use Distributive Property to write binomials.

Factor:

1) 2n2 + 11n + 5

2) 5x2 + 34x + 24

3) 2y2 – 23y + 60

4) 4y2 + 62y + 30

5) 8t2 + 26t + 15

Factoring ax2 + bx +cFactoring ax2 + bx +c

Factoring when ac is negative

The sums of positive and negative numbers

gives us our b.

Can we apply the same steps we have learned to

factor trinomials that contain negative numbers?

Yes. Your goal is still to find factors of ac that have sum b. Because ac < 0 (less than), the

factors must have different signs.

We need to use all combinations of factors.

Ex: factors of -15.1•(-15) (-1)•15 3•(-5) (-3)•5

1+(-15)=-14 (-1)+15=14 3+(-5)=-2 (-3)+5 =2

Factors of -45 1, -45 -1, 45 3, -15 -3, 15 5, -9 -5, 9

Sums of factors -44 44 -12 12 -4 4

Factor 3x2 + 4x – 15

Find factors of ac with the sum b.Since ac = -45 and b = 4, find factors of -45.

3x2 -5x +9x – 15 Rewrite bx: 4x = -5x + 9x.

(3x2 -5x) + (9x – 15) Group terms together to factor.

x(3x – 5) + 3(3x– 5) Factor GCF of each pair of terms.

(x + 3)(3x – 5) Use Distributive to rewrite binomials.

Not all expressions of the form ax2 + bx – c can be factored.

This is especially common when the polynomial contains

subtraction. Try this.-10x2 + 21x - 5

Factor:

1) 3k2 + 4k – 4

2) 5x2 + 4x – 1

3) 10y2 – 11y – 6

4) 6q2 – 7q – 49

5) 2y2 + 11y – 90

TRY THIS

Geometry The area of a rectangle is 2y2 – 13y – 7.What are the possible dimensions of the rectangle? Use factoring.

2y2 – 13y – 7 =

Simplifying before factoring

Simplifying before factoring

Some polynomials can be factored repeatedly.

Some polynomials can be factored repeatedly. This means you can continue the process of

factoring until there are no common factors other than 1. If a trinomial has a common monomial

factor, factor it out before trying to find binomial factors.

Take for example:(6h + 2)(h + 5) and (3h + 1)(2h + 10)

Find each product.

6h2 + 32h + 10 and 6h+ 32h + 10

What do you notice about these two

polynomials?Can you factor a monomial before

factoring a binomial?

NEXT

Step 2:Rewrite bx using

factors.

Factor 20x2 + 80x + 35 completely.

20x2 + 80x + 35 = Factor out GCF monomial.

5(4x2 + 16x + 7) = Rewrite bx: 16x = 2x + 14x.

5[4x2 + 2x + 14x + 7)] = Factor GCF of each pair of terms.

5[2x(2x + 1) + 7(2x + 1)] = 5(2x + 7)(2x + 1) Rewrite using the

Distributive Property.

Step 1:Find factors of ac with

sum b.

Factors of 28 1 x 28 2 x 14 4 x 7

Sum of factors

29 16 11

Solving Polynomial Equations by

Factoring

Solving Polynomial Equations by

Factoring

Finding x

If you remember graphing linear equations, many times the line crossed the x-axis. You could find the x- and y-

intercepts of these lines.Polynomials have the same characteristics,

but quadratics can have no x-intercept, one x-intercept or two x-intercepts.

x-intercept

y-intercept

x-interceptx-intercept

y-intercept

Make a table of the polynomial shown below. y = x2 + 2x – 3

Identify the numbers that appear to be the x-intercepts.

Rewrite the equation by factoring the right side.[y = (x + 3)(x – 1)]

If you notice the x-intercept are solutions to the equation (x + 3)(x – 1) = 0

x-intercept = -3

x-intercept = 1

Remember the Standard Form of a Quadratic Equation ax2 + bx + c, where a ≠ 0. The value of the variable in a standard form equation is called the solution, or the

root, of the equation.

Let’s discuss the Multiplication Property of Zero which states that if a = 0 or b = 0, then

ab = 0. We can use the Zero-Product Property to solve quadratic equations once the quadratic expression has been factored into a

product of two linear factors.Let’s see how we can use this property to

solve quadratic expressions.

Apply the Zero-Product Property.

Solve (4x + 5)(3x – 2) = 0

(4x + 5)(3x – 2) = 0 The quadratic expression has already been factored.

4x + 5 = 0 or 3x – 2 = 0 If (4x + 5)(x – 2) = 0, then (4x + 5) = 0 or 3x – 2) = 0.

4x = -5 3x = 2 x = -5/4 x = 2/3 Solve each equation for x.

x-interceptx-intercept

TRY THIS

Solve (x + 5)(2x – 6) = 0

(x + 5)(2x – 6) = 0 x + 5 = 0 or 2x – 6 = 0 Use the Zero-Product Property.

2x = 6 Solve for x.

x = -5 or x = 3 Substitute – 5 for x. Substitute 3 for x.(x + 5)(2x – 6) = 0 (x + 5)(2x – 6) = 0(-5 + 5)[2(-5) – 6] = 0 (3 + 5)[2(3) – 6) = 0 (0)(-16) = 0 (8)(0) = 0

Solve

1) (x + 7)(x – 4) = 0

2) (x – 3)(x – 7) = 0

3) (x + 4)(2x – 9) = 0

4) (2x + 3)(x – 4) = 0

5) (x + 3)(x + 5) = 0

You can also use the Zero-Product Property to solve equations of the form ax2 + bx + c = 0, if the quadratic expression ax2 + bx + c can be factored.

Solve x2 – 8x – 48 = 0(x – 12)(x + 4) = 0 Factor x2 – 8x – 48.

x – 12 = 0 or x + 4 = 0 Use the Zero-Product Property.

x = 12 or x = -4 Solve for x.

Try This x2 + x – 12 = 0

Before solving a quadratic equation, you may need to add or subtract terms in order to write the equation in standard form. Then factor the quadratic expression.

Solve 2x2 – 5x = 88

2x2 – 5x – 88 = 0 Subtract 88 from each side.

(2x + 11)(x – 8) = 0 Factor 2x2 – 5x – 88.

2x + 11 = 0 or x – 8 = 0 Use the Zero=Product Property.

2x = -11 or x = -8 Solve for x.

x = -5.5

Try This x2 – 12x = -36

Solve

1) b2 + 3b – 4 = 0

2) y2 – 3y – 10 = 0

3) 2z2 – 10z = -12

4) n2 + n – 12 = 0

5) x2 + 8x = -15

6) 4y2 = 25

MORE

Write each equation in standard form. Then solve.

1) 2q2 + 22q = -60

2) 3a2 + 4a = 2a2 – 2a – 9

3) 4x2 + 20 = 10x + 3x2 – 4

4) 3t2 + 8t = t2 – 3t – 12

Solving Cubic Equations by

Factoring

Solving Cubic Equations by

Factoring

Cubic Equationsax3 +bx2 + cx + d = 0

The standard form of a cubic equation in x is any equation that can be written in the form

ax3 + bx2 + cx + d = 0, where a ≠ 0.Using an extension of the Zero-Product Property,

you can solve many cubic equations.If a, b, and c represent real numbers and abc =

0,[(5)(4)(0) = 0]

then a = 0, b = 0 or c = 0.For example,

if x(x – 2)(3x + 4) = 0, you can write the following.

x = 0 or x – 2 = 0 or 3x + 4 = 0

Solving each for x, then x = 0, 2, and -4/3

Solve 2n3 + 8n2 – 42n = 0

2n3 + 8n2 – 42n = 0 Factor the GCF, 2n, from each term.

2n(n2 + 4n – 21) = 0 Factor n2 + 4n – 21.

2n(n + 7)(n – 3) = 0 Apply the Zero-Product Property.

2n = 0 or (n + 7) = 0 or (n – 3) = 0 Solve for n.

n = 0 or n = - 7 or n = 3

The solutions are 0, -7, and 3.

Try 10k3 – 13k2 + 4k =0

Solve m3 + 22m2 + 121m = 0

m3 + 22m2 + 121m = 0 Factor for GCF, m, from each term.

m(m2 + 22m + 121) = 0 Factor m2 + 22m + 121.

m(m + 11)(m + 11) = 0 Apply the Zero-Product Property.

m = 0 or m + 11 = 0 or m + 11 = 0 Solve for n.

m = 0 m = -11 m = - 11

The solution are 0 and -11.

Try z3 – 14z2 + 49z = 0

The steps for solving a polynomial by factoring are:

Step 1: Write the equation in standard form.Step 2: Factor the GCF, if one exists, from each term

in the equation.Step 3: Factor the polynomial

Step 4: Apply the Zero-Product Property and set each factor equal to zero.

Step 5: Solve for the variable.Step 6: Check you solution(s) in the original

equation.