6.4 Factoring and Solving Polynomial Equations

Factor Polynomial ExpressionsIn the previous lesson, you factored various polynomial expressions.

Such as:x3 – 2x2 =x4 – x3 – 3x2 + 3x =

= =

Common Factor

x2(x – 2)

x[x2(x – 1) – 3(x – 1)]

x(x2 – 3)(x – 1)

x(x3 – x2 – 3x + 3)

Grouping – common factor the first two terms and then the last two terms.

Common Factor

Solving Polynomial Equations

The expressions on the previous slide are now equations:

y = x3 – 2x2 and y = x4 – x3 – 3x2 +3x

To solve these equations, we will be solving for x when y = 0.

Solve

y = x3 – 2x2 0 = x3 – 2x2

0 = x2(x – 2)

x2 = 0 or x – 2 = 0 x = 0 x = 2

Therefore, the roots are 0 and 2.

Let y = 0

Common factor

Separate the factors and set them equal to zero.

Solve for x

Solve

y = x4 – x3 – 3x2 + 3x 0 = x4 – x3 – 3x2 + 3x 0 = x(x3 – x2 – 3x + 3)

0 =x[x2(x – 1) – 3(x – 1)]0 = x(x – 1)(x2 – 3)

x = 0 or x – 1 = 0 or x2 – 3 = 0

x = 0 x = 1 x =

Therefore, the roots are 0, 1 and ±1.73

Let y = 0

Common factor

Separate the factors and set them equal to zero.

Solve for x

3

Group

The Quadratic Formula

02

42

awherea

acbbx

For equations in quadratic form: ax2 + bx + c = 0, we can use the quadratic formula to solve for the roots of the equation.

This equation is normally used when factoring is not an option.

Using the Quadratic Formula

Solve the following cubic equation:

y = x3 + 5x2 – 9x

0 = x(x2 + 5x – 9)

x = 0 x2 + 5x – 9 = 0

We can, however, use the quadratic formula.

YES it can – YES it can – common factor.common factor.

Can this equation be Can this equation be factored?factored?

We still need to solve for x We still need to solve for x here. Can this equation be here. Can this equation be factored?factored?

No. There are no two No. There are no two integers that will multiply integers that will multiply to -9 and add to 5.to -9 and add to 5.a = 1

b = 5

c = -941.1,41.6

2

615

)1)(2(

)9)(1(4)5()5( 2

x

x

x

Therefore, the roots are 0, 6.41 and -1.41.

Remember, the root 0 came from an earlier step.

Factoring Sum or Difference of Cubes

If you have a sum or difference of cubes such as a3 + b3 or

a3 – b3, you can factor by using the following patterns.

3 3 2 2

3 3 2 2

Sum of Two Cubes

( )( )

Difference of Two Cubes

( )( )

a b a b a ab b

a b a b a ab b

Note: The first and last term are cubed and these are binomials.

Example

Factor x3 + 343.

Note: This is a binomial. Are the first and last terms cubed? 7343 33 3 xx

))(( 2233 babababa

x3 + 343 = (x)3 + (7)3

= ( + )( - + ) x 7 x2 7x 49

ExampleFactor 64a4 – 27a

= a(64a3 – 27)

Note: Binomial. Is the first and last terms cubes?

= a( (4a)3 – (3)3) Note: 3 3 2 2( )( )a b a b a ab b

= a( - )( + + )4a 3 16a2 12a 9

Factor by Grouping

Some four term polynomials can be factor by grouping.

Example. Factor 3x3 + 7x2 +12x + 28

Step 1 Pair the terms.)2812()73( 23 xxx

Step 2 Factor out common factor from each pair.

)73(4)73(2 xxx

Identical factors

Step 3 Factor out common factor from each term.

)4)(73( 2 xx

ExampleFactor 3x3 + 7x2 -12x - 28

Step 1 )2812()73( 23 xxx Note: Subtraction is the same as adding a negative

Step 2 )73)(4()73(2 xxx

Step 3 2(3 7)( 4)x x

Note: This factor can be further factored(3 7)( 2)( 2)x x x

Solving Polynomial Equations

Solve 3 22 9 18x x x

Set equation equal to zero.3 22 9 18 0x x x

Factor.3 2

2

2

( 2 ) ( 9 18) 0

( 2) ( 9)( 2) 0

( 9)( 2) 0

( 3)( 3)( 2) 0

x x x

x x x

x x

x x x

Set each factor equal to zero and solve.( 3) 0 ( 3) 0 ( 2) 0

3 3 2

x x x

x x x