# 6.4 Factoring and Solving Polynomial Equations

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6.4 Factoring and Solving Polynomial Equations

Factor Polynomial Expressions
In the previous lesson, you factored various polynomial expressions. Such as: x3 – 2x2 = x4 – x3 – 3x2 + 3x = = Grouping – common factor the first two terms and then the last two terms. Common Factor x2(x – 2) x(x3 – x2 – 3x + 3) x[x2(x – 1) – 3(x – 1)] Common Factor x(x2 – 3)(x – 1)

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
Let y = 0 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. Common factor Separate the factors and set them equal to zero. Solve for x

Solve Let y = 0 y = x4 – x3 – 3x2 + 3x 0 = x4 – x3 – 3x2 + 3x
x = 0 or x – 1 = 0 or x2 – 3 = 0 x = x = x = Therefore, the roots are 0, 1 and ±1.73 Let y = 0 Common factor Group Separate the factors and set them equal to zero. Solve for x

The Quadratic Formula 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.

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. Can this equation be factored? We still need to solve for x here. Can this equation be factored? YES it can – common factor. Remember, the root 0 came from an earlier step. No. There are no two integers that will multiply to -9 and add to 5. a = 1 b = 5 c = -9 Therefore, the roots are 0, 6.41 and

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

Example Factor x Note: This is a binomial. Are the first and last terms cubed? x = (x)3 + (7)3 = ( )( ) x 7 x2 7x 49

Example Factor 64a4 – 27a = a(64a3 – 27)
Note: Binomial. Is the first and last terms cubes? = a( (4a)3 – (3)3) Note: = 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. Step 2 Factor out common factor from each pair. Identical factors Step 3 Factor out common factor from each term.

Example Factor 3x3 + 7x2 -12x - 28 Step 1
Note: Subtraction is the same as adding a negative Step 2 Step 3 Note: This factor can be further factored

Solving Polynomial Equations
Solve Set equation equal to zero. Factor. Set each factor equal to zero and solve.