2Factor 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 Factorx2(x – 2)x(x3 – x2 – 3x + 3)x[x2(x – 1) – 3(x – 1)]Common Factorx(x2 – 3)(x – 1)
3Solving Polynomial Equations The expressions on the previous slide are now equations:y = x3 – 2x2 and y = x4 – x3 – 3x2 +3xTo solve these equations, we will be solving for x when y = 0.
4Solve y = x3 – 2x2 0 = x3 – 2x2 0 = x2(x – 2) x2 = 0 or x – 2 = 0 Let y = 0y = x3 – 2x20 = x3 – 2x20 = x2(x – 2)x2 = 0 or x – 2 = 0x = 0 x = 2Therefore, the roots are 0 and 2.Common factorSeparate the factors and set them equal to zero.Solve for x
5Solve Let y = 0 y = x4 – x3 – 3x2 + 3x 0 = x4 – x3 – 3x2 + 3x x = 0 or x – 1 = 0 or x2 – 3 = 0x = x = x =Therefore, the roots are 0, 1 and ±1.73Let y = 0Common factorGroupSeparate the factors and set them equal to zero.Solve for x
6The Quadratic FormulaFor 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.
7Using the Quadratic Formula Solve the following cubic equation:y = x3 + 5x2 – 9x0 = x(x2 + 5x – 9)x = 0 x2 + 5x – 9 = 0We 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 = 1b = 5c = -9Therefore, the roots are 0, 6.41 and
8Factoring Sum or Difference of Cubes If you have a sum or difference of cubes such as a3 + b3 ora3 – b3, you can factor by using the following patterns.Note: The first and last term are cubed and these are binomials.
9ExampleFactor xNote: This is a binomial. Are the first and last terms cubed?x = (x)3 + (7)3= ( )( )x7x27x49
10Example Factor 64a4 – 27a = a(64a3 – 27) Note: Binomial. Is the first and last terms cubes?= a( (4a)3 – (3)3)Note:= a( )( )4a316a212a9
11Factor by GroupingSome four term polynomials can be factor by grouping.Example Factor 3x3 + 7x2 +12x + 28Step 1 Pair the terms.Step 2 Factor out common factor from each pair.Identical factorsStep 3 Factor out common factor from each term.
12Example Factor 3x3 + 7x2 -12x - 28 Step 1 Note: Subtraction is the same as adding a negativeStep 2Step 3Note: This factor can be further factored
13Solving Polynomial Equations SolveSet equation equal to zero.Factor.Set each factor equal to zero and solve.