2 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 Factorx2(x – 2)x(x3 – x2 – 3x + 3)x[x2(x – 1) – 3(x – 1)]Common Factorx(x2 – 3)(x – 1)
3 Solving 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.
4 Solve 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
5 Solve 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
6 The 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.
7 Using 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
8 Factoring 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.
9 ExampleFactor xNote: This is a binomial. Are the first and last terms cubed?x = (x)3 + (7)3= ( )( )x7x27x49
10 Example Factor 64a4 – 27a = a(64a3 – 27) Note: Binomial. Is the first and last terms cubes?= a( (4a)3 – (3)3)Note:= a( )( )4a316a212a9
11 Factor 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.
12 Example Factor 3x3 + 7x2 -12x - 28 Step 1 Note: Subtraction is the same as adding a negativeStep 2Step 3Note: This factor can be further factored
13 Solving Polynomial Equations SolveSet equation equal to zero.Factor.Set each factor equal to zero and solve.