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6.4 Factoring and Solving Polynomial Equations

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Factor Polynomial Expressions In the previous lesson, you factored various polynomial expressions. Such as: x 3 – 2x 2 = x 4 – x 3 – 3x 2 + 3x = = = Common Factor x 2 (x – 2) x[x 2 (x – 1) – 3(x – 1)] x(x 2 – 3)(x – 1) x(x 3 – x 2 – 3x + 3) Grouping – common factor the first two terms and then the last two terms. Common Factor

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Solving Polynomial Equations The expressions on the previous slide are now equations: y = x 3 – 2x 2 and y = x 4 – x 3 – 3x 2 +3x To solve these equations, we will be solving for x when y = 0.

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Solve y = x 3 – 2x 2 0 = x 3 – 2x 2 0 = x 2 (x – 2) x 2 = 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

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Solve y = x 4 – x 3 – 3x 2 + 3x 0 = x 4 – x 3 – 3x 2 + 3x 0 = x(x 3 – x 2 – 3x + 3) 0 =x[x 2 (x – 1) – 3(x – 1)] 0 = x(x – 1)(x 2 – 3) x = 0 or x – 1 = 0 or x 2 – 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 Group

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The Quadratic Formula For equations in quadratic form: ax 2 + 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.

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Using the Quadratic Formula Solve the following cubic equation: y = x 3 + 5x 2 – 9x 0 = x(x 2 + 5x – 9) x = 0x 2 + 5x – 9 = 0 We can, however, use the quadratic formula. YES it can – common factor. Can this equation be factored? We still need to solve for x here. Can this equation be factored? 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 Remember, the root 0 came from an earlier step.

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Factoring Sum or Difference of Cubes If you have a sum or difference of cubes such as a 3 + b 3 or a 3 – b 3, you can factor by using the following patterns. Note: The first and last term are cubed and these are binomials.

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Example Factor x Note: This is a binomial. Are the first and last terms cubed? x = (x) 3 + (7) 3 = ( + )( - + )x7x2x2 7x49

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Example Factor 64a 4 – 27a = a(64a 3 – 27) Note: Binomial. Is the first and last terms cubes? = a( (4a) 3 – (3) 3 ) Note: = a( - )( + + )4a316a 2 12a9

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Factor by Grouping Some four term polynomials can be factor by grouping. Example. Factor 3x 3 + 7x 2 +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.

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

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

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