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Polynomial Division and Factoring: Examples and Solutions

Learn how to divide polynomials using synthetic division and factor them completely with examples and step-by-step solutions.

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Polynomial Division and Factoring: Examples and Solutions

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  1. – 3 2 1 – 8 5 – 6 15 – 21 2x3 + x2– 8x + 5 16 2 – 5 7 – 16 = 2x2– 5x + 7 – ANSWER x + 3 x + 3 EXAMPLE 3 Use synthetic division Dividef(x)= 2x3 + x2– 8x + 5byx +3using synthetic division. SOLUTION

  2. – 2 3 – 4 – 28 – 16 – 6 20 16 3 – 10 – 8 0 EXAMPLE 4 Factor a polynomial Factorf (x) = 3x3– 4x2– 28x – 16completely given that x + 2is a factor. SOLUTION Because x + 2 is a factor of f (x), you know that f (–2)= 0. Use synthetic division to find the other factors.

  3. EXAMPLE 4 Factor a polynomial Use the result to write f (x) as a product of two factors and then factor completely. f (x) = 3x3– 4x2– 28x – 16 Write original polynomial. = (x + 2)(3x2– 10x – 8) Write as a product of two factors. = (x + 2)(3x + 2)(x – 4) Factor trinomial.

  4. x3 + 4 x2–x – 1 – 3 1 4 – 1 – 1 11 = x2+ x – 4 + ANSWER x + 3 x + 3 – 3 – 3 12 3 1– 4 11 for Examples 3 and 4 GUIDED PRACTICE Divide using synthetic division. 3. (x3 + 4x2–x – 1)  (x + 3) SOLUTION (x3 + 4x2–x – 1)  (x + 3)

  5. 4x3 + x2– 3x + 1 1 4 1 – 3 7 9 = 4x2+ 5x + 2 + ANSWER x – 1 x – 1 45 2 4 52 9 for Examples 3 and 4 GUIDED PRACTICE 4. (4x3 + x2– 3x + 7)  (x – 1) SOLUTION (4x3 + x2– 3x + 7)  (x – 1)

  6. 4 1 – 6 512 4– 8 –12 1 – 2 – 3 0 for Examples 3 and 4 GUIDED PRACTICE Factor the polynomial completely given that x –4 is a factor. 5. f (x) = x3– 6x2 + 5x + 12 SOLUTION Because x – 4 is a factor of f (x), you know that f (4)= 0. Use synthetic division to find the other factors.

  7. for Examples 3 and 4 GUIDED PRACTICE Use the result to write f (x) as a product of two factors and then factor completely. f (x) = x3– 6x2+ 5x + 12 Write original polynomial. = (x – 4)(x2– 2x – 3) Write as a product of two factors. = (x – 4)(x –3)(x + 1) Factor trinomial.

  8. 4 1 4– 22 40 412–40 1 3– 10 0 for Examples 3 and 4 GUIDED PRACTICE 6. f (x) = x3 –x2– 22x + 40 SOLUTION Because x – 4 is a factor of f (x), you know that f (4)= 0. Use synthetic division to find the other factors.

  9. for Examples 3 and 4 GUIDED PRACTICE Use the result to write f (x) as a product of two factors and then factor completely. f (x) = x3–x2– 22x + 40 Write original polynomial. = (x – 4)(x2+ 3x – 10) Write as a product of two factors. = (x – 4)(x –2)(x +5) Factor trinomial.

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