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EXAMPLE 1

EXAMPLE 1. Find a common monomial factor. Factor the polynomial completely. a. x 3 + 2 x 2 – 15 x. = x ( x 2 + 2 x – 15). Factor common monomial. = x ( x + 5)( x – 3 ). Factor trinomial. = 2 y 3 ( y 2 – 9). b. 2 y 5 – 18 y 3. Factor common monomial.

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EXAMPLE 1

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  1. EXAMPLE 1 Find a common monomial factor Factor the polynomial completely. a. x3 + 2x2 – 15x = x(x2 + 2x – 15) Factor common monomial. = x(x + 5)(x – 3) Factor trinomial. = 2y3(y2 – 9) b. 2y5 – 18y3 Factor common monomial. = 2y3(y + 3)(y – 3) Difference of two squares = 4z2(z2 – 4z + 4) c. 4z4 – 16z3 + 16z2 Factor common monomial. = 4z2(z – 2)2 Perfect square trinomial

  2. = 2z2 (2z)3 – 53 EXAMPLE 2 Factor the sum or difference of two cubes Factor the polynomial completely. = x3 + 43 a. x3 + 64 Sum of two cubes = (x + 4)(x2 – 4x + 16) = 2z2(8z3 – 125) b. 16z5 – 250z2 Factor common monomial. Difference of two cubes = 2z2(2z – 5)(4z2 + 10z + 25)

  3. EXAMPLE 3 Factor by grouping Factor the polynomial x3 – 3x2 – 16x + 48 completely. = x2(x – 3) –16(x – 3) x3– 3x2– 16x + 48 Factor by grouping. = (x2– 16)(x – 3) Distributive property = (x+ 4)(x – 4)(x – 3) Difference of two squares

  4. EXAMPLE 4 Factor polynomials in quadratic form Factor completely:(a) 16x4 – 81and(b) 2p8 + 10p5 + 12p2. a. 16x4 – 81 = (4x2)2 – 92 Write as difference of two squares. = (4x2 + 9)(4x2 – 9) Difference of two squares = (4x2 + 9)(2x + 3)(2x – 3) Difference of two squares Factor common monomial. b. 2p8 + 10p5 + 12p2 = 2p2(p6 + 5p3 + 6) Factor trinomial in quadratic form. =2p2(p3 + 3)(p3 + 2)

  5. EXAMPLE 5 Standardized Test Practice SOLUTION 3x5 + 15x = 18x3 Write original equation. 3x5 – 18x3 + 15x = 0 Write in standard form. 3x(x4 – 6x2 + 5) = 0 Factor common monomial.

  6. x = 0, x = – 1, x = 1, x = 5 ,or x = – 5 The correct answer is D. ANSWER EXAMPLE 5 Standardized Test Practice Factor trinomial. 3x(x2 – 1)(x2 – 5) = 0 3x(x + 1)(x – 1)(x2– 5) = 0 Difference of two squares Zero product property

  7. EXAMPLE 1 Use polynomial long division Divide f (x) = 3x4 – 5x3 + 4x – 6 byx2 – 3x + 5. SOLUTION Write polynomial division in the same format you use when dividing numbers. Include a “0” as the coefficient of x2 in the dividend. At each stage, divide the term with the highest power in what is left of the dividend by the first term of the divisor. This gives the next term of the quotient.

  8. quotient ) x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6 3x4 – 9x3 + 15x2 4x3– 12x2 + 20x –3x2 + 9x – 15 remainder EXAMPLE 1 Use polynomial long division 3x2+ 4x – 3 Multiply divisor by 3x4/x2 = 3x2 Subtract. Bring down next term. 4x3–15x2+ 4x Multiply divisor by4x3/x2 = 4x Subtract. Bring down next term. – 3x2 – 16x – 6 Multiply divisor by– 3x2/x2 = – 3 – 25x + 9

  9. 3x4 – 5x3 + 4x – 6 – 25x + 9 ANSWER = 3x2 + 4x – 3 + x2 – 3x + 5 x2 – 3x + 5 = 3x4–5x3+ 4x –6 EXAMPLE 1 Use polynomial long division CHECK You can check the result of a division problem by multiplying the quotient by the divisor and adding the remainder. The result should be the dividend. (3x2 + 4x – 3)(x2 – 3x + 5) + (– 25x + 9) = 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9 = 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9

  10. quotient ) x – 2 x3 + 5x2 – 7x + 2 x3 – 2x2 7x2– 14x 7x – 14 remainder x3 + 5x2– 7x +2 16 ANSWER = x2 + 7x + 7 + x – 2 x – 2 EXAMPLE 2 Use polynomial long division with a linear divisor Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2. x2+ 7x + 7 Multiply divisor byx3/x = x2. 7x2 – 7x Subtract. Multiply divisor by7x2/x = 7x. 7x + 2 Subtract. Multiply divisor by7x/x = 7. 16

  11. – 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

  12. – 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.

  13. 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.

  14. 3 1 – 2 – 23 60 3 3 – 60 1 1 – 20 0 EXAMPLE 5 Standardized Test Practice SOLUTION Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division.

  15. The correct answer is A. ANSWER EXAMPLE 5 Standardized Test Practice Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x3– 2x2– 23x + 60 = (x – 3)(x2 + x – 20) = (x – 3)(x + 5)(x – 4) The zeros are3, – 5, and4.

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