Algebra-Trigonometry-CynthiaYoung-3ed8.pdf - c00b.qxd 11:17 AM Page 41 0.4 Factoring Polynomials EXAMPLE 5 41 Factoring the Difference of Two Cubes

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0.4 Factoring Polynomials 41 Answer: a. ( x 2)( x 2 2 x 4) b. ( x 4)( x 2 4 x 16) Factoring a Trinomial as a Product of Two Binomials The first step in factoring is to look for a common factor. If there is no common factor, look to see whether the polynomial is a special form for which we know the factoring formula. If it is not of such a special form and if it is a trinomial, then we proceed with a general factoring strategy. We know that ( x 3)( x 2) x 2 5 x 6, so we say the factors of x 2 5 x 6 are ( x 3) and ( x 2) . In factored form we have x 2 5 x 6 ( x 3)( x 2). Recall the FOIL method from Section 0.3. The product of the last terms (3 and 2) is 6, and the sum of the products of the inner terms (3 x ) and the outer terms (2 x ) is 5 x . Let’s pretend for a minute that we didn’t know this factored form but had to work with the general form: The goal is to find a and b . We start by multiplying the two binomials on the right. Compare the expression we started with on the left with the expression on the far right x 2 5 x 6 x 2 ( a b ) x ab . We see that ab 6 and ( a b ) 5 . Start with the possible combinations of a and b whose product is 6, and then look among those for the combination whose sum is 5. ab 6 a , b : 1, 6 1, 6 2, 3 2, 3 a b 7 7 5 5 All of the possible a , b combinations in the first row have a product equal to 6, but only one of those has a sum equal to 5. Therefore the factored form is x 2 + 5 x + 6 = ( x + a )( x + b ) = ( x + 2)( x + 3) x 2 + 5 x + 6 = ( x + a )( x + b ) = x 2 + ax + bx + ab = x 2 + ( a + b ) x + ab x 2 + 5 x + 6 = ( x + a )( x + b ) EXAMPLE 5 Factoring the Difference of Two Cubes Factor x 3 125. Solution: Rewrite as the difference of two cubes. x 3 125 x 3 5 3 Write the difference of two cubes formula. a 3 b 3 ( a b )( a 2 ab b 2 ) Let a x and b 5. x 3 125 x 3 5 3 ( x 5)( x 2 5 x 25) YOUR TURN Factor: a. x 3 8 b. x 3 64
42 CHAPTER 0 Prerequisites and Review Answer: ( x 4)( x 5) Answer: ( x 6)( x 3) In Example 6, all terms in the trinomial are positive. When the constant term is negative, then (regardless of whether the middle term is positive or negative) the factors will be opposite in sign, as illustrated in Example 7. EXAMPLE 7 Factoring a Trinomial Factor x 2 3 x 28. Solution: Write the trinomial as a product of two binomials in general form. x 2 3 x 28 ( x )( x ) Write all of the integers whose product is 28. Integers whose product is 28 1, 28 1, 28 2, 14 2, 14 4, 7 4, 7 Determine the sum of the integers. Integers whose product is 28 1, 28 1, 28 2, 14 2, 14 4, 7 4, 7 Sum 27 27 12 12 3 3 Select 4, 7 because the product is 28 (last term of the trinomial) and the sum is 3 (middle term coefficient of the trinomial). x 2 3 x 28 ( x 4)( x 7) Check: ( x 4)( x 7) x 2 7 x 4 x 28 x 2 3 x 28 YOUR TURN Factor x 2 3 x 18. n n EXAMPLE 6 Factoring a Trinomial Factor x 2 10 x 9. Solution: Write the trinomial as a product of two binomials in general form. x 2 10 x 9 ( x n )( x n ) Write all of the integers whose product is 9.

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• Summer '17
• juan alberto
• Fractions, Elementary arithmetic, Greatest common divisor