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6.2,6.3 Factoring Trinomials ü Trinomials of the Form x 2 + b x + c If a polynomial of the form x2 + b x + c is factorable then it can be written as Hx + uL Hx + vL. Note: Hx + uL Hx + vL = x2 + v x + u x + u ÿ v = x2 + Hu + vL x + u ÿ v Looking at this carefully we see that if a trinomial of the form x2 + b x + c is factorable then b = u + v and c = u ÿ v, for some numbers u and v. So to factor this form of a polynomial we need to find two numbers that have product c and that have sum b. ü Examples Example 1: Factor x2 + 7 x + 12 Solution: We need to find two numbers that multiply to be 12, but add to be 7. These two numbers are 3 and 4. This then gives x2 + 7 x + 12 = Hx + 3L Hx + 4L the factorization Example 2: Factor x2 - 12 x + 35 Solution: We need to find two numbers that multiply to be 35, but add to be -12. These two numbers are -7 and - 5. This then x2 - 12 x + 35 = Hx - 7L Hx - 5L gives the factorization Example 3: Factor x2 + 5 x - 24 Solution: We need to find two numbers that multiply to be -24, but add to be 5. These two numbers are -3 and 8. This then x2 + 5 x - 24 = Hx - 3L Hx + 8L gives the factorization ü Trinomials of the Form a x 2 + b x + c If a polynomial of the form a x2 + b x + c is factorable then it can be written as Hm x + uL Hn x + vL. Hm x + uL Hn x + vL = m ÿ n x2 + m ÿ v x + n ÿ u x + u ÿ v Note: = m ÿ n x2 + Hm ÿ v + n ÿ uL x + u ÿ v a ÿ c = m ÿ n ÿ u ÿ v = Hm ÿ vL Hn ÿ uL, for some numbers m, n, u and v. So to factor this form of a polynomial we need to find two Looking at this carefully we see that if a trinomial of the form a x2 + b x + c is factorable then b = m ÿ v + n ÿ u and numbers that have product a ÿ c and that have sum b. We can then rewrite the polynomial as four terms (similar to the first step above) and then factor by grouping. ü Examples Example 1: Factor 6 x2 + 7 x + 2 Solution: First we find the product of a and c. In this case it's 6 ÿ 2 = 12. So we need two numbers that multiply to be 12, but add to be 7. These numbers are 3 and 4. We use these numbers to rewrite the middle term as 3 x + 4 x and then factor by grouping as follows: 6 x2 + 7 x + 2 = 6 x2 + 3 x + 4 x + 2 = I6 x2 + 3 xM + H4 x + 2L = 3 xH2 x + 1L + 2 H2 x + 1L = H2 x + 1L H3 x + 2L Example 2: Factor 10 x2 - 23 x + 12 Solution: First we find the product 10 ÿ 12 = 120. So we need two numbers that multiply to be 120 and add to be -23. These two numbers are -15 and - 8. So the middle term can be rewritten as -15 x - 8 x. Now factor by grouping. 10 x2 - 23 x + 12 = 10 x2 - 15 x - 8 x + 12 = I10 x2 - 15 xM + H-8 x + 12L = 5 xH2 x - 3L - 4 H2 x - 3L = H2 x - 3L H5 x - 4L Example 3: Factor 24 x2 - 14 x - 3 Solution: First we find the product 24 ÿ -3 = -72. So we need two numbers that multiply to be -72 and add to be -14. These two numbers are -18 and 4. So the middle term can be rewritten as -18 x + 4 x. Now factor by grouping. 24 x2 - 14 x - 3 = 24 x2 - 18 x + 4 x - 3 = I24 x2 - 18 xM + H4 x - 3L = 6 xH4 x - 3L + 1 H4 x - 3L = H4 x - 3L H6 x + 1L

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rational expressions, factoring trinomials, final exam, real numbers, quadratic equations, radical expressions, solving equations, linear equations in two variables, problem solving, intermediate algebra, chapter 6, factoring by grouping, rational exponents, solving quadratic equations, word problems

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posted: | 12/23/2009 |

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