Question: A: I will use factor by grouping, A 8k2 - 6k - 9 The coefficient of k2 is 8 and the constant term os -9.

A: I will use factor by grouping, A 8k2 - 6k - 9 The coefficient of k2 is 8 and the constant term os -9. The product of 8 and -9 is -72. The factors of -72 which sum to -6 are 6 and -12. This equation cannot by factored by grouping, No factor of 72 sum up to 6. So 8k2 - 6k - 9 = 8k2 - 12k + 6k - 9 = 2k (4k + 3) - 3 (4k + 3) 2k (4k + 3) - 3 (4k + 3) Since the two quantities in parentheses are now identical that means we can factor out the common factors. (4k + 3) (2k - 3) B: Factor by grouping, Using factor by grouping again, B7x' +7 - yx2 - y *Polynomial Degree:3 *Leading Term -xy *Leading Coefficient -1 Factor out 7 from 7x2+7 7 (x2 + 1) Factor out -y from -x y - y - y (x2 + 1) =7 (x2 +1) -y(x2 +1) Factor out the common term x2 + 1 (x2 + 1) (7 - y)

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