Math  /  Algebra

Question218w23w28w33w27w612w2+80w28\frac{2-18 w^{2}}{3 w^{2}-8 w-3} \cdot \frac{3 w^{2}-7 w-6}{12 w^{2}+80 w-28}

Studdy Solution
Our **final simplified expression** is: (3w+1)(3w+2)2(3w+1)(w+7)=(3w+2)2(w+7) \frac{-(3w + 1)(3w + 2)}{2(3w + 1)(w + 7)} = \frac{-(3w + 2)}{2(w + 7)} Remember that ww cannot be a value that makes the denominator zero, so w13w \ne -\frac{1}{3}, w13w \ne \frac{1}{3}, w3w \ne 3 and w7w \ne -7.

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