Multiply: $$ \left(\frac{4x^3-64x}{(2x^2-9x)-5}\right) \times \left(\frac{(2x^2-7x)-4}{(4x^2+32x)+64}\right) = $$ Your answer must be in fully factored, simplified form, but you do not need to state any necessary restrictions on the value of $x$.
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The expression is: $$ \left(\frac{4x^3-64x}{(2x^2-9x)-5}\right) \times \left(\frac{(2x^2-7x)-4}{(4x^2+32x)+64}\right) $$ First, let's simplify the numerator and denominator of the first fraction. Numerator 1: $4x^3 - 64x$ Factor out $4x$: $4x(x^2 - 16)$ Recognize Show more…
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