Question

In Exercises 15–32, multiply or divide as indicated. (x3−25x)/4x2 ⋅ (2x2−2)/(x2−6x+5) ÷ (x2+5x)/(7x+7)

Accepted Answer

Factorize all the polynomials in the given expression. For the numerator and denominator of the first fraction: $$x^3 - 25x$$ can be factored as $$x(x^2 - 25)$$, and $$x^2 - 25$$ can be further factored as $$(x - 5)(x + 5). $$The denominator $$4x^2$$ remains as is. For the second fraction: $$2x^2 - 2$$ can be factored as $$2(x^2 - 1)$$, and $$x^2 - 1$$ can be further factored as $$(x - 1)(x + 1). $$The denominator $$x^2 - 6x + 5$$ factors as $$(x - 5)(x - 1). $$For the third fraction (which is part of the division): $$x^2 + 5x$$ can be factored as $$x(x + 5)$$, and $$7x + 7$$ can be factored as $$7(x + 1). $$Rewrite the division problem as multiplication by the reciprocal of the third fraction. This means flipping the numerator and denominator of the third fraction. The expression becomes: $$ \frac{x(x - 5)(x + 5)}{4x^2} \cdot \frac{2(x - 1)(x + 1)}{(x - 5)(x - 1)} \cdot \frac{7(x + 1)}{x(x + 5)}. $$ Simplify the expression by canceling out common factors in the numerators and denominators. Look for terms such as $$x$$, $$x - 5$$, $$x + 5$$, $$x - 1$$, and $$x + 1$$ that appear in both the numerator and denominator. After canceling, multiply the remaining terms to get the simplified result.

In Exercises 15–32, multiply or divide as indicated. (x³−25x)/4x² ⋅ (2x²−2)/(x²−6x+5) ÷ (x²+5x)/(7x+7)

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