What is (x^4+8x^3)/(8x)*(x^2-9)/(x^2+11x+24) ?

The solution to (x^4+8x^3)/(8x)*(x^2-9)/(x^2+11x+24) is (x^2(x-3))/8

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simplify (x^4+8x^3)/(8x)*(x^2-9)/(x^2+11x+24)

simplify

\frac{x ^{4} + 8 x ^{3}}{8 x} \cdot \frac{x ^{2} - 9}{x ^{2} + 11 x + 24}

\frac{x ^{2} ( x - 3 )}{8}

Solution steps

\frac{x ^{4} + 8 x ^{3}}{8 x} \cdot \frac{x ^{2} - 9}{x ^{2} + 11 x + 24}

Cancel

\frac{x ^{4} + 8 x ^{3}}{8 x} : \frac{x ^{2} ( x + 8 )}{8}

= \frac{x ^{2} ( x + 8 )}{8} \cdot \frac{x ^{2} - 9}{x ^{2} + 11 x + 24}

Cancel

\frac{x ^{2} - 9}{x ^{2} + 11 x + 24} : \frac{x - 3}{x + 8}

= \frac{x ^{2} ( x + 8 )}{8} \cdot \frac{x - 3}{x + 8}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{x ^{2} ( x + 8 ) ( x - 3 )}{8 ( x + 8 )}

Cancel the common factor:

x + 8

= \frac{x ^{2} ( x - 3 )}{8}

\frac{42}{8}

\frac{x ^{2} - 1}{x ^{2} - 3 x + 2} \cdot \frac{x ^{3} - 2 x ^{2}}{7 x + 7}

2 \frac{5}{7}

lo g_{10} (0.002)

(- 3 x^{4} - 4 x^{3} - 1) + (- x^{4} + 4 x^{3} + 2 x^{2} + x + 1)

What is (x^4+8x^3)/(8x)*(x^2-9)/(x^2+11x+24) ?
The solution to (x^4+8x^3)/(8x)*(x^2-9)/(x^2+11x+24) is (x^2(x-3))/8