Question

If f(x)=x-6 and g(x)=x^1/2(x+3) what is f(x) times g(x

Answer 1

$\bf \begin{cases} f(x)=x-6\\ g(x)=x^{\frac{1}{2}}(x+3) \end{cases} \\\\\\ f(x)\cdot g(x)\implies (x-6)\cdot \left[ x^{\frac{1}{2}}(x+3) \right]\implies (x-6)\cdot \left[ \stackrel{distributing}{x^{\frac{3}{2}}+3x^{\frac{1}{2}}} \right]$

$\bf x\left( x^{\frac{3}{2}}+3x^{\frac{1}{2}} \right)-6\left( x^{\frac{3}{2}}+3x^{\frac{1}{2}} \right) \\\\\\ x^{\frac{5}{2}}+3x^{\frac{3}{2}}~~-~~6x^{\frac{3}{2}}-18x^{\frac{1}{2}}\implies \stackrel{\textit{adding like-terms}}{x^{\frac{5}{2}}-3x^{\frac{3}{2}}-18x^{\frac{1}{2}}} \\\\\\ \stackrel{\textit{common factoring}}{x^{\frac{1}{2}}\left( x^{\frac{4}{2}}-3x^{\frac{2}{2}}-18 \right)}\implies x^{\frac{1}{2}}(x^2-3x-18)\implies x^{\frac{1}{2}}(x-6)(x+3)$

Answer 2

This  is (x - 6)*x^1/2(x + 3)

= x^1/2 (x^2 - 3x - 18)

= x^(5/2) - 3x^(3/2) - 18x^(1/2)