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^{(1)/(2)}(x+3) \end{cases} \\\\\\ f(x)\cdot g(x)\implies (x-6)\cdot \left[ x^{(1)/(2)}(x+3) \right]\implies (x-6)\cdot \left[ \stackrel{distributing}{x^{(3)/(2)}+3x^{(1)/(2)}} \right]`


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