SAT Mathematics Practice Test 4

In $\mathit{f}\left(\mathit{x}\right)$, factoring out the greatest common factor, $2\mathit{x}$, yields $\mathit{f}\left(\mathit{x}\right)=2\mathit{x}\left({\mathit{x}}^2+3\mathit{x}+2\right)$.It is given that $\mathit{g}\left(\mathit{x}\right)={\mathit{x}}^2+3\mathit{x}+2$,so using substitution,$\mathit{f}\left(\mathit{x}\right)$ can be rewritten as $\mathit{f}\left(\mathit{x}\right)=2\mathit{x}·\mathit{g}\left(\mathit{x}\right)$.In the equation $\mathit{p}\left(\mathit{x}\right)=\mathit{f}\left(\mathit{x}\right)+3\mathit{g}\left(\mathit{x}\right),$ substituting $2\mathit{x}·\mathit{g}\left(\mathit{x}\right)\mathit{f}\mathit{o}\mathit{r}\mathit{f}\left(\mathit{x}\right)$ yields $\mathit{p}\left(\mathit{x}\right)=2\mathit{x}·\mathit{g}\left(\mathit{x}\right)+3·\mathit{g}\left(\mathit{x}\right).$ In $\mathit{p}\left(\mathit{x}\right)$, factoring out the greatest common factor,$\mathit{g}\left(\mathit{x}\right)$, yields $\mathit{p}\left(\mathit{x}\right)=\left(\mathit{g}\left(\mathit{x}\right)\right)\left(2\mathit{x}+3\right)$. Because $2\mathit{x}+3$ is a factor of $\mathit{p}\left(\mathit{x}\right)$, it follows that $\mathit{p}\left(\mathit{x}\right)$ is divisible by $2\mathit{x}+3.$
Choices A, C, and D are incorrect because $2\mathit{x}+3$ is not a factor of the polynomials $\mathit{h}\left(\mathit{x}\right),\mathit{r}\left(\mathit{x}\right),$ or $\mathit{s}\left(\mathit{x}\right)$.Using the substitution $\mathit{f}\left(\mathit{x}\right)=2\mathit{x}·\mathit{g}\left(\mathit{x}\right),$ and factoring further, $\mathit{h}\left(\mathit{x}\right),\mathit{r}\left(\mathit{x}\right),$ and $\mathit{s}\left(\mathit{x}\right)$ can be rewritten as follows:$\mathit{h}\left(\mathit{x}\right)=\left(\mathit{x}+1\right)\left(\mathit{x}+2\right)\left(2\mathit{x}+1\right)$
$\mathit{r}\left(\mathit{x}\right)=\left(\mathit{x}+1\right)\left(\mathit{x}+2\right)\left(4+3\right)$
$\mathit{s}\left(\mathit{x}\right)=2\left(\mathit{x}+1\right)\left(\mathit{x}+2\right)\left(3\mathit{x}+1\right)$Because $2\mathit{x}+3$ is not a factor of $\mathit{h}\left(\mathit{x}\right),\mathit{r}\left(\mathit{x}\right)$, or $\mathit{s}\left(\mathit{x}\right)$,it follows that $\mathit{h}\left(\mathit{x}\right)$, $\mathit{r}\left(\mathit{x}\right)$, and $\mathit{s}\left(\mathit{x}\right)$ are not divisible by $2\mathit{x}+3.$