Question

$f\left(x\right)=2x^3+6x^2+4x$
$g\left(x\right)=x^2+3x+2$
The polynomials $f\left(x\right)$ and $g\left(x\right)$ are defined above. Which of the following polynomials is divisible by $2x+3$ ?

• A. $h\left(x\right)=f\left(x\right)+g\left(x\right)$
• B. $p\left(x\right)=f\left(x\right)+3g\left(x\right)$
• C. $r\left(x\right)=2f\left(x\right)+3g\left(x\right)$
• D. $s\left(x\right)=3f\left(x\right)+2g\left(x\right)$

Answer 1

The correct option is B $p\left(x\right)=f\left(x\right)+3g\left(x\right)$

Put $2x+3=0$, we get $x=-\frac{3}{2}$.

Substitute $x=-\frac{3}{2}$ in $f\left(x\right)=2x^3+6x^2+4x$ as follows:

$f\left(x\right)=2x^3+6x^2+4x$

$f(-\frac{3}{2})=2{(-\frac{3}{2})}^3+6{(-\frac{3}{2})}^2+4(-\frac{3}{2})$

$f(-\frac{3}{2})=2(-\frac{27}{8})+6\left(\frac{9}{4}\right)-6$

$f(-\frac{3}{2})=-\frac{27}{4}+\frac{54}{4}-6$

$f(-\frac{3}{2})=\frac{27}{4}-6$

$f(-\frac{3}{2})=\frac{3}{4}$

Now substitute $x=-\frac{3}{2}$ in $g\left(x\right)=x^2+3x+2$ as follows:

$g\left(x\right)=x^2+3x+2$

$g(-\frac{3}{2})={(-\frac{3}{2})}^2+3(-\frac{3}{2})+2$

$g(-\frac{3}{2})=\frac{9}{4}-\frac{9}{2}+2$

$g(-\frac{3}{2})=\frac{9-18+8}{4}g(-\frac{3}{2})=-\frac{1}{4}$

Finally substituting $x=-\frac{3}{2}$ in the polynomial $p\left(x\right)=f\left(x\right)+3g\left(x\right)$ we get,

$p\left(x\right)=f\left(x\right)+3g\left(x\right)$

$p(-\frac{3}{2})=f(-\frac{3}{2})+3g(-\frac{3}{2})$

$p(-\frac{3}{2})=\frac{3}{4}+3(-\frac{1}{4})$

$p(-\frac{3}{2})=\frac{3}{4}-\frac{3}{4}$

$p(-\frac{3}{2})=0$

Hence, the polynomial $p\left(x\right)=f\left(x\right)+3g\left(x\right)$ is divisible by $2x+3$.