Question

If $2x+1$ is a factor of $(3b+2)x^3+(b-1)$, then find the value of b.

• A. $b=-4$
• B. $b=1$
• C. $b=-5$
• D. $b=2$

Answer 1

The correct option is D $b=2$

Let $f\left(x\right)=(3b+2)x^3+(b-1)$
If $(2x+1)$ is a factor of $f\left(x\right)$, then $f(-\frac{1}{2})=0$
Now,
$f(-\frac{1}{2})=0$
$=>(3b+2)(\frac{-1}{2}{)}^3+(b-1)=0$
$=>(3b+2)\left(\frac{-1}{8}\right)+(b-1)=0$
$=>\frac{-3b}{8}-\frac{1}{4}+b-1=0$
$=>\frac{-3b+8b}{8}-\frac{5}{4}=0$
$=>\frac{5b}{8}=\frac{5}{4}$
$=>b=\frac{8}{4}$
$=>b=2$