Question

If $a{x}^3-5x^2+x+p$ is divisible by $x^2-3x+2$ then find the values of $a$ and $p$.

• A. $a=2,p=2$
• B. $a=2,p=3$
• C. $a=3,p=1$
• D. $a=1,p=3$

Answer 1

The correct option is A $a=2,p=2$

$x^2-3x+2=(x-1)(x-2)$
Since $a{x}^3-5x^2+x+p$ is divisible by $x^2-3x+2$, or by each of $(x-1)$ and $(x-2)$ we have $f\left(1\right)=0$ and also $f\left(2\right)=0$
$f\left(1\right)=a{\left(1\right)}^3-5{\left(1\right)}^2+1+p=0$
$=>a-5+1+p=0$
$=>a+p=4$ --- (1)
And
$f\left(2\right)=a{\left(2\right)}^3-5{\left(2\right)}^2+2+p=0$
$=>8a-20+2+p=0$
$=>8a+p=18$ --- (2)
Subtracting eqn 1 from eqn 2, we get
$7a=14$
$=>a=2$
Substituting $a=2$ in eqn 1, we get
$2+p=4$
$=>p=2$