Question

If $(x^2+px+1)$ is a factor of $(a{x}^3+bx+c)$, then

• A. $a^2+c^2=-ab$
• B. $a^2-c^2=-ab$
• C. $a^2-c^2=ab$
• D. None of these

Answer 1

The correct option is C

$a^2-c^2=ab$

Explanation for the correct option:

Applying basic concept of polynomial:

According to the given information, when we will divide $(a{x}^3+bx+c)$ by $(x^2+px+1)$, the remainder will be $0$.

$x^2+px+1ax+apa{x}^3+bx+c\underline{a{x}^3+ap{x}^2+ax}-ap{x}^2-ax+bx+c\underline{ap{x}^2+a{p}^{2}x+ap}\left(b-a+a{p}^2\right)x+\left(c+ap\right)$

$\begin{array}{rcl}& \Rightarrow & \left(b-a+a{p}^2\right)x+\left(c+ap\right)=0\\ \Rightarrow \left(c+ap\right)& =& 0...\left(1\right)\\ and\left(b-a+a{p}^2\right)& =& 0...\left(2\right)\end{array}$

From $\left(1\right)$, we get

$$\begin{gathered} c+ap=0 \\ \Rightarrow p=-\frac{c}{a} \end{gathered}$$

By substituting the value of $p$ in $\left(2\right)$, we get

$$\begin{gathered} b-a+a\left(\frac{c^2}{a^2}\right)=0 \\ \Rightarrow \frac{ab-a^2+c^2}{a}=0 \\ \Rightarrow a^2-c^2=ab \end{gathered}$$

Hence, option C is correct.