Question

Let $p\left(x\right)=x^2+bx+c$ where $b,c\in I$. If $p\left(x\right)$ is a factor of both $x^4+6x^2+25$ and $3x^4+4x^2+28x+5$, then $P\left(2\right)$ equals

• A. $0$
• B. $5$
• C. $4$
• D. $2$

Answer 1

The correct option is B $5$

Given, $p\left(x\right)$ is a factor of $x^4+6x^2+25=g\left(x\right)$ (say) and $3x^4+4x^2+28x+5=h\left(x\right)$ (say)
$\Rightarrow p\left(x\right)$ must be the factor of $h\left(x\right)-kg\left(x\right)\forall k\in R$
This is also valid for $k=3$, then
$\therefore (3x^4+4x^2+28x+5)-3(x^4+6x^2+25)=-14(x^2-2x+5)$
$\therefore p\left(x\right)=x^2-2x+5(\because$ Given $p\left(x\right)$ is of the form $x^2+bx+c$)
$\therefore p\left(2\right)=4-4+5=5$