Question

Let $P\left(x\right)=x^2+bx+c$, where $b$ and $c$ are integers. If $P\left(x\right)$ is a facter of both $x^4+6x^2+25$ and $3x^4+4x^2+28x+5$, then value of $P\left(1\right)$ is -

• A. $10$
• B. $8$
• C. $12$
• D. $4$

Answer 1

The correct option is D $4$

$P\left(x\right)=x^2+bx+c$ is a factor of$x^4+6x^2+25.$

$x^4+6x^2+25=x^4+10x^2-4x^2+25=(x^4+10x^2+25)-4x^2=(x^2+5{)}^2-(2x{)}^2$

$\Rightarrow x^4+6x^2+25=(x^2+2x+5)(x^2-2x+5)$

One of them is a factor of $3x^4+4x^2+28x+5.$

$3x^4+4x^2+28x+5=3x^4-6x^3+15x^2+6x^3-11x^2+28x+5$

$\Rightarrow 3x^2(x^2-2x+5)+6x^3-12x^2+30x+x^2-2x+5=3x^2(x^2-2x+5)+6x(x^2-2x+5)+x^2-2x+5$

$\Rightarrow 3x^4+4x^2+28x+5=(x^2-2x+5)(3x^2+6x+1)$

So, $P\left(x\right)=x^2-2x+5$
$\therefore P\left(1\right)=1-2+5=4$