Question

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

• A. $P\left(x\right)=0$ has imaginary roots
• B. $P\left(x\right)=0$ has roots of opposite sign
• C. $P\left(1\right)=4$
• D. $P\left(1\right)=6$

Answer 1

Answer: (A), (C)

$P\left(x\right)=0$ has imaginary roots
$P\left(1\right)=4$

$P\left(x\right)=x^2+bx+c$ 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 $3x^4+4x^2+28x+5$.

$3x^4+4x^2+28x+5=3x^4-6x^3+15x^2+6x^3-11x^2+28x+5=3x^2(x^2-2x+5)+6x^3-12x^2+30x+x^2-2x+5=3x^2(x^2-2x+5)+6x(x^2-2x+5)+x^2-2x+5$

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

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

$P\left(x\right)=0\Rightarrow \Delta =2^2-4\left(5\right)=-16<0$ has imaginary roots.

$P\left(1\right)=1-2+5=4$