Question

Let $P\left(x\right)=x^2+bx+c$, where $b$ and $c$ are integers. 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)

$P\left(1\right)=4$

Since $P\left(x\right)$ divides both the expressions, hence $P\left(x\right)$ also divides $(3x^4+4x^2+28x+5)-3(x^4+6x^2+25)$
$=-14x^2+28x-70$
$=-14(x^2-2x+5)$ which is a quadratic.
Hence, $P\left(x\right)=x^2-2x+5$
$\therefore P\left(1\right)=4$

$D=4-20=-16<0$
$\therefore P\left(x\right)=0$ has imaginary roots.