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 $3x^4+4x^2+28x+5$, what is $p\left(1\right)$ ?

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (D)

$4$

If $p\left(x\right)$ is the common factor of both $x^4+6x^2+25$ and $3x^4+4x^2+28x+5$ then it means that $p\left(x\right)$ is the HCF of both the expressions. So, we divide $3x^4+4x^2+28x+5$ by $x^4+6x^2+25$ as shown in the above image:

Therefore, the remainder is $-14x^2+28x-70$ that is $R\left(x\right)=-14(x^2-2x+5)$.

Thus, the HCF is $x^2-2x+5$ which means that $p\left(x\right)=x^2-2x+5$

Now, substitute $x=1$ in $p\left(x\right)=x^2-2x+5$ as shown below:

$p\left(1\right)=1^2-(2\times 1)+5=1-2+5=6-2=4$

Hence, $p\left(1\right)=4$.