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

Question

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

1. $P(x)=0$ has imaginary roots
2. $P(x)=0$ has roots of opposite sign
3. $P(1)=4$
4. $P(1)=6$

My Attempt:

$f(x)=x^4+6x^2+25$ is always positive, i.e. no roots.

This implies that, $P(x)=0$ has no real roots either.

$\therefore$ Option (1) is correct.

I have no idea how to calculate Option (3) or (4)

All I could see was $f(1)=32$, $g(1)=40$, hinting me at (3).

Any hints would be really helpful thanks.

Answer 1

Hint: It is quite easy to factor $x^4+6x^2+25$ into irreducible factors (over $\mathbb{Z}$): $$x^4+6x^2+25=(x^2+5)^2-4x^2\,.$$ So, what should you do?

The two irreducible factors of $x^4+6x^2+25$ are $x^2-2x+5$ and $x^2+2x+5$. (By the way, they are obviously irreducible over $\mathbb{R}$, whence also over $\mathbb{Z}$, as they are positive polynomials, i.e., without real roots.) Since the coefficient of $x$ in either of these polynomials is not $0$, (1) and (2) cannot be the answer. The values at $x=1$ of these polynomials are $4$ and $8$, respectively. Thus, (3) is the only possible choice. Indeed, $$3x^4+4x^2+28x+5=(x^2-2x+5)(3x^2+6x+1)\,.$$

Answer 2

The quartic $f(x)$ has $4$ roots: $1\pm2i$ and $-1\pm2i$. Of these, only the first $2$ are roots of $g(x)$. Therefore, $P(x)=(x-1-2i)(x-1+2i)=x^2-2x+5$. Now, note that $P(1)=4$.

Answer 3

We have $$f(x)= (x^2+3)^2+16 $$ $$= (x^2+3+4i)(x^2+3-4i)$$ $$ = (x^2-(1-2i)^2)(x^2-(1+2i)^2) $$ $$= \color{red}{(x-1+2i)}(x+1-2i)\color{red}{(x-1-2i)}(x+1+2i)$$ $$ = \big(\color{red}{(x-1)^2-4i^2}\big)\big((x+1)^2-4i^2)$$ $$ = (x^2-2x+5)(x^2+2x+5)$$

Now it is easy to see $$g(x)= (x^2-2x+5)(3x^2+6x+1)$$

So $P(x)= x^2-2x+5$.

Answer 4

since $P(x)$ is a factor

$f(x)=x^4+6x^2+25=P(x)*q(x)$ and $g(x)=3x^4+4x^2+28x+5=P(x)*q'(x)$

where $q,q'$ quadratic polynomials

$3f(x)-g(x)=14x^2-28x+70=P(x)*(3q(x)-q'(x))$

$P(x)*(3q(x)-q'(x))=14(x^2-2x+5)$

since $P(x)$ is a quadratic $(3q(x)-q'(x))$ must be a constant$(14)$

thus $$P(x)=x^2-2x+5$$