Find values $a,b\in\Bbb{R}$ so the polynomial $$P(x)=6x^4-7x^3+ax^2+3x+2$$
is divisible by the polynomial
$$Q(x)=x^2-x+b$$
So what I know and how I do these problems most of the time, is since:
$$P(x)=Q(x)D(x)$$
I would know that by plugging the roots of $Q(x)$ in $P(x)$ should give me enough equations for me to solve this.
So I tried finding the roots of $Q(x)$:
$$x^2-x+b=0\\x_{1/2}={1\pm\sqrt{1-4b}\over2}$$
Okay so this $b$ value is giving me a headache here. The only thing I gathered from this is that (probably) $b\le0$. I tried now plugging this in $P(x)$, as I know that
$P(x_1)=0$ and $P(x_2)=0$
The exponents on everything made this a real pain and I'm pretty certain that it shouldn't be done this way. I'm stuck.
Best Answer
Let $D(x) = 6x^2+cx+d$ (to fix the coefficient of $x^4$). Then multiply out:
$$Q(x)D(x) = (x^2-x+b)(6x^2+cx+d) = 6x^4+(c-6)x^3+(6b+d-c)x^2+(bc-d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
Equate coefficients:
$$c-6=-7 \Longrightarrow c=-1$$
$$6x^4-7x^3+(6b+d+1)x^2-(b+d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
$$bd = 2 \Longrightarrow d=\dfrac{2}{b} \\ b+d=-3 \Longrightarrow b+\dfrac{2}{b}=-3 \Longrightarrow b^2+3b+2=0 \Longrightarrow b\in \{-1,-2\}$$
Case 1: $b=-1$
$$d=\dfrac{2}{b}=\dfrac{2}{-1}=-2 \\ 6b+d+1 = 6(-1)+(-2)+1 = -7 = a$$
Case 2: $b=-2$
$$d=\dfrac{2}{b} = -1 \\ 6b+d+1 = 6(-2)+(-1)+1=-12 = a$$
So, the options are $(a,b) \in \{(-7,-1),(-12,-2)\}$.