A polynomial is exactly divided by $x+1$, and when it is divided by $3x-1$, the remainder is $4$. Given that the polynomial gives a remainder $hx+k$ when divided by $3x^2+2x-1$, find $h$ and $k$
I've been having a little bit of trouble with this question because I'm not entirely sure what to do when the polynomial isn't given.
Givens:
$P(x) = Q_1(x)\cdot (x+1)$ [note here that $P(-1) = 0$]
and
$P(x) = Q_2(x)\cdot (3x-1) + 4$ [note here that $P(\frac 13) = 4$]
Need to determine $h,k$ where:
$P(x) = Q_3(x)\cdot (x+1)(3x-1) + hx + k$
$P(-1) = 0 \implies -h + k = 0 \implies h = k$.
$P(\frac 13)= 4 \implies \frac h3 + k = 4 \implies \frac {4h}3 = 4 \implies h = k = 3$.
$\ \overbrace{f\!-\!4\bmod (3x\!-\!1)(x\!+\!1)}^{\Large {\rm factor\ out}\,\ 3x\,-1\ \mid\ f-4\ \ \ \rightarrow\!\!\!\!\!\!\!} =\, (3x\!-\!1)\!\!\underbrace{\overbrace{\left[\dfrac{\color{#c00}f\!-\!4}{\color{#0a0}{3x}\!-\!1}\bmod x\!+\!1\right]}^{\Large\ \color{#c00}f\bmod x+1\ =\ \color{#c00}0\ \ \,\rightarrow\!\!\!\!\!\!\!\!\!\!\!}}_{\Large \color{#0a0}g\bmod x+1\, =\, \color{#0a0}{g(-1)}\ \ \rightarrow\!\!\!\!\!\!}\!\! =\, (3x\!-\!1)\left[\dfrac{\ \ \color{#c00}0\!-\!4}{\color{#0a0}{-3}\!-\!1}\right] =\, 3x\!-\!1$
