Polynomial long division - find the value of p and q

Question

When the polynomial $P(x)=4x^3+px^2+qx+1$ is divided by $x-1$, the remainder is -2. When $P(x)$ is divided by $(2x-1)$, the remainder is $\frac{13}{4}$.

Result of first division seems to be $4x^2+(p+4)x+(p+4+q)+\frac{p+4+q+1}{x-1}$, the second one gives $2x^2+\frac{2+p}{2}x+\frac{1+p+q}{2}+\frac{\frac{1+p+q}{2}}{2x-1}$.

So I have two equations, $p+4+q+1=-2$ and $\frac{1+p+q}{2}=\frac{13}{4}$. But this doesn't make any sense. If I try to put $p$ from the first one to the second one, I get $-3=\frac{13}{4}$...

I tried to do the divisions three times and I don't think there is a mistake, so do you have any idea where the problem might be?

Answer 1

I think you made a mistake in your second division.

First step: $$ (4x^3 + px^2 + qx + 1)/(2x - 1) = 2x^2 + ((p + 2)x^2 + qx + 1)/(2x - 1) $$

Second step: $$ ((p + 2)x^2 + qx + 1)/(2x - 1) = \frac{p + 2}{2}x + ((q + \frac{p + 2}{2})x + 1)/(2x - 1) $$

And $q + \frac{p + 2}{2} = \frac{p + 2q + 2}{2}$

Note the $2$ in front of $q$.

So you end up with two equations: $$ p + q = -7 \\ p + 2q = 7 $$

And from here the problem is solvable ($p = -21$, $q = 14$).

Answer 2

Hint:

Use remainder theorem or Factor theorem.

It states that the remainder of the division of a polynomial f(x) by a linear polynomial ${\displaystyle x-r}$ is equal to ${\displaystyle f(r).}$ In particular, ${\displaystyle x-r}$ is a divisor of f(x) if and only if, ${\displaystyle f(r)=0,}$ a property known as the factor theorem.

So, from this, you're given:

$P(1)=-2$ and $P(1/2)=13/4$

Can you take it from here now?

Answer 3

By the Remainder Theorem $\,P(a) = P(x)\bmod x\!-\!a.\,$ So evaluating $P(x)$ at $a = 1,\,1/2\,$ yields $\,2\,$ linear equations for $p,q$. To avoid errors with fractions we can eliminate them by scaling by $4$, viz.

$\begin{align} \!\!\!\bmod\ \ x-1\!:\qquad\qquad\ \ \ P(x) &= 4\,\color{#c00}x^3+p\color{#c00}x^2+q\color{#c00}x+1\\ \color{#c00}{x\equiv 1}\,\Rightarrow\ \ -2 \equiv P(1) &\equiv 4\,\color{#c00}1^3 + p\,\color{#c00}1^2 + q\,\color{#c00}1 + 1\, \equiv\, p+q+5\quad\ \ \iff\quad\ \ \bbox[5px,border:1px solid orange] { p+q = -7} \end{align}$

$\begin{align} \!\!\!\bmod{\ 2x - 1}\!:\qquad\qquad 4 P(x) &= 2(\color{#c00}{2x})^3\!\!+p(\color{#c00}{2x})^2\!\! + 2q(\color{#c00}{2x})\!+\!4\\ \color{#c00}{2x\equiv 1}\,\Rightarrow\,13 = 4P(\frac{1}2) &\equiv 2(\color{#c00}1)^3 +\ p(\color{#c00}1)^2 +\ 2q(\color{#c00}1)\ + 4\,\equiv\, p+2q+6\!\iff\! \bbox[5px,border:1px solid orange]{p+2q = 7} \end{align}$