Finding $a$ and $b$ values in a polynomial function

Question

If $f(x) = ax^3 + bx^2 -x + 3$ is divided by $(x+1)$, the remainder is $3$. When $f(x)$ is divided by $(x+2)$, the remainder is $-7$.

Answer 1

We have $$f(x)=ax^3+bx^2-x+3$$

We can say, using the Polynomial Remainder Theorem$^{*}$, that

\begin{align}f(-1)&=3\\ f(-2)&=-7\end{align}

Therefore, we have

\begin{align}a(-1)^3+b(-1)^2-(-1)+3&=3\\ -a+b+1+3&=3\\ a-b&=1\\ \\ a(-2)^3+b(-2)^2-(-2)+3&=-7\\ -8a+4b+2+3&=-7\\ 2a-b&=3\end{align}

We can solve these simultaneous equations:

\begin{align}a&=1+b\\ &\Downarrow\\ 2(1+b)-b&=3\\ 2+2b-b&=3\\ 2+b&=3\\ b&=1\\ &\Downarrow\\ a&=1+1\\ a&=2\end{align}

Therefore, the asnwer is $a=2$, $b=1$ and thus $$f(x)=2x^3+x^2-x+3$$

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$^*$ The Polynomial Remainder Theorem states that the remainder when a polynomial $f(x)$ is divided by $(x-a)$ is equal to $f(a)$ ie, we could say that $$f(x)=k(x-a)+f(a)$$

The proof can be seen by clicking on the link

Answer 2

Perhaps the issue is a problem with division. Consider $23\div 7 = 3R2$ because $23 = 3\times 7 + 2$. We do something similar with polynomials. Consider the result of dividing $3x^2+2x+7$ by $x+2$. We have $$\begin{align} 3x^2+2x+7 &= 3x^2\color{red}{+6x-6x}+2x+7\\ &= 3x(x+2)-6x+2x+7\\ &= 3x(x+2)-4x+7\\ &= 3x(x+2)-4x\color{red}{-8+8}+7\\ &= 3x(x+2)-4(x+2)+8+7\\ &= 3x(x+2)-4(x+2)+15\\ &= (3x-4)(x+2)+15\\ \end{align}$$

Thus $(3x^2+2x+7)\div (x+2) = (3x-4) \text{ }R\text{ }15$

The above could also have been done in fractional form. $$\frac{3x^2+2x+7}{x+2} = \frac{3x^2\color{red}{+6x-6x}+2x\color{red}{-8+8}+7}{x+2}=\dots$$

Just use this (or a similar method) to find the remainders in terms of $a$ and $b$ in the original problem.