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Given that $f(x) = x^3 - x^2 - ax - b$ has a factor $x - 3$ but leaves a remainder of $13x - 11$ when divided by $x + 4$, find $a$ and $b$.

I get that in order for $x-3$ to be a factor then according to the remainder theorem, to find remainder of $f(x)$ divided by $x-a$, calculate $f(a)$ which in this case $f(3)=0$.

So, $f(3)= 18-3a-b=0$

$3a+b=18$ is the first equation found.

It starts to get tricky when dividing $f(x)$ by $x+4$ and finding values of $a$ and $b$ such that the remainder = $13x-11$ to find our second equation according to the answer , $a=5$ , $b=3$

Could someone please help me with this. This is supposed to be grade 10 math and i've got a degree in mathematics yet I still can't get it.

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    $\begingroup$ Are you sure it is $x+4$ and not $x^2+4$? The remainder has degree less than the divisor, so the remainder of the division by $x+4$ cannot be $13x-11$. $\endgroup$ – egreg Jul 29 '15 at 14:29
  • $\begingroup$ How about applying Long Division? $\endgroup$ – Hetebrij Jul 29 '15 at 14:32
  • $\begingroup$ if the remainder is of degree 1 then the denominator must be at least of degree 2 $\endgroup$ – Sepideh Abadppour Jul 29 '15 at 14:38
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$g(x)=f(x)-(13x-11)$ has a factor of $x+4$ so $$g(-4)=f(-4)-(13(-4)-11)=0$$ $$f(-4)=-64-16+4a-b=13(-4)-11$$ $$ \implies\left\{ \begin{array}{c} 4a-y=17 \\ 3a+b=18 \end{array} \right.\implies \left\{ \begin{array}{c} a=5 \\ b=3 \end{array} \right. $$

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