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If $(x - 2)$ is a factor of $x^3 + ax^2 - 6x - 4$, then find $a$. This is regarding polynomials. The answer is $a = 2$. Could someone please provide the working out and help me out on this please.

Thank you!

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4 Answers 4

If $x-2$ is a factor of the polynomial, it means that when you plug $x=2$ into the polynomial, you get $0$. Do you see how this helps you?

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Factorize the polynomial:

$x^3+ax^2-6x-4=(x-2)(x^2+(a+2)x+2(a-1))+4a-8$

Therefore $x^3+ax^2-6x-4 \pmod {x-2} \equiv 4a-8$

Because $x-2$ is the factor of $x^3+ax^2-6x-4$, then $4a-8=0$.

$\therefore a=2$

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How do you type x to the power of 3 so it looks the way you did it. –  Orbit Jan 20 at 2:50
    
@Orbit enclose math in dollar signs. For example, $x^3$ makes $x^3$. –  Goos Jan 20 at 3:12
    
Thank you so much for that –  Orbit Jan 20 at 7:09

So the expression should be $0$ when you put $x=2$. Thus, $8+4a-12-4=0$. So $a=2$

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You should not give out answers like this. Especially when the OP has not provided work. –  Cameron Williams Jan 20 at 2:34
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Why does it matter. I just want an answer. –  Orbit Jan 20 at 2:35
    
@CameronWilliams: Point noted. –  voldemort Jan 20 at 2:37

Recall that if $(x - a)$ is a factor of a polynomial $f(x)$, then $f(a) = 0$.

Since $(x - 2)$ is a factor of $x^3 + ax^2 - 6x - 4$, the unfactored polynomial (substituted with $x = 2$) must be $0$. Therefore...

$$\begin{aligned} (2)^3 + a(2)^2 - 6(2) - 4 &= 0\\ 8 + 4a - 12 - 4 &= 0\\ 4a - 8 &= 0\\ 4a &= 8\\ a &= 2 \end{aligned}$$

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