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$f(x)=2x^3-5x^2=kx-20$ Help is appreciated.

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Welcome to MSE! It helps to share what you have tried, and where you are stuck, so people can take it from there. –  gnometorule Mar 3 '13 at 3:46
    
I see you have some answers, but I cannot understand really what your question is. –  Pedro Tamaroff Mar 3 '13 at 4:10
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3 Answers

$$ \begin{align} f(x) & = 2x^3 - 5x^2 = kx - 20 \\ \\ & = 2x^3 - 5x^2 - kx + 20 = 0 \tag{1}\\ \\ &= (x+2)(2x^2 -9x + 10)=0\tag{2}\\ \\ \end{align} $$

Note that $(-18 +10)x = -kx \implies k = 8.$

Determining what the remaining factors must be in $(2)$: we know $2x^2$ must lead, and it must end in $10$ to obtain $2x^3$ and the constant $20$. We must also then have a term of $-9x$ because we need for $4x^2 - 9x^2 = - 5x^2$. And we see that we can argue that $k = 8$.

You can try doing this using polynomial division, dividing $(1)$ by $(x + 2)$, or you can use trial an error to determine what the remaining term in the second factor must be. To make this work, you'll see $k$ must equal 8.

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Thank you . That clears up things a little. –  Robert Mar 3 '13 at 3:43
    
You're welcome, Robert! –  amWhy Mar 3 '13 at 4:11
    
You're also Welcome, Amy. hAVE A NICE SLEEP –  B. S. Mar 3 '13 at 5:40
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Hint $\ $ By the Factor Theorem, $\rm\: x\!+\!2\:$ is a factor of $\rm\:g(x)\iff g(-2) = 0.\:$ Applying this to your polynomial $\rm\ g(x) = 2x^3-5x^2-kx+20,\ $ the criterion is: $\rm\ g(-2) = 2k-16 = 0\iff k =\: \ldots $

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If $x+2$ is a factor of $2x^3-5x^2-kx+20$, then $-2$ is a zero of that polynomial. So, $(-2)^3-5(-2)^2-k(-2)+20=0$. Now you just have to do the arithmetic to find $k$.

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