Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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
add comment

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.

share|improve this answer
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
add comment

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 $

share|improve this answer
add comment

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 $k$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.