4
$\begingroup$

How do I factor the followinig polynomial? $$x^6-14x^4+49x^2$$ I can't find a way to get this to work. The previous problems were all over either a difference of squares or sum/difference of cubes, but this one is different, it has three terms and cannot be grouped and split as a four termed one could be. I tried to get it down to what a 3 term quadratic is, but simply cant work it out. Factored it to $x^2(x-7)(x-7)$. The middle term, $14x^2$ does not work out when it is foiled.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ If you had factored out the $x^2$ to begin with... $\endgroup$ – J. M. isn't a mathematician Apr 20 '11 at 0:33
  • $\begingroup$ First factor out $x^2$. The other factor is a perfect square, a quadratic in the variable $x^2$ $\endgroup$ – Rita the dog Apr 20 '11 at 0:41
  • $\begingroup$ Please include all the information in the body of your posts; don't rely on the title, which is meant to be more an indexing feature and informational than an intrinsic part of your post. Thank you. $\endgroup$ – Arturo Magidin Apr 20 '11 at 1:40
4
$\begingroup$

Once you factor out the $x^2$ you will have $(x^2)(x^4-14x^2+49)$. Notice that the polynomial on the right is $((x^2)^2-(14x^2)+49)$ so you can factor this the way you would $t^2-14t+49$ (where I am using $t$ in place of $x^2$). Just make sure your final answer involves only the variable $x$ (and not $t$).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Many thanks, I was used to rewriting everything as either ^2 or ^2 cubed,so I rewrote 49 as 7^2, that was my mistake. $\endgroup$ – Neal Apr 20 '11 at 0:44
  • $\begingroup$ This works, but I don't get how you can just write (x^4)as (x^2)^2 without rewriting everything and then just factor the whole thing, what happens to the ^2? $\endgroup$ – Neal Apr 20 '11 at 1:09
  • $\begingroup$ @Neal: Once you've factored $t^2-14t+49$, replace the $t$ with $x^2$ and find stuff you can factor further... $\endgroup$ – J. M. isn't a mathematician Apr 20 '11 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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