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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.

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    $\begingroup$ If you had factored out the $x^2$ to begin with... $\endgroup$ Apr 20, 2011 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, 2011 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. $\endgroup$ Apr 20, 2011 at 1:40

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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 $t$).

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  • $\begingroup$ 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, 2011 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, 2011 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... $\endgroup$ Apr 20, 2011 at 1:42

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