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I missed a day of math class and didn't have time today to go over this with my teacher.

In math we covered factoring and solving higher order polynomials yesterday, but I wasn't there.

I've figured out difference and sum of cubes, which are pretty easy. I've also figured out difference of squares such as $4x^4-16$, but I'm having trouble with some other polynomiAls.

Take for example, $x^4-9x^2+14$. I recognize that the first part is difference of squares, so I can get $(2x^2-3x)(2x^2+3x)+14$, but then what do I do with the 14?

For another example, take $6x^4+24x^2-72$. I can take out the common factor of $6$, and get $x^4+4x^2-12$, but then what do I do from here? I see a sum of squares, but can that be factored? Is that the next reasonable step or do I do something different?

On the same page is also $m^4-8m^2+7$, and I'm not even sure where to start in this one. There's no common factor, so I can't do that, and it's not sum or difference of squares or cubes, so what do I do?

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  • $\begingroup$ Let $u=m^2$ and use quadratic formula or factor if you can (hint: you can). $\endgroup$ – user223391 Mar 3 '17 at 0:51
  • $\begingroup$ @ZacharySelk I'm not quite sure what you mean, care to expand a little bit? $\endgroup$ – Travis Mar 3 '17 at 0:52
  • $\begingroup$ Can you factor $u^2-8u+7$ ? $\endgroup$ – user223391 Mar 3 '17 at 0:53
  • $\begingroup$ The general method is to find roots through guess and check (plug into the function, check it comes out to zero) then use polynomial division to pull the linear factor out of the polynomial, and get a new smaller polynomial to factor. Of course, as you've shown above, there is better methods for some polynomials. $\endgroup$ – Kaynex Mar 3 '17 at 0:53
  • $\begingroup$ @ZacharySelk oh, thanks, that particular equation is meant to be facotring, not solving, but I'm sure that will come in handy at some time $\endgroup$ – Travis Mar 3 '17 at 0:54
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In general, multiplication is easy, but undoing it (factoring) is hard, both for numbers and for polynomials.

In the particular case of the polynomials you're looking at, where all the exponents are even, you can make the substitution $u = x^2$. So $x^4 - 9x^2 + 14$ becomes $u^2 - 9u + 14$. You can factor this as $(u-2)(u-7)$, either by "inspection" (that is, looking at it until you recognize the answer) or by using the quadratic formula to find that this has solutions $u = (9 \pm \sqrt{25})/2 = (9 \pm 5)/2 = 2, 7$. Then you can remember that $u = x^2$ to get $x^4 - 9x^2 + 14 = (x^2-2)(x^2-7)$. This same idea will work for the other polynomials you're looking at.

Noticing that you have a sum or difference of squares or cubes, when you're just looking at part of the polynomial, is a red herring in this case. Say you have the number 9900. You notice that it's $100^2-10^2$ and so you can factor it as $(100-10)(100+10) = 90 \times 110$. Does this tell you anything about the factorization of 9901? (No.). Polynomials work the same way.

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If all the terms have even powers, you have a polynomial of half the degree in the square of the variable. $5x^4+24x^2-72=6((x^2)^2+4(x^2)-12)$, which is a quadratic that you should know how to factor. It might be easier to see if you define $y=x^2$ and write the polynomial as $6(y^2+4y-12)$. Once you have factored that, plug the $x^2$s back in and you will have two quadratic factors that you can factor.

Generally and here it is not useful to factor part of an expression. In your first example, you can write it as $(x^2+3x)(x^2-3x)+14$ (you have extra factors of $2$) but you are right that you don't know what to do with the $14$. You could complete the square to get $x^4-9x^2+14=(x^2-\frac 92)^2-\frac {25}4$ and then use the difference of squares.

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