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Factor: -8x^4-14x^2+4

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2 Answers

First, one can consider the rational root theorem.  It states that if a polynomial (an*x^n+...+a0) has a rational root (p/q), then p is a factor of a0 and q is a factor of an.  Thus, given our polynomial, p(x) = -8x^4-14*x^2+4, we have an=4 and a0=8.  Thus, p=+-1, 2, 4 and q=+-1,2,4,8.  So our possible roots are +-1, +-1/2, +-1/4, +-2, +-4.

Then, we can try our possible roots and find p(1/2)=0 and p(-1/2)=0.  Thus, 2x+1 and 2x-1 are roots.

Then, we can divide our polynomial by 2x+1.  This gives -4 x^3+2 x^2-8 x+4.  Now, we can divide this by 2x-1 giving -2x^2-4.  Finally, we can factor out a multiple of -2 giving us x^2+2.

Putting this all together we have our final factorization: p(x)=-2*(2x+1)*(2x-1)*(x^2+2).

Hi Kacie,
There are 2 solutions for factoring this polynomial. First, let's look just at the factors of the 8 and factors of 4. Factors of 8= 8x1 and 4x2  Factors of 4= 4x1 and 2x2 If you choose to use the factors 4x2 for 8 then this is your solution process:

(-4x2 + __)(2x2 + __)   {one of the factors of 8 will have to be negative since you have it is actually -8}
Now looking at the factors of 4 you have to decide which one you can use to multiply and get 4 at the end as well as add to get -14 in the middle. So let's use 4x1. Because 4x4=16 and 2x1=2
Your final factorization looks like this:
(-4x2+1)(2x2+4)

If you chose to use 8x1 in the beginning you would have to adjust your factors of 4 and use 2x2.

Hope this helps! If you have any further questions please let me know!

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