Solving higher-degree polynomials by factoring

foahchon

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Nov 12, 2006
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Hi, I have a couple of questions concerning factoring higher-degree polynomials.

The first one comes from an example in the book:

\(\displaystyle 2x^2 - 16x^2 = 0\)

\(\displaystyle 2x^2(x^3 - 8) = 0\)

\(\displaystyle 2x^2(x - 2)(x^2 + 2x + 4) = 0\)

I guess my question is, how does it get from step 2 to step 3? Where does the (x - 2) come from, and where does the quadratic equation in step 3 come form?

The second question is this:

\(\displaystyle w^4 + 8w = 0\)

The problem specifies to find all of the real and imaginary solutions for the equation. I guess I'm supposed to solve this by factoring, but I don't see how factoring could be useful in this problem. The back of the book says the solutions are 0, 2, and +- i*sqrt(3).

Thanks.
 
foahchon said:
how does it get from step 2 to step 3?
Review "factoring sums and differences of cubes". (The formulas may be in the endpapers of your textbook.)

Where does the (x - 2) come from, and where does the quadratic equation in step 3 come form?

foahchon said:
The second question is this: w<sup>4</sup> + 8w = 0

...I don't see how factoring could be useful in this problem.
Take the common factor of "w" out front, and then factor the resulting sum of cubes, w<sup>3</sup> + 2<sup>3</sup>. Then apply the Quadratic Formula to find the (complex) roots of the quadratic factor.

Eliz.
 
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