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Section 1.5 : Factoring Polynomials

16. Factor the following polynomial.

\[{x^6} + 3{x^3} - 4\]

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Start Solution

Don’t let the fact that this polynomial is not quadratic worry you. Just because it’s not a quadratic polynomial doesn’t mean that we can’t factor it.

For this polynomial we can see that \({\left( {{x^3}} \right)^2} = {x^6}\) and so it looks like we can factor this into the form,

\[\left( {{x^3} + \underline {\,\,\,\,\,} } \right)\left( {{x^3} + \underline {\,\,\,\,\,} } \right)\]

At this point all we need to do is proceed as we did with the quadratics we were factoring above.

Show Step 2

After writing down the factors of -4 we can see that we need to have the following factoring.

\[{x^6} + 3{x^3} - 4 = \left( {{x^3} + 4} \right)\left( {{x^3} - 1} \right)\] Show Step 3

Now, we need to be careful here. Sometimes these will have further factoring we can do. In this case we can see that the second factor is a difference of perfect cubes and we have a formula for factoring a difference of perfect cubes.

Therefore, the factoring of this polynomial is,

\[{x^6} + 3{x^3} - 4 = \left( {{x^3} + 4} \right)\left( {{x^3} - 1} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left( {{x^3} + 4} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\]