Algebra - Factoring Polynomials

Hide Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 1.5 : Factoring Polynomials

Back to Problem List

16. Factor the following polynomial.

${x^6} + 3{x^3} - 4$

Show All Steps Hide All Steps

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,

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