# Finding the factors and roots of a polynomial equation

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## What do we mean by "polynomial?"

Polynomial expressions are ones that have different powers of the same variable in, for example **quadratics** (or **trinomials** to Americans?) have an x^{2} term, an x term and a constant - **cubics** have an x^{3}, an x^{2}, an x and a constant. And so on...

The roots of a polynomial expression f(x) are where the graph of y = f(x) crosses the x-axis, or the values of x that make the expression equal to zero.

The simplest case of this would be factorising a quadratic equation in order to solve it, but the principles apply to all polynomials.

## Factorising cubics

Cubic expressions, having x^{3} as the largest power of x, can be thought of as the product of a quadratic and a linear expression:

(x + n)(ax^{2 }+ bx + c) would give a cubic expression as the highest possible power of x would be given by multiplying the x term in the first bracket by the x^{2} term in the second.

You would generally be given the linear expression in the question, so the question becomes: how do we find the values of a, b, and c in the quadratic?

- Notes: University of Warwick

An excellent PowerPoint that breaks down the method described here, and shows an alternative too. - Alternative: Long division

You could also divide the cubic by the linear expression using "algebraic long division" - here's how! - Practice: Cool Math

Practice your polynomial skills online - bags and bags of activities here!

## Comparing coefficients

If we know the linear expression that is a factor of our cubic, and the cubic expression we are trying to find the roots of, we can simply expand the brackets above and compare our coefficients - for example:

*"Given that (x + 1) is a factor of f(x) = x*^{3}* + 4x*^{2}* - 15x - 18, solve f(x) = 0"*

We multiply our (x + 1) by a "dummy" quadratic *ax*^{2}* + bx + c* to see how many of each power of x we get - we then compare the coefficients to f(x) in order to find the values of a, b, and c in our quadratic. See image!

Personally, I find it easier to multiply through by each term in the linear factor separately, and write the x^{3}, x^{2}, x and constant terms together - but that's just me.

We can then factorise the quadratic normally to find the missing roots.

In this example, the quadratic we need would be x^{2} + 3x - 18 (given the values of a, b, and c in the image above) which factorises to (x - 3)(x + 6), so our three factors are **(x +1)(x - 3)(x + 6)**.

Our roots are whatever x values make f(x) = 0, which we can do by making each bracket zero - just like solving a quadratic. This gives x = -1, x = 3, and x = -6. This is where the graph of y = f(x) crosses the x-axis, if you want to check!

## What if you're not given the linear factor?

Any linear factor of a polynomial can be found using the Factor Theorem, which is really just a special case of the Remainder Theorem - but that's another story!

## Alternative method to comparing coefficients:

## 2 comments

You beat me to it, answering this question! I also teach math. You mentioned in your profile that you are using hubpages to build a gcse and math revision site. I'm not familiar with that terminology - I assume it's a British thing? What does it mean?