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Meaning of term factoring polynomials

factoring polynomials

Factoring polynomials is an action in which a polynomial or polynomials are represented as a product of simpler polynomials that no longer can be factored.

Some factored polynomials you know from before:

\ x^2 + 2xy + y^2 = (x + y)^2

\ x^3 + 3x^2y + 3xy^2 + y^3 = (x + y)^3 and so on.

How do you factorize? This is very simple, you just take the common multiple of all your terms and take them out in front of braces.

factoring polynomials

Example 1: Factorize: x^2 + 4x. Their common multiple is obviously x so we’ll extract them in front of braces, and multiply by a polynomial divided by x.

\ x^2 + 4x = x(x + 4)
 

How do you check it? Simply multiply those two factors you got:

\ x(x + 4) = x^2 + 4x
Now we can move on to a bit more complicated examples.

Example 2: Factor the following polynomial 4x^4 + 4x^3 - 24x^2

Here we can factor \ 4x^2 from every term. 4x^2 (x^2 + x - 6). Here we got a quadratic polynomial which we can factorize using its zeros. We got that zeros of this quadratic polynomial are -3 and 2. Now we can write the final factorized polynomial.

4x^4 + 4x^3 - 24x^2 = 4x^2 (x + 3)(x - 2)

What are the zeros and their multiplicity? We have first factor x^2, its zero is 0 with multiplicity 2, zero of (x + 3) is -3 , and zero of (x – 2) is 2 both with multiplicity 1. Remember that the number of all these zeros must add up to the leading exponent of given polynomial.

Example 3:
Factorize \ x^2 y + xy^2 + 4xy. The common multiple of these terms is xy. \ x^2 y + xy^2 + 4xy = xy(x + y + 4). If you ever get unsure about what is your full common multiple you can always solve it in few steps, first extract x to simplify, and then you’ll easily notice more common multiples.

Example 4:

Factorize \ 0.4xa + 0.2x^3 ab + \frac{1}{5}a
 

\ 0.4 xa + 0.2x^3 ab + 4a = \frac{2}{5}xa + \frac{1}{5}ab + \frac{1}{5}a = \frac{1}{5}a (2x + b + 1)
 

Example 5: Factorize \ x^2 + 2x + 1. Here the factors aren’t seeable at first, but you can use all your knowledge you have to solve this.

You already know that \ (x + y)^2 = x^2 + 2xy + y^2. And our polynomial can be written like this: \ x^2 + 2 * 1 * x. In this case \ y = 1, and \ x = x. So it is easy to conclude that \ x^2 + 2x + 1 = (x + 1)^2

But what with cases you can’t conclude the solution so easily?

Example 6: Factorize \ x^5 + x - 2x^4 - 2.

You do this by grouping certain terms, and factorize them individually and then trying to factorize whole polynomial:

\ (x^5 + x) - (2x^4 + 2) = x(x^4 + 1) - 2(x^4 + 1)

And now you should notice that your polynomial has two terms that have one factor the same, so you can extract him, and get your factor:

\ x(x^4 + 1) - 2(x^4 + 1) = (x^4 + 1)(x - 2)

You could also group differently:

\ x^5 + x - 2x^4 - 2 = (x^5 - 2x^4) + (x - 2) = x^4(x - 2) + (x - 2) = (x^4 + 1)(x - 2)

There is another way in factorizing. We’ll show it on quadratic polynomial (whose largest exponent equals 2 ). That is finding zeros.

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