Overview

This section will be an introduction into factoring polynomials and some crucial factoring identities. Factoring is a fundamental skill needed if you want to go anywhere in competition math. There are ofcourse more advanced techniques which will be taught in later sections.

Key Ideas

Factoring is the process of simplifying expressions into a product of several terms called factors, as opposed sums of terms. This can be used to solve for the roots of polynomials or simplify expressions to extract information.

Factoring identities

using basic algebra, we can derive the following identities.

If n is a positive integer, then

Basic polynomial factoring

The first thing you should always look at if you want to factor a polynomial is to see if there is a possible GCF (Greatest Common Factor) which you can factor from the polynomial. The GCF is simply the largest expression which can be divided from every term in the polynomial. For example in the polynomial $4x^{3} - 2x^{2} + 2x$, the GCF is $2x$, since 2x can be divided from each term in the polynomial.

Example 1: Factor the polynomial:

First, find the GCF of the terms: GCF of $6$ and $9$ is $3$

GCF of $x^3$ and $x^2$ is $x^2$

$\implies$ the GCF is $3x^2$: $\boxed{6x^3 + 9x^2 = 3x^2(2x + 3)}$

Example 2: Factor the polynomial:

Find the GCF: GCF of $8$ and $12$ is $4$ GCF of $x^2$ and $x$ is $x$

So the GCF is $4x$: $\boxed{8x^2 - 12x = 4x(2x - 3)}$

However, not all polynomial are able to factored this way; some polynomials may require more insight rather than simply looking for a general GCF. There are some polynomials which must have their terms "grouped" and then factored so that it may be reduced further. To see what I mean, let's look at an example.

Example 1: Factor the polynomial.

First we group the terms. $(2x^3 + 4x^2) + (3x + 6)$

then we can factor each group: $2x^2(x + 2) + 3(x + 2)$

notice that each of the terms share a common factor, so we can factor out the common binomial $(x + 2)$: $(2x^2 + 3)(x + 2)$

Factoring polynomials like so is colloquilly known as factoring by "grouping". This most commonly appears in problems where you must factor a 4 term polynomial, however many polynomials may be factored this way and it appears a fair amount of times in AMC-like problems.

Now lets look at ways to factor a more specific kind of polynomial: quadratics. Quadratics are a second degree polynomial, meaning that no term will have an x raised to a power greater than 2. This also means that quadratics will have at most 3 terms. Using this definition, we can see that all quadratics are of the form

$ax^{2} + bx + c$

Ofcourse, you can factor this by simply solving for the roots using the quadratic formula and rewriting. But for some quadratics, there are quicker ways. One of these methods is called completing the square. This method involves using the factoring identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$ by adding the square of half of the coefficient of the x term. To see what I mean, let's look at an example.

Example 1: Factor the quadratic: $x^2 + 6x + 5$

Add the square of half the coefficient of the x term (i.e 9 in this case), then complete the square: $x^2 + 6x + 9 - 4 = (x + 3)^2 - 4$

Factor using difference of squares: $(x + 3)^2 - 4 = (x + 1)(x + 5)$

Example 2: Factor the quadratic: $x^2 + 4x - 5$

add the square of half of the coefficient of x (i.e 4) to complete the square: $x^2 + 4x + 4 - 9 = (x + 2)^2 - 9$

Factor: $(x + 2)^2 - 9 = (x - 1)(x + 5)$

By using the difference of squares, and adding the square of half the coefficient of the x term (i.e $(\frac{b}{2a})^{2})$), we can rearrange the polynomial without having to use the quadratic formula.

One of the ways we can find roots to a polynomial is by using the Rational Root Theorem. The Rational Root Theorem states that for a polynomial with integer coefficients, any possible rational root $\frac{q}{p}$ (in lowest terms) must have p dividing the constant term and q dividing the leading coefficient. For example

Example 1 : Factorise

from the rational root theorem we can see that the possible rational roots are \( \pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2} \)

Testing $( x=2 )$, we get \( f(2)=0 \), so $( x-2 )$ is a factor.

Dividing gives
$f(x)=(x-2)(2x^2+x-6)=(x-2)(2x-3)(x+2).$

Example 2 Factor $(g(x)=x^3+2x^2-5x-6 )$.

From the rational root theorem, we can see that the possible rational roots are $( \pm1,\pm2,\pm3,\pm6 ).$ Testing ( x=2 ), we get \( g(2)=0 \), so \( x-2 \) is a factor.

Dividing gives
$g(x)=(x-2)(x^2+4x+3)=(x-2)(x+1)(x+3).$

If n is an odd integer, then $a^{n} + b^{n} = (a+b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$

Worked Example

Example 1 (AMC 10A 2021 #9) Find the least possible value of: $(x - y)^2 + (x + 2y - 3)^2$

Expand the expression: $(x - y)^2 = x^2 - 2xy + y^2$ $(x + 2y - 3)^2 = x^2 + 4y^2 + 9 + 4xy - 6x - 12y$

Add: $2x^2 + 5y^2 + 2xy - 6x - 12y + 9$

Group terms: $(2x^2 + 2xy - 6x) + (5y^2 - 12y) + 9$

Factor by grouping: $2x(x + y - 3) + y(5y - 12) + 9$

Rewrite as sums of squares: $(x - y)^2 + (x + 2y - 3)^2$

To minimize the expression, set each square equal to $0$: $x - y = 0$ $x + 2y - 3 = 0$

Solve: $x = y$ $y + 2y - 3 = 0 \Rightarrow 3y = 3 \Rightarrow y = 1$ $x = 1$

Minimum value: $0$

Example 2 Find all rational roots of: $6x^3 + x^2 - 19x + 6$

Possible rational roots (Rational Root Theorem): $\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}$

Test $x = 2$: $6(2)^3 + (2)^2 - 19(2) + 6 = 48 + 4 - 38 + 6 = 20 \neq 0$

Test $x = -2$: $6(-2)^3 + (-2)^2 - 19(-2) + 6 = -48 + 4 + 38 + 6 = 0$

So $(x + 2)$ is a factor. Divide:

$6x^3 + x^2 - 19x + 6 = (x + 2)(6x^2 - 11x + 3)$

Factor the quadratic: $6x^2 - 11x + 3 = (3x - 1)(2x - 3)$

Final factorization: $(x + 2)(3x - 1)(2x - 3)$

Rational roots: $x = -2, \frac{1}{3}, \frac{3}{2}$

Common Pitfalls

• Forgetting to include the negative sign when completing the square.
• Forgetting to factor out the GCF first.
• Treating $a^{2}+b^{2}$ as $a^{2}-b^{2}$

Practice Problems