Factorising Polynomials (Optional)

This section here will serve as a reminder of some of the material you saw in S4 along with an important skill that could become more relevant in S7, when we discuss oblique asymptotes.

Table of Contents

Motivation

We know that the solutions to a quadratic equation are given by the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

However, using this formula can take more time than ideal, and often students make calculation mistakes. Compare solving a quadratic equation using the formula as opposed to using factorisation:

Example: Solve the quadratic equation $x^2 + 5x + 4 = 0$ using factorisation.

Example: Solve the quadratic equation $x^2 + 5x + 4 = 0$ using the quadratic formula.

Factorisation, when possible, offers us a fast way to solve quadratic equations, typically avoiding the biggest places where miscalculations are possible.

Factorising Quadratic Expressions

In S4, you saw the following three basic identities of factorisation:

$ax^2+bx=x(ax+b)$
$a^2-b^2=(a+b)(a-b)$
$x^2+(a+b)x+a\cdot b=(x+a)\cdot (x+b)$

The last identity is the one used in the example above. Simply put, if the coefficient multiplying the $x^2$ is 1, we can factorise the expression quickly by thinking of two numbers that multiplied yield the independent term and added give the coefficient of the term with $x$ .

In the example above, we had $x^2+5x+4$ so we needed two numbers $a$ and $b$ such that $a+b=5$ and $a\cdot b=4$ , the answer being $a=1,b=4$ . If the $x^2$ has a coefficient different from 1, we could try to take common factor to possible arrive at an expression that is nice enough to factorise using this third identity. For example:

$2x^2+10x+8=0\implies 2(x^2+5x+4)=0\implies2\cdot(x+1)(x+4)=0$

Thus having the same solutions. Of course, if taking a common factor makes the other coefficients into decimals, it may be harder to factorise.

The reason why factorisation is so useful, is that it changes a quadratic equation into two linear ones, which are significantly easier. We can apply this logic to higher degree polynomials.

Factorising Polynomials of Degree 3 or more

For this method, we need to use a fundamental theorem of algebra:

Theorem (The Factor Theorem): A polynomial has a linear binomial factor associated with each of its distinct or repeated roots across the complex field.

$P(k) = 0 \iff (x - k) \text{ is a factor of } P(x)$

This theorem provides the algebraic foundation for completely factorising higher-degree polynomials into linear components, directly linking the geometric cross-points of a function’s graph to its symbolic root terms.

A polynomial of degree $n$ can be written as: $P(x)=a_n\cdot x^n+a_{n-1}\cdot x^{n-1}+...+a_2\cdot x^2+a_1\cdot x+a_0$ . The theorem states that if $r_1,r_2,r_3,...r_n$ are all the roots of the polynomials, then:

$P(x)=a_n\cdot(x-r_1)\cdot(x-r_2)\cdot...\dcot(x-r_n)$

That is, if we know all the solutions, we can factorise any polynomial easily using those.

Example: Factorise the cubic polynomial $P(x) = 2x^3 - 14x + 12$ given that its roots are $1$ , $2$ , and $-3$ .

The question becomes how do we use this technique without knowing all the solutions. There are 3 possible roots that are called obvious roots or inspection roots. These are the ones that are very easy to check if they are solutions. These are $x=1$ , $x=-1$ , and $x=0$ . If we inspect that our polynomial has one of these roots, we already know one of its factors. If it is a polynomial of degree 3, and we have one of its factors, we can write the other factor as a quadratic polynomial. Then, opening brackets and comparing coefficients, we can conclude what the quadratic factor is. Then, we can possibly factorise that quadratic factor using the rules above.

Example: Factorise the cubic polynomial $P(x) = x^3 + 7x^2 + 14x + 8$ completely by checking for an initial root at $x = -1$ and comparing coefficients.

This could be repeated for any root that is given to us. If none are given, we can always check 0, 1, and -1 as they are quickly verified.

Exercise

Find all real solutions for each of the following five quadratic equations. You may use factorisation, completing the square, or the quadratic formula. Simplify any radical terms completely.

$x^2 - 7x + 12 = 0$

$x^2 + 6x + 9 = 0$

$2x^2 + 5x - 3 = 0$

$x^2 - 4x + 1 = 0$

$3x^2 + 2x + 4 = 0$

Factorise each of the following five cubic polynomials completely into real linear factors. For the first three expressions, inspect the constant and coefficients to discover an obvious integer root.

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

$P(x) = x^3 + 4x^2 - 7x - 10$

$P(x) = 2x^3 - 5x^2 + 3x$

$P(x) = x^3 - 4x^2 - 11x + 30 \quad (\text{given that } x = 2 \text{ is a root})$

$P(x) = 2x^3 + 3x^2 - 11x - 6 \quad (\text{given that } x = -3 \text{ is a root})$

Factorise the following fourth-degree polynomial expression completely into linear real factors:

$P(x) = x^4 + 3x^3 - 6x^2 - 8x$