Tag Archives: Factorising

April 30, 2025 · 6:08 am

Factorising Non-Monic Quadratics

The general equation of a quadratic is ax^2+bx+c

Let’s explore different methods of factorising a non-monic quadratic (the a term is not 1)

Factorise 6x^2+x-12

We need to find two numbers that add to 1 and multiply to -72 (i.e. add to b and multiply to a\times c)

The two numbers are 9 and -8

Method 1 – Splitting the middle term

This is the method I teach the most often

    \begin{equation*}6x^2+x-12\end{equation}

Split the middle term (the b term) into the two numbers

    \begin{equation*}6x^2-8x+9x-12\end{equation}

The order doesn’t matter.

Find a common factor for the first term terms, and then for the last two terms.

    \begin{equation*}2x(3x-4)+3(3x-4)\end{equation}

There is a common factor of (3x-4), factor it out.

    \begin{equation*}(3x-4)(2x+3)\end{equation}

Method two – Fraction

    \begin{equation*}6x^2+x-12\end{equation}

Put 6x into both factors and divide by 6

    \begin{equation*}\frac{(6x-8)(6x+9)}{6}\end{equation}

Factorise

    \begin{equation*}\frac{2(3x-4)3(2x+3)}{6}\end{equation}

    \begin{equation*}\frac{6(3x-4)(2x+3)}{6}\end{equation}

    \begin{equation*}(3x-4)(2x+3)\end{equation}

Method 3 – Monic to non-monic

    \begin{equation*}y=6x^2+x-12\end{equation}

Multiply both sides of the equation by a

    \begin{equation*}6y=6(6x^2+x-12)\end{equation}

    \begin{equation*}6y=6^2x^2+6x-72\end{equation}

    \begin{equation*}6y=(6x)^2+6x-72\end{equation}

Let A=6x

    \begin{equation*}6y=A^2+A-72\end{equation}

Factorise

    \begin{equation*}6y=(A+9)(A-8)\end{equation}

Replace the A with 6x

    \begin{equation*}6y=(6x+9)(6x-8)\end{equation}

    \begin{equation*}6y=3(2x+3)2(3x-4)\end{equation}

    \begin{equation*}6y=6(2x+3)(3x-4)\end{equation}

    \begin{equation*}y=(2x+3)(3x-4)\end{equation}

Method 4 – Cross Method

    \begin{equation*}6x^2+x-12\end{equation}

Place the two numbers in the cross

Place the two numbers that add to 1 and multiply to -72 in the other parts of the cross.

Divide these two numbers by 6 (i.e a)

Simplify

Hence,

    \begin{equation*}(x-\frac{4}{3})(x+\frac{3}{2})\end{equation}

Which is

    \begin{equation*}(3x-4)(2x+3)\end{equation}

Method 5 – By Inspection

This is my least favourite method – although students get better with practice

    \begin{equation*}6x^2+x-12\end{equation}

The factors of a are 1, 2, 3, and 6 and the factors of 12 are 1, 2, 3, 4, 6, 12

We know one number is positive and one number negative.

Which give us all of these possibilities

Possible factorisationsb term of expansion
(x-1)(6x+12)12x-6x=6xNo
(x-2)(6x+6)6x-12x=-6xNo
(x-3)(6x+4)4x-18x-14xNo
(2x-1)(3x+12)24x-3x=21xNo
(2x-2)(3x+6)12x-6x=6xNo
(2x-3)(3x+4)8x-9x=-1xAlmost, switch the signs
(2x+3)((3x-4)-8x+9x=1xYes

    \begin{equation*}(2x+3)(3x-4)\end{equation}

With a bit of practice you don’t need to check all of the possibilties, but I find students struggle with this method.

Method 6 – Grid

    \begin{equation*}6x^2+x-12\end{equation}

Create a grid like the one below

6x^2
-12

Find the two numbers that multiply to -72 and add to 1 and place them in the other grid spots (see below)

6x^2-8x
9x-12

Find the HCF (highest common factor) of each row and put in the first column.

Row 1 HCF=2x, Row 2 HCF=3

2x6x^2-8x
39x-12

For the columns, calculate what is required to multiple the HCF to get the table entry.

For example, what do you need to multiple 2x and 3 by to get 6x^2 and 9x? In this case it is 3x. It’s always going to be the same thing, so just use one value to calculate it,

3x-4
2x6x^2-8x
39x-12

The factors are column 1 and row 1
(2x+3)(3x-4)

The two methods I use the most are splitting the middle term, and the cross method, but I can see value in the grid method.

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September 19, 2024 · 9:43 am

Solving Cubic Functions

I have been thinking about cubics a bit lately because some of my students are solving and then sketching cubics. Plus I am reading An Imaginary Tale by Paul Nahin, which talks about solving cubics and complex numbers.

Cubics must have at least one real root. If one of the roots is a rational number, then we can use the Factor and Remainder Theorem.

For example,

Solve 2x^3-3x^2-3x+2=0

We know the root(s) must be a factor of 2\times3=6.
I always start with 1 or -1
2(1)^3-3(1)^2-3(1)+2=2-3-3+2=-2 \therefore x\neq=1
Try -1
2(-1)^3-3(-1)^2-3(-1)+2=-2-3+3+2=0 \therefore x=-1 and (x+1) is a factor.
Then we can do polynomial long division.

Now we know that 2x^3-3x^2-3x+2=(x+1)(2x^2-5x+2)
And we can factorise the quadratic (or using the quadratic equation formula)
2x^2-5x+2=2x^2-4x-x+2
=2x(x-2)-1(x-2)
=(2x-1)(x-2)
\therefore x=-1, \frac{1}{2}, 2

But what if it is not factorisable?

For example,

Solve 2x^3+5x^2-2x+4=0

How many roots does this equation have?

We could find the derivative and find out how many stationary points the function has.

f'(x)=6x^2+10x-2

This is a quadratic function. Find the discriminant to determine the number of roots.

\Delta=b^2-4ac=100-4(6)(-2)=148

As \Delta>0, there are two stationary points, which means we could have 1, 2 (one root is repeated) or three roots, depending on if the function crosses the x-axis between stationary points. So not much use.

We could try the discriminant of a cubic.

\Delta=18abcd-4b^3d+b^2c^2-4ac^2-27a^2d^2

\Delta=18(2)(5)(-2)(4)-4(5^3)(4)+(5^2)(-2)^2-4(2)(-2)^2-27(2)^2(4)^2=-5100

The discriminant is negative so there is one real root.

From my reading, we need to turn the cubic into a depressed cubic (cubics of the form x^3+px+q=0).

We can do this by using a change of variable.

Let x=t-\frac{5}{6}
2(x^3+\frac{5}{2}x^2-x+2)=0
\therefore x^3+\frac{5}{2}x^2-x+2=0
Substitute t-\frac{5}{6} into the cubic.
(t-\frac{5}{6})^3+\frac{5}{2}(t-\frac{5}{6})^2-(t-\frac{5}{6})+2
(t^3-3(\frac{5}{6})t^2+3(\frac{5}{6})^2t-(\frac{5}{6})^3+\frac{5}{2}(t^2-2(\frac{5}{6})t+(\frac{5}{6})^2)-t+\frac{5}{6}+2
t^3-\frac{5}{2}t^2+\frac{25}{12}t-\frac{125}{216}+\frac{5}{2}t^2-\frac{25}{6}t+\frac{125}{72}-t+\frac{5}{6}+2
t^3-\frac{37}{12}t+\frac{431}{108}

We can then use Cardano’s formula

x=\sqrt[3]{-\frac{q}{2}+\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}

p=-\frac{37}{12} and q=\frac{431}{108}
\frac{p}{3}=\frac{-37}{36}
\frac{q}{2}=\frac{431}{216}
(\frac{431}{216})^2+(\frac{-37}{36})^3=\frac{139}{48}
t=\sqrt[3]{\frac{-431}{216}+\sqrt{\frac{139}{48}}}+\sqrt[3]{\frac{-431}{216}-\sqrt{\frac{139}{48}}}
t=-2.21095...
\therefore x=-2.21095...-\frac{5}{6}=-3.044

We can see from the sketch below that there is only one solution and it is about -3.

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