4. Factorisation
The idea of expansion was introduced in the previous chapter. It involved removing brackets. Factorisation is the opposite process and introduces brackets. It is one of the most important processes in algebra.
Factors are whole numbers that multiply together to get a particular number. For example, the factors of [latex]6[/latex] are [latex]3[/latex] and [latex]2[/latex] or [latex]1[/latex] and [latex]6[/latex] and so
\begin{align*}
6&=2 \times 3 = 3 \times 2 \\
6&=6 \times 1 = 1 \times 6.
\end{align*}
Factorising an algebraic expression involves writing it as a product of numbers or numbers and expressions. For example
\begin{align*}
5x+10xy+15x^{2}y&=5x\left( 1+2y+3xy \right)\\
\end{align*}
or
\begin{align*}
x^2-25&=\left(x-5 \right)\left(x+5 \right).\\
\end{align*}
Do you need this chapter?
Here are some questions on the topics in this chapter. If you can do these you may skip this chapter.
Quiz: Factorisation
1. What are the common factors of the terms [latex]9xy[/latex], [latex]12x^2y[/latex] and [latex]6xy^2[/latex].
2. Factorise [latex]6y+12[/latex].
3. Factorise [latex]4xy+12xy^2+8x^2y [/latex].
4. Factorise [latex]x^2-14x+49[/latex].
5. Factorise [latex]x^2-5x+6[/latex].
6. Factorise [latex]3x^2-8x-3[/latex].
7. Factorise [latex]m^2-36[/latex].
8. Simplify
\begin{align*}
\frac{2}{x^{2}+x-12}-\frac{1}{x^{2}-9}.
\end{align*}
Answers
1) [latex]3,[/latex] [latex]x[/latex] and [latex]y[/latex].
2) [latex]6\left( y+2 \right)[/latex].
3) [latex]4xy\left( 1+3y+2x \right)[/latex].
4) [latex]\left(x-7 \right)\left(x+7 \right)[/latex].
5) [latex]\left(x-3 \right)\left(x-2 \right)[/latex].
6) [latex]\left(3x+1 \right)\left(x-3 \right)[/latex].
7) [latex]\left(m-6 \right)\left(m+6 \right)[/latex].
8) [latex]\dfrac{x+2}{\left(x+4 \right)\left(x-3 \right)\left(x+3 \right)}.[/latex]
Factorisation by removing a common factor
Here is an introductory video on factorisation using common factors.
Factorisation: Common factors (4:43 min)
Factorisation: Common factors (4:43 min) by RMIT University Library Videos (YouTube)
The steps for using common factors are:
- Search each term in the expression for a common factor (every term must have this factor)
- There may be several common factors. Search until you have found all of them
- If there is more than one common factor multiply them to give Highest Common Factor (HCF)
- The HCF is placed before the bracket.
- The terms inside the bracket are found by dividing each term by the HCF.
These steps are shown in the following examples.
Examples: Factorising using a common factor
[latex]\begin{align*} \mathbf{1.}\ 5y+10 & =5\times y+5\times2 \quad \text{common factor is }5\\ & =5(y+2) \quad \text{take common factor outside bracket. }\\ \end{align*}[/latex]
[latex]\begin{align*}\mathbf{2.}\ 3x+3y & =3\times x+3\times y \quad \text{common factor is }3\\ & =3(x+y) \quad \text{take common factor outside bracket. }\\ \end{align*}[/latex]
[latex]\begin{align*}\mathbf{3.}\ p^{2}+p & =p\times p+p\times1 \quad \text{common factor is }p\\ & =p(p+1) \quad \text{take common factor outside bracket. }\\ \end{align*}[/latex]
[latex]\begin{align*} \mathbf{4.}\ 7y^{2}+7y & =7y\times y+7y\times1 \quad \text{common factors are }7 \text{and }y. \text{ HCF is }7y\\ & =7y(y+1) \quad \text{take HCF outside bracket. }\\ \end{align*}[/latex]
[latex]\begin{align*} \mathbf{5.}\ 2abc-12ac & =2a\times bc-2a\times6c \quad \text{common factor is }2a\\ & =2ac\times b-2ac\times6 \quad \text{highest common factor is }2ac\\ & =2ac(b-6) \quad \text{take HCF outside bracket. }\\ \end{align*}[/latex]
[latex]\begin{align*} \mathbf{6.}\ -2a-2b & =(-2)\times a+(-2)\times b \quad \text{common factor is }-2\\ & =-2(a+b) \quad \text{take common factor outside bracket. }\\ \end{align*}[/latex]
[latex]\begin{align*} \mathbf{7.}\ -3x+6xy & =(-3x)\times1-(-3x)\times2y \quad \text{highest common factor is }-3x\\ & =-3x(1-2y) \quad \text{take HCF outside bracket }\\ & =3x(-1+2y) \quad \text{rearrange to get +ve sign outside bracket.} \\ & \qquad \qquad \qquad \qquad \text{It looks better but is not necessary}\\ & =3x(2y-1)\quad \text{rearrange to get positive term first in bracket.}\\ & \qquad \qquad \qquad \qquad \text{It looks better but is not necessary.} \end{align*}[/latex]
Please try these exercises.
Exercises: Factorising using a common factor
Factorise where possible.
[latex]\begin{array}{lllll} \quad \mathbf{1.}\ 3x+3y & \mathbf{2.}\ 2a-2b & \mathbf{3.}8a-8b+8c\\ \quad \mathbf{4.} \ xy-5x & \mathbf{5.}\ x^{2}-x & \mathbf{6.}\ 7x+21y\\ \quad \mathbf{7.}\ 5m-2n & \mathbf{8.}\ c^{2}-2bc-3c & \mathbf{9.}\ 5mn-10n\\ \quad \mathbf{10.}\ 3m^{2}-3mnp & \mathbf{11.}\ 7x+21x^{2} &\mathbf{12.}\ 12m^{2}-18mn\\ \quad \mathbf{13.}\ 5xy-10xz & \mathbf{14.}\ 5pq-pq^{2}-3pqr & \mathbf{15.}\ 2ab^{2}c+6abc^{2}\\ \quad \mathbf{16.}\ rst+5rst^{2}-2rs & \mathbf{17.}\ 5mn+10m-pqr & \mathbf{18.}\ 5xyz-x^{2}yz^{2}+10x\\ \end{array}[/latex]
Answers
[latex]\begin{array}{lllll} \quad \mathbf{1.}\ 3\left(x+y\right) & \mathbf{2.}\ 2\left(a-b\right) & \mathbf{3.}\ 8\left(a-b+c\right)\\ \quad \mathbf{4.}\ x\left(y-5\right) & \mathbf{5.}\ x\left(x-1\right) & \mathbf{6.}\ 7\left(x+3y\right)\\ \quad \mathbf{7.}\ \text{no factors} & \mathbf{8.}\ c\left(c-2b-3\right)& \mathbf{9.}\ 5n\left(m-2\right)\\ \quad \mathbf{10.}\ 3m\left(m-np\right) & \mathbf{11.}\ 7x\left(1+3x\right) &\mathbf{12.}\ 6m\left(2m-3n\right)\\ \quad \mathbf{13.}\ 5x\left(y-2z\right) & \mathbf{14.}\ pq\left(5-q-3r\right) & \mathbf{15.}\ 2abc\left(b+3c\right)\\ \quad \mathbf{16.}\ rs\left(t+5t^{2}-2\right) & \mathbf{17.}\ \text{no factors} & \mathbf{18.}\ x\left(5yz-xyz^{2}+10\right) \\ \end{array}[/latex]
Factorisation using perfect squares
In the previous chapter, expressions like [latex]\left(a+b\right)^2[/latex] and [latex]\left(a-b\right)^2[/latex] were expanded and two important rules were introduced:
\begin{align*}
\left(a+b\right)^2&=a^2+2ab+b^2\\
\left(a-b\right)^2&=a^2-2ab+b^2.\\
\end{align*}
The right-hand sides of these formulas are called perfect squares because they can be written as the square of a binomial expression. We also refer to numbers like [latex]9[/latex] and [latex]25[/latex] as perfect squares because [latex]9=3^2[/latex] and [latex]25=5^2[/latex].
The rules for expanding perfect squares may be used in reverse to factorise algebraic expressions as shown in the following examples.
Examples: Factorising using perfect squares
1. Factorise [latex]x^{2}+14x+49[/latex].
Solution:
Note that [latex]49=7^{2}[/latex] and [latex]14=2\times7.[/latex] Applying the rule for [latex]\left(a+b\right)^2[/latex] we have
\begin{align*}
x^{2}+14x+49 & =\left(x+7\right)^{2}.
\end{align*}
2. Is [latex]y^{2}-20y+25[/latex] a perfect square?
Solution:
Note that [latex]25=5^{2}[/latex] but [latex]-20\neq2\times5[/latex] hence [latex]y^{2}-20y+25[/latex] is not a perfect square.
3. Is [latex]4a^{2}-12a-9[/latex] a perfect square?
Solution:
The last term is negative and so [latex]4a^{2}-12a-9[/latex] is not a perfect square.
4. Is [latex]100x^{2}-180x+81[/latex] a perfect square?
Solution:
\begin{align*}
100x^{2} & =\left(10x\right)^{2}\\81 & =9^{2}\end{align*}and [latex]180=2\times10\times9[/latex]. Hence [latex]\begin{align*}100x^{2}-180x+81 & =\left(10x-9\right)^{2}\end{align*}[/latex] and is a perfect square.
5. Factorise [latex]50x^{2}+80x+32[/latex].
Solution:
At first sight, the expression is not a perfect square because neither [latex]50x^{2}[/latex] nor [latex]32[/latex] is a perfect square. However, if we divide the expression by [latex]2[/latex] we have
\begin{align*}
50x^{2}+80x+32 & =2\left(25x^{2}+40x+16\right)
\end{align*}
The expression [latex]25x^{2}+40x+16[/latex] is a perfect square because
\begin{align*}
25x^{2}+40x+16 & =\left(5x+4\right)^{2}
\end{align*}
and so
\begin{align*}
50x^{2}+80x+32 & =2\left(5x+4\right)^{2}.
\end{align*}
So while [latex]50x^{2}+80x+32[/latex] is not a perfect square, it can be factorised as [latex]50x^{2}+80x+32=2\left(5x+4\right)^{2}[/latex].
Please try these exercises.
Exercises: Factorising using perfect squares
Check each of the following expressions. If it is a perfect square, write the expression as a perfect square.
[latex]\begin{array}{llll} \quad \mathbf{1.}\ a^{2}+2a+1 && \mathbf{2.}\ x^{2}-4x+4 && \mathbf{3.}\ 25x^{2}-10x+1 \\ \quad \mathbf{4.}\ 4y^{2}-6y+9 && \mathbf{5.}\ 81x^{2}+108x+36 && \mathbf{6.}\ 9a^{2}-24a-16\\ \quad \mathbf{7.}\ 16x^{2}-40xy+25y^{2} && \mathbf{8.}\ 121z^{2}+88z+64 && \mathbf{9.}\ 2x^{2}+8x+8 \end{array}[/latex]
Answers
[latex]\begin{array}{lllll} \quad \mathbf{1.}\ \left(a+1\right)^{2} \\ \quad \mathbf{2.}\ \left(x-2\right)^{2} \\ \quad \mathbf{3.}\ \left(5x-1\right)^{2} \\ \quad \mathbf{4.}\ \text{Not a perfect square} \\ \quad \mathbf{5.}\ \left(9x+6\right)^{2} \\ \quad \mathbf{6.}\ \text{Not a perfect square} \\ \quad \mathbf{7.}\ \left(4x-5y\right)^{2} \\ \quad \mathbf{8.}\ \text{Not a perfect square} \\ \quad \mathbf{9.}\ \text{Not a perfect square but }2x^{2}+8x+8=2\left(x+2\right)^2 \end{array}[/latex]
Factorisation using the difference of two squares
Consider the following expansion:
\begin{align*}
(x+5)(x-5) & =x^{2}-5x+5x-25\\
& =x^{2}-25\\
& =x^{2}-5^{2}
\end{align*}
In general,
\begin{alignat*}{1}
(x+a)(x-a) & =x^{2}-ax+ax-a^{2}\\
& =x^{2}-a^{2}.
\end{alignat*}
Note that:
- The terms in the brackets differ only in the sign of the second term;
- The expansion is the difference of two terms, both of which are perfect
squares.
More generally:
\begin{alignat*}{1}
(a+b)(a-b) & =a^{2}-b^{2}
\end{alignat*}
so
\begin{alignat*}{1}
a^{2}-b^{2} & =(a+b)(a-b),
\end{alignat*}
and is known as the DOTS (Difference Of Two Squares) rule.
This can be used to factorise expressions of the form a^{2}-b^{2}. It is an important rule and you should memorize it.
The following video shows how to factorise using the difference of two squares.
Factorisation: Difference of two squares (DOTS) (4:43 min)
Factorisation: Difference of two squares (DOTS) (4:43 min) by RMIT University Library Videos (YouTube)
Here are some examples of the use of the DOTS rule.
Examples: Factorising using the difference of two squares
1. Factorise [latex]a^{2}-36[/latex].
Solution:
\begin{align*}
a^{2}-36 & =a^{2}-\left(6\right)^{2}\ \text{expression is the difference of two squares}\\
& =\left(a+6\right)\left(a-6\right)\ \text{using the DOTS rule}\\
& =\left(a-6\right)\left(a+6\right)\ \text{order doesn't matter, either is correct.}
\end{align*}
2. Factorise [latex]4^{2}-y^{2}[/latex].
Solution:
\begin{align*}
4^{2}-y^{2} & =\left(2\right)^{2}-y^{2}\ \text{expression is the difference of two squares}\\
& =\left(2+y\right)\left(2-y\right)\ \text{using the DOTS rule}.
\end{align*}
3. Factorise [latex]3x^{2}-48[/latex].
Solution:
At first sight, we cannot use DOTS. But taking out a common factor
of [latex]3[/latex] we have
\begin{align*}
3x^{2}-48 & =3\left(x^{2}-16\right)\\
& =3\left(x^{2}-4^{2}\right)\\
& =3\left(x+4\right)\left(x-4\right)\ \text{using the DOTS rule}.
\end{align*}
4. Factorise [latex]\left(x+2\right)^{2}-9[/latex].
Solution:
\begin{align*}
\left(x+2\right)^{2}-9 & =\left(x+2\right)^{2}-3^{2}\ \text{expression is the difference of two squares}\\
& =\left(x+2+3\right)\left(x+2-3\right)\ \text{using the DOTS rule}\\
& =\left(x+5\right)\left(x-1\right).
\end{align*}
5. Factorise [latex]y^{2}+36[/latex].
Solution:
This expression is the sum of two squares, not the difference, hence
the DOTS rule cannot be applied. There are no real factors for this
expression. There are complex factors but we do not consider them in this chapter.
Please try these exercises.
Exercises: Factorising using the difference of two squares
[latex]\begin{array}{lllll} \quad \mathbf{1.}\ \ x^{2}-4 && \mathbf{21.}\ a^{2}-100 &&\mathbf{3.}\ 49-x^{2} \\ \quad \mathbf{4.}\ 64x^{2}-1 && \mathbf{5.}\ 121x^{2}-49y^{2} && \mathbf{6.}\ a^{2}b^{2}-25 \\ \quad \mathbf{7.}\ 5x^{2}-20 && \mathbf{8.}\ a^{2}+100 && \mathbf{9.}\ x^{2}y^{3}-36y\\ \quad \mathbf{10.} \ \left(x+2\right)^{2}-y^{2} && \mathbf{11.}\ \left(x-5\right)^{2}-36 && \mathbf{12.}\ \left(a+1\right)^{2}-\left(b-2\right)^{2} \end{array}[/latex]
Answers
[latex]\begin{array}{lllll} \quad \mathbf{1.}\ \left(x+2\right)\left(x-2\right) && \mathbf{2.}\ \left(a+10\right)\left(a-10\right) \\ \quad \mathbf{3.}\ \left(7+x\right)\left(7-x\right) && \mathbf{4.}\ \left(8x+1\right)\left(8x-1\right) \\ \quad \mathbf{5.}\ \left(11x+7y\right)\left(11x-7y\right) && \mathbf{6.}\ \left(ab+5\right)\left(ab-5\right) \\ \quad \mathbf{7.}\ 5\left(x+2\right)\left(x-2\right) && \mathbf{8.}\ \text{Does not factorise.} \\ \quad \mathbf{9.}\ y\left(xy+6\right)\left(xy-6\right) && \mathbf{10.}\ \left(x+2+y\right)\left(x+2-y\right) \\ \quad \mathbf{11.}\ \left(x-11\right)\left(x+1\right) && \mathbf{12.}\ \left(a+b-1\right)\left(a-b+3\right) \end{array}[/latex]
Factorisation of quadratics of form: [latex]x^2+bx+c[/latex]
The following video gives an introduction to factorising quadratics.
The general form of a quadratic expression is: [latex]ax^{2}+bx+c[/latex], [latex]a\neq0[/latex], where [latex]a[/latex], [latex]b[/latex] and [latex]c[/latex] are real constants and [latex]x[/latex] is a variable. Quadratics frequently occur in maths and science. Being able to factorise them is an essential skill.
Quadratics with [latex]a=1[/latex] are considered first. In the following section, the general case is considered. To factorise [latex]x^{2}+bx+c[/latex] you have to find two factors such that
\begin{align*}
x^{2}+bx+c&=\left(x+m\right)\left(x+n\right)
\end{align*}
where [latex]m[/latex] and [latex]n[/latex] are numbers to be found.
It is easier to consider a case with some numbers in it. To factorise [latex]x^{2}+5x+6[/latex] we need two numbers [latex]m[/latex] and [latex]n[/latex] such that
\begin{align*}
x^{2}+5x+6 & =\left(x+m\right)\left(x+n\right).\\
\end{align*}
Expanding the right-hand side,
\begin{align*}
x^{2}+5x+6 & =\left(x+m\right)\left(x+n\right)\\
&=x^2+xn+mx+mn\\
&=x^2+\left(n+m\right)x+mn.
\end{align*}
Comparing the left-hand side with the right-hand side, [latex]5=n+m[/latex] and [latex]6=nm[/latex]. The factors of six are 1 and 6 or 2 and 3. Only 2 and 3 add to 5 so
\begin{align*}
x^{2}+5x+6 & =\left(x+2\right)\left(x+3\right).\\
\end{align*}
Note that the order of factors on the right-hand side does not matter and
\begin{align*}
x^{2}+5x+6 & =\left(x+3\right)\left(x+2\right),\\
\end{align*}
is also correct.
This can be generalized to give the rule:
To factorise [latex]x^{2}+bx+c[/latex], find two numbers [latex]m[/latex] and [latex]n[/latex] such that
\begin{align*}
x^{2}+bx+c & =\left(x+m\right)\left(x+n\right)
\end{align*}
where [latex]m\times n=c[/latex] and [latex]m+n=b[/latex].
The following examples show how the rule is used.
Examples: Factorising quadratics I
1. Factorise [latex]x^{2}+9x+14[/latex]
Solution:
\begin{align*}
x^{2}+9x+14 & =\left(x+m\right)\left(x+n\right)
\end{align*}
where according to the rule above,
\begin{align*}
m\times n & =14\ \text{and }\\
m+n & =9.
\end{align*}The factors of [latex]14[/latex] are\begin{align*}
1.\ m & =1 & 2.\ m & =-1 & 3.\ m & =2 & 4.\ m & =-2\\
n & =14 & n & =-14 & n & =7 & n & =-7.
\end{align*}
Of these, only the factors in [latex]3[/latex] satisfy the requirement that [latex]m+n=9[/latex]. So
\begin{align*}
x^{2}+9x+14 & =\left(x+2\right)\left(x+7\right).
\end{align*}
2. Factorise [latex]y^{2}-7y+12[/latex].
Solution:
We want to write
\begin{align*}
y^{2}-7x+12 & =\left(y+m\right)\left(y+n\right)
\end{align*}
where according to the rule above,
\begin{align*}
m\times n & =12\ \text{and }\\
m+n & =-7.
\end{align*}The factors of [latex]12[/latex] are\begin{align*}
1.\ m & =3 & 2.\ m & =-3 & 3.\ m & =2 & \\
n & =4 & n & =-4 & n & =6\\
4.\ m & =-2 & 5.\ m & =12 & 6.\ m & =-12 \\
n & =-6 & n & =1 & n & =-1.
\end{align*}
Of these, only the factors in [latex]2[/latex] satisfy the requirement that [latex]m+n=-7[/latex].
So
\begin{align*}
y^{2}-7x+12 & =\left(y-3\right)\left(y-4\right).
\end{align*}
3. Factorise [latex]p^{2}-5p-14[/latex].
Solution:
We want to write
\begin{align*}
p^{2}-5p-14 & =\left(p+m\right)\left(p+n\right)
\end{align*}
where according to the rule above, [latex]m\times n =14[/latex] and [latex]m+n =-5.[/latex]
The factors of [latex]-14[/latex] are\begin{align*}
1.\ m & =1 & 2.\ m & =-1 & 3.\ m & =-2 & 4.\ m & =2\\
n & =-14 & n & =14 & n & =7 & n & =-7.
\end{align*}
Of these, only the factors in [latex]4[/latex] satisfy the requirement that [latex]m+n=-5[/latex].
So
\begin{align*}
p^{2}-5x-14 & =\left(p+2\right)\left(p-7\right).
\end{align*}
4. Factorise [latex]a^{2}+6a-7[/latex].
Solution:
We want to write
\begin{align*}
a^{2}+6a-7 & =\left(a+m\right)\left(a+n\right)
\end{align*}
where according to the rule above, [latex]m\times n=-7[/latex] and [latex]m+n =6[/latex].
The factors of [latex]-7[/latex] are
\begin{align*}
1.\ m & =1 & 2.\ m & =-1\\
n & =-7 & n & =7.
\end{align*}Of these, only the factors in [latex]2[/latex] satisfy the requirement that [latex]m+n=6[/latex].
So
\begin{align*}
a^{2}+6a-7 & =\left(a-1\right)\left(a+7\right).
\end{align*}
5. Factorise [latex]a^{2}+3a+6[/latex].
Solution:
We want to write
\begin{align*}
a^{2}+3a+6 & =\left(a+m\right)\left(a+n\right)
\end{align*}
where according to the rule above, [latex]m\times n =6[/latex] and [latex]m+n =3[/latex].
The factors of 6 are
\begin{align*}
1.\ m & =1 & 2.\ m & =-1 & 3.\ m & =-2 & 4.\ m & =2\\
n & =6 & n & =-6 & n & =-3 & n & =3.
\end{align*}None of these factors satisfy the requirement that [latex]m+n=3[/latex]. So it is not possible to factorise the expression [latex]a^{2}+3a+6[/latex]. In this case, we say there are no real factors. Geometrically this means that the graph of [latex]y=a^{2}+3a+6[/latex] does not touch or intersect the [latex]a-\text{axis}[/latex]
Please try the following exercises.
Exercises: Factorising quadratics 1
Factorise the following quadratic expressions:
1. [latex]x^2+7x+10[/latex]
2. [latex]x^2-5x+6[/latex]
3. [latex]x^2-2x-24[/latex]
Answers
1. [latex]\left(x+5 \right)\left(x+2 \right)[/latex]
2. [latex]\left(x-3 \right)\left(x-2 \right)[/latex]
3. [latex]\left(x-6 \right)\left(x+4 \right)[/latex]
Factorisation of quadratics of form: [latex]ax^2+bx+c[/latex]
In this section factorisation of expressions of the form
\begin{align*}
ax^{2}+bx+c
\end{align*}
where [latex]a\neq 0[/latex] or [latex]1[/latex].
Expressions of the type [latex]ax^{2}+bx+c[/latex] can be factorised using a technique similar to that used for expressions of the type [latex]x^{2}+bx+c[/latex].
In this case, the coefficient of [latex]x[/latex], in at least one bracket, will not equal [latex]1[/latex].
Consider the following product.
\begin{align*}
(3x+2)(2x+1) & =(3x)(2x)+(3x)(1)+(2)(2x)+(2)(1)\\
& =6x^{2}+3x+4x+2\\
& =6x^{2}+7x+2.
\end{align*}
Note that
- multiplying the first terms in each bracket gives the [latex]x^{2}[/latex] term.
In this case [latex]6x^2[/latex]; - multiplying the last term in each bracket gives the constant term,
in this case [latex]2[/latex]; - the coefficient of the [latex]x-[/latex]term is the sum of the [latex]x[/latex] terms in the expansion. In this case [latex]3x+4x=7x[/latex].
We can use these ideas to factorise expressions like [latex]ax^{2}+bx+c[/latex]
as shown in the following examples.
Examples: Factorising quadratics of form: [latex]ax^2+bx+c[/latex]
1. Factorise [latex]2x^{2}+7x+6[/latex].
Solution:
The only factors of the coefficient of the [latex]x^{2}[/latex] term are [latex]2[/latex] and [latex]1[/latex]. So we are looking for a factorisation like
[latex]\begin{align*} 2x^{2}+7x+6 & =\left(2x+m\right)\left(x+n\right)\text{}\\ & =2x^{2}+2nx+mx+nm\\ & =2x^{2}+\left(2n+m\right)x+nm \end{align*}[/latex]
where [latex]mn=6[/latex] and [latex]2n+m=7[/latex]. We have the following possibilities:
\begin{align*}
1.\ m & =3 & 2.\ m & =2 & 3.\ m & =6 & 4.\ m & =1\\
n &=2 & n & =3 & n & =1 & n & =6.
\end{align*}
(Note that negative factors like [latex]m=-6,\,n=-1[/latex] and [latex]m=-1,\,n=-6[/latex] don't need to be considered as they don't satisfy the condition [latex]2n+m=7[/latex].)
Of these possibilities, the only one that satisfies [latex]2n+m=7[/latex] is number [latex]1.[/latex] That is [latex]m=2[/latex] and [latex]n=2.[/latex] Hence the solution is:
\begin{align*}
2x^{2}+7x+6 & =\left(2x+3\right)\left(x+2\right).
\end{align*}
2. Factorise [latex]2x^{2}-10x+12[/latex].
Solution:
At first this looks like a case where [latex]a=2[/latex] but a factor of [latex]2[/latex] can be taken out to get:
\begin{align*}
2x^{2}-10x+12 & =2\left(x^{2}-5x+6\right)
\end{align*}
Now\begin{align*}
2x^{2}-10x+12 & =2\left(x+m\right)\left(x+n\right)
\end{align*}
where [latex]mn=6[/latex] and [latex]m+n=-5[/latex]. This implies [latex]m=-2[/latex] and [latex]n=-3[/latex] so the solution is:
\begin{align*}
2x^{2}-10x+12 & =2\left(x^{2}-5x+6\right)\\
& =2\left(x-2\right)\left(x-3\right).
\end{align*}
3. Factorise [latex]6x^{2}+13x-8[/latex].
Solution:
In this case, there are no numbers that divide into each term as in the previous example. The coefficient of the [latex]x^{2}[/latex] term is [latex]6[/latex] which has factors [latex]6[/latex] and [latex]1[/latex], and [latex]2[/latex] and [latex]3[/latex]. So we are looking for a factorisation such as:
\begin{align*}
6x^{2}+13x-8 & =\left(6x+m\right)\left(x+n\right)\\
& =6x^{2}+6xn+mx-8\\
& =6x^{2}+\left(6n+m\right)x-8 & \left(3.1\right)
\end{align*}
or
\begin{align*}
6x^{2}+13x-8 & =\left(3x+m\right)\left(2x+n\right)\\
& =6x^{2}+3xn+2mx-8\\
& =6x^{2}+\left(3n+2m\right)x-8. & \left(3.2\right)
\end{align*}
In both cases, [latex]mn=-8[/latex]. So we have the possibilities
\begin{align*}
m & =-8 & n & =1\\
m & =8 & n & =-1\\
m & =4 & n & =-2\\
m & =-4 & n & =2.
\end{align*}
For eqn [latex]\left(3.1\right)[/latex] we know that [latex]6n+m=13[/latex]. This is not satisfied by any of the [latex]m[/latex] and [latex]n[/latex] values above.
For eqn [latex]\left(3.2\right)[/latex] we know that [latex]3n+2m=13[/latex]. This is satisfied by [latex]m=8[/latex] and [latex]n=-1[/latex] so the solution is
\begin{align*}
6x^{2}+13x-8 & =\left(3x+8\right)\left(2x-1\right).
\end{align*}
4. Factorise [latex]4x^{2}+4x+1.[/latex]
Solution:
In this case, there are no numbers that divide into each term. The coefficient of the [latex]x^{2}[/latex] term is [latex]4[/latex] which has factors [latex]4[/latex] and [latex]1[/latex], and [latex]2[/latex] and [latex]2[/latex]. So we are looking for a factorisation such as:
\begin{align*}
4x^{2}+4x+1 & =\left(4x+m\right)\left(x+n\right)\\
& =4x^{2}+4xn+mx+mn\\
& =4x^{2}+\left(4n+m\right)x+1 & \left(4.1\right)
\end{align*}
or
\begin{align*}
4x^{2}+4x+1 & =\left(2x+m\right)\left(2x+n\right)\\
& =4x^{2}+2xn+2mx+mn\\
& =4x^{2}+\left(2n+2m\right)x+1 & \left(4.2\right)
\end{align*}
where [latex]mn=1[/latex]. That means
\begin{align*}
m & =1 & n= & 1 & \left(4.3\right)\\
m & =-1 & n= & -1. & \left(4.4\right)
\end{align*}
Suppose eqn [latex]\left(4.1\right)[/latex] is correct then [latex]\left(4n+m\right)=4[/latex]. But this is not possible with the choices for [latex]m[/latex] and [latex]n[/latex] in [latex]\left(4.3\right)[/latex] and [latex]\left(4.4\right)[/latex]. Hence the factorisation must be as in eqn [latex]\left(4.2\right)[/latex] with [latex]2n+2m=4[/latex]. The latter is achieved with [latex]\left(4.3\right)[/latex] and so the solution is
\begin{align*}
4x^{2}+4x+1 & =\left(2x+1\right)\left(2x+1\right).
\end{align*}
The approach given in these examples is useful, provided there are not too many factors for [latex]a[/latex] and [latex]c[/latex]. If the number of factors is excessive, or not whole numbers, we can employ other methods such as completing the square or the quadratic formula. These are discussed in Chapter 8: Quadratic Functions, Graphs and Equations.
Please try the following exercises.
Exercises: Factorising quadratics 2
Factorise the following quadratic expressions:
1. [latex]\ 2x^2+3x+1\ .[/latex]
2. [latex]\ 6x^2+19x+3\ .[/latex]
3. [latex]\ 6x^2+11x+3\ .[/latex]
4. [latex]\ 12x^2+5x-3\ .[/latex]
5. [latex]\ 6x^2+13x+6\ .[/latex]
6. [latex]\ 6x^2-5x-6\ .[/latex]
Answers
1. [latex]\left(2x+1 \right) \left(x+1 \right)[/latex]
2. [latex]\left(6x+1 \right) \left(x+ 3 \right)[/latex]
3. [latex]\left(3x+1 \right) \left(2x+ 3 \right)[/latex]
4. [latex]\left(4x-3 \right) \left(3x+1 \right)[/latex]
5. [latex]\left(3x+2 \right) \left(2x+ 3\right)[/latex]
6. [latex]\left(3x+2 \right) \left(2x- 3 \right)[/latex]
Simplification of algebraic fractions involving quadratics
More complicated algebraic fractions can involve polynomials of any order. However, it is unlikely that you will have to deal with polynomials higher than quadratics.
In this section, algebraic fractions involving quadratics are considered. There is no general theory for this so we proceed via examples. Please note that algebraic fractions may exclude certain values of [latex]x[/latex]. This is to ensure we never get a zero denominator. You should point out any excluded values.
Examples: Algebraic fractions involving quadratics
1. Simplify
\begin{align*}
\frac{2}{x^{2}+x-12}-\frac{1}{x^{2}-9}.
\end{align*}
Solution:
You could use [latex]\left(x^{2}+x-12\right)\times\left(x^{2}-9\right)[/latex] as a common denominator and work from there. However, that would give you a quartic, a polynomial involving x^{4} as a denominator which may be a bit difficult to reduce to the lowest terms. Instead, we note that the denominators may be factorised. In particular,
\begin{align*}
x^{2}+x-12 & =\left(x+4\right)\left(x-3\right)
\end{align*}
and
\begin{align*}
x^{2}-9 & =\left(x-3\right)\left(x+3\right).
\end{align*}
So we can write
\begin{align*}
\frac{2}{x^{2}+x-12}-\frac{1}{x^{2}-9} & =\frac{2}{\left(x+4\right)\left(x-3\right)}-\frac{1}{\left(x-3\right)\left(x+3\right)}.
\end{align*}
A common denominator is
\begin{align*}
\left(x+4\right)\left(x-3\right)\left(x+3\right)
\end{align*}
so
\begin{align*}
\frac{2}{x^{2}+x-12}-\frac{1}{x^{2}-9} & =\frac{2\left(x+3\right)}{\left(x+4\right)\left(x-3\right)\left(x+3\right)}-\frac{1\left(x+4\right)}{\left(x-3\right)\left(x+3\right)\left(x+4\right)}\\
& =\frac{2\left(x+3\right)-1\left(x+4\right)}{\left(x+4\right)\left(x-3\right)\left(x+3\right)}\\
& =\frac{2x+6-x-4}{\left(x+4\right)\left(x-3\right)\left(x+3\right)}\\
& =\frac{x+2}{\left(x+4\right)\left(x-3\right)\left(x+3\right)}\text{ where } x\neq-4,-3,3.
\end{align*}
2. Simplify
\begin{align*}
\dfrac{\dfrac{-3}{x^{2}+2x-3}+\dfrac{1}{x-1}}{\dfrac{3}{x-1}+3}. &
\end{align*}
Solution:
The numerator is:
\begin{align*}
\frac{-3}{x^{2}+2x-3}+\frac{1}{x-1} & =\frac{-3}{\left(x+3\right)\left(x-1\right)}+\frac{1}{x-1}\\
& =\frac{-3}{\left(x+3\right)\left(x-1\right)}+\frac{1}{x-1}.
\end{align*}
Using a common denominator, we have
\begin{align*}
\frac{-3}{x^{2}+2x-3}+\frac{1}{x-1} & =\frac{-3}{\left(x+3\right)\left(x-1\right)}+\frac{1}{x-1}\\
& =\frac{-3}{\left(x+3\right)\left(x-1\right)}+\frac{\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}\\
& =\frac{x}{\left(x+3\right)\left(x-1\right)}.
\end{align*}Now consider the denominator:
\begin{align*}
\frac{3}{x-1}+3 & =\frac{3}{x-1}+\frac{3\left(x-1\right)}{x-1}\\
& =\frac{3x}{x-1}.
\end{align*}
Hence\begin{align*}
\dfrac{\dfrac{-3}{x^{2}+2x-3}+\dfrac{1}{x-1}}{\dfrac{3}{x-1}+3} & =\frac{x}{\left(x+3\right)\left(x-1\right)}\div\frac{3x}{x-1}\\
& =\frac{x}{\left(x+3\right)\left(x-1\right)}\times\frac{\left(x-1\right)}{3x}\\
& =\frac{x}{3x\left(x+3\right)}\\
& =\frac{1}{3\left(x+3\right)}\text{ where } x\neq-3.
\end{align*}
Please try the following exercises.
Exercises: Algebraic fractions involving quadratics
Simplify the following
[latex]\begin{array}{lllll} \qquad \mathbf{1.}\ \dfrac{2}{x^{2}+2x}+\dfrac{1}{x^{2}-4} && \mathbf{2.}\ \dfrac{\dfrac{1}{x^{2}-4}+\dfrac{1}{2x+4}}{1+\dfrac{2}{x-2}}\\ \end{array}[/latex]
Answers
[latex]\begin{array}{lllll} \qquad \mathbf{1.}\ \dfrac{3x-4}{x\left(x+2\right)\left(x-2\right)} \ \text{ where }x\neq-2,0,2. && \mathbf{2.}\ \dfrac{1}{2\left(x+2\right)} \ \text{ where } x\neq-2. \end{array}[/latex]
Key takeaways
1. Factorising an algebraic expression involves writing it as a product of numbers or numbers and expressions.
2. Factorisation is the opposite of expansion and requires insertion of brackets.
3. A common factor is a number or pronumeral that divides into every term in an algebraic expression. For example [latex]2xy[/latex] is a common factor of every term in the expression
\begin{align*}
4x^{2}yz-6x^{2}y^{2}+30xyz.&
\end{align*}
4. When factorising an expression, look for common factors and take them outside a pair of brackets. For example, for the expression above,
\begin{align*}
4x^{2}yz-6x^{2}y^{2}+30xyz&=2xy\left( 2xz-3xy+15z \right) .\\
\end{align*}
5. The difference of two squares formula is useful and should be remembered:
\begin{align*}
x^{2}-y^{2}&=\left( x-y \right)\left( x+y \right).
\end{align*}
6. To factorise a quadratic [latex]x^{2}+bx+c[/latex], find two numbers [latex]m[/latex] and [latex]n[/latex] such that
\begin{align*}
x^{2}+bx+c & =\left(x+m\right)\left(x+n\right)
\end{align*}
where [latex]m\times n=c[/latex] and [latex]m+n=b[/latex].
