Given a quadratic equation of the following form,

$a{x}^{2}+bx+c=0$

we can find the roots by first converting the left hand side of the equation into factor form (product of factors).

${f}_{1}\cdot {f}_{2}\cdot \dots \cdot {f}_{n}=0$

This is because if ${f}_{1}\cdot {f}_{2}\cdot \dots \cdot {f}_{n}=0$ , then one of the factors (${f}_{1}$, ${f}_{2}$, or ${f}_{n}$) must be $0$.

Examples:

1. Find the roots of the quadratic equation

${x}^{2}-4x+3=0$

We convert the left hand side (LHS) of the equation into factor form:

$\left(x-3\right)\left(x-1\right)=0$

This implies that either $\left(x-3\right)=0$ or $\left(x-1\right)=0$

To find the solution (roots) of the quadratic equation, we can solve for the above two equations: $\begin{array}{rl}x-3=& 0\\ x=& 3\end{array}$ $\begin{array}{rl}x-1=& 0\\ x=& 1\end{array}$

Hence, the roots are: ${x}_{1}=3$ ${x}_{2}=1$

2. Find the roots of the quadratic equation

${x}^{2}+5x=0$

We convert the left hand side (LHS) of the equation into factor form:

$x\left(x+5\right)=0$

This implies that either $x=0$ or $\left(x+5\right)=0$

Hence, the roots are: ${x}_{1}=0$ ${x}_{2}=-5$

#### Methods/Techniques to factorise a quadratic expression

Solving a quadratic equation by factorisation depends on whether we can transform the quadratic expression on the left hand side to factor form. Not all quadratic equations can be factorised. Below are a few techniques/methods to factorise a quadratic expression:

##### Removal of common factors

Look at the terms in the expression, and take out the Highest Common Factor (HCF or GCD). You should always try this method first before applying any other methods.

Examples:

• $2{x}^{2}+5x$
the HCF of the terms is $x$.
So the expression can be factorised as
$x\left(2x+5\right)$
• $3{x}^{2}+9x-12$
the HCF of the terms is $3$.
So the expression can be factorised as
$3\left({x}^{2}+3x-4\right)$
The expression in the bracket can then be factorised further using another method. (see below)
$3\left(x+4\right)\left(x-1\right)$
##### Factorisation of perfect square

This technique uses the binomial expansion of perfect squares: ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$ ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$

${\left(a+b\right)}^{2}$ and ${\left(a-b\right)}^{2}$ are perfect squares.

So, we can have the following factorisations: ${x}^{2}+2ax+{a}^{2}={\left(x+a\right)}^{2}$ ${x}^{2}-2ax+{a}^{2}={\left(x-a\right)}^{2}$

Examples:

• ${x}^{2}+8x+16$
is equal to ${x}^{2}+2\cdot 4x+{4}^{2}$
and can be factorised as ${\left(x+4\right)}^{2}$
• $25-10x+{x}^{2}$
is equal to ${5}^{2}-2\cdot 5x+{x}^{2}$
and can be factorised as ${\left(5-x\right)}^{2}$
##### Difference of two squares

This technique uses the fact that: ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

So, if we have a quadratic expression that is in the form of a difference of two squares, we can factorise it using the above.

Examples:

• $25-{x}^{2}$
is a difference of two squares: ${5}^{2}-{x}^{2}$
and it can be factorised as
$\left(5+x\right)\left(5-x\right)$
• $4{x}^{2}-9$
is a difference of two squares: ${\left(2x\right)}^{2}-{3}^{2}$
and it can be factorised as
$\left(2x+3\right)\left(2x-3\right)$
##### Splitting the $x$-term

This technique splits the $x$-term (i.e. $bx$) into two terms $px+qx$ such that $pq=ac$ and $p+q=b$. It is usually used when $a\ne 1$.

First, we convert a quadratic expression in the form of $a{x}^{2}+bx+c$ into this form: $a{x}^{2}+px+qx+c$

Then we group the terms in pairs: $\left(a{x}^{2}+px\right)+\left(qx+c\right)$

Next, we factorise each group, so that we get this form: $kx\left(mx+n\right)+l\left(mx+n\right)$

And finally, we take out common factor from each group, to get the final factor form: $\left(kx+l\right)\left(mx+n\right)$

Examples:

• Factorise this quadratic expression: $3{x}^{}+13x+4$
First, we need to split $13x$ into two terms $px+qx$ such that $pq=3×4=12$ and $p+q=13$.
We need to find factors of $12$ that add to $13$. In other words, two numbers with a product of $12$ and a sum of $13$.
These two numbers are $12$ and $1$.
So,
$3{x}^{2}+13x+4$
 $=3{x}^{2}+12x+1x+4$ (split the $x$-term) $=\left(3{x}^{2}+12x\right)+\left(1x+4\right)$ (group the terms in pairs) $=3x\left(x+4\right)+1\left(x+4\right)$ (factorise each group) $=\left(x+4\right)\left(3x+1\right)$ (take out common factor from each group)
• Factorise this quadratic expression: $2{x}^{}+11x-21$
First, we need to split $11x$ into two terms $px+qx$ such that $pq=2×-21=-42$ and $p+q=11$.
We need to find factors of $-42$ that add to $11$. In other words, two numbers with a product of $-42$ and a sum of $11$.
These two numbers are $14$ and $-3$.
So,
$2{x}^{2}+11x-21$
 $=2{x}^{2}+14x-3x-21$ (split the $x$-term) $=\left(2{x}^{2}+14x\right)-\left(3x+21\right)$ (group the terms in pairs) $=2x\left(x+7\right)-3\left(x+7\right)$ (factorise each group) $=\left(x+7\right)\left(2x-3\right)$ (take out common factor from each group)
##### Trial and error

A quadratic expression in the form of $a{x}^{2}+bx+c$ can be factorised into this form: $\left(px+q\right)\left(rx+s\right)$

This technique uses trial and error to find the values of $p$, $q$, $r$, and $s$, such that $p$ and $r$ are factors of $a$, (i.e. $pr=a$) while $q$ and $s$ are factors of $c$, (i.e. $qs=c$). In addition, $ps+qr=b$.

It's easier to see how this method works through examples:

• Factorise this quadratic expression: $2{x}^{}+7x+5$
In this case, possible values for $p$ and $r$ are: $2$ and $1$.
Possible values for $q$ and $s$ are: $1$ and $5$.
We can list all of these possible values and check which one satisfy the condition $ps+qr=b$.
Since we know that $b$ is positive and $c$ is also positive, $q$ and $s$ must be both positive.  $\left(2x+1\right)\left(x+5\right)=2{x}^{2}+11x+5$ ✗ $\left(2x+5\right)\left(x+1\right)=2{x}^{2}+7x+5$ ✓
• Factorise this quadratic expression: $3{x}^{}-17x+10$
In this case, possible values for $p$ and $r$ are: $3$ and $1$ or $-3$ and $-1$.
Possible values for $q$ and $s$ are all the factors of $10$ (i.e. $1$, $2$, $5$, $10$ and their negatives).
We can list all of these possible values and check which one satisfy the condition $ps+qr=b$.
Since we know that $b$ is negative and $c$ is positive, $q$ and $s$ must be both negative.  $\left(3x-1\right)\left(x-10\right)=3{x}^{2}-31x+10$ ✗ $\left(3x-2\right)\left(x-5\right)=3{x}^{2}-17x+10$ ✓ $\left(3x-5\right)\left(x-2\right)=3{x}^{2}-11x+10$ ✗ $\left(3x-10\right)\left(x-1\right)=3{x}^{2}-13x+10$ ✗
• Factorise this quadratic expression: $8{x}^{}+34x+15$
In this case, possible values for $p$ and $r$ are: $8$ and $1$ or $4$ and $2$.
Possible values for $q$ and $s$ are $3$ and $5$ or $1$ and $15$.
We can list all of these possible values and check which one satisfy the condition $ps+qr=b$.
Since we know that $b$ is positive and $c$ is positive, $q$ and $s$ must be both positive.  $\left(8x+3\right)\left(x+5\right)=8{x}^{2}+43x+15$ ✗ $\left(4x+3\right)\left(2x+5\right)=8{x}^{2}+26x+15$ ✗ $\left(8x+5\right)\left(x+3\right)=8{x}^{2}+29x+15$ ✗ $\left(4x+5\right)\left(2x+3\right)=8{x}^{2}+22x+15$ ✗ $\left(8x+1\right)\left(x+15\right)=8{x}^{2}+121x+15$ ✗ $\left(4x+1\right)\left(2x+15\right)=8{x}^{2}+62x+15$ ✗ $\left(8x+15\right)\left(x+1\right)=8{x}^{2}+23x+15$ ✗ $\left(4x+15\right)\left(2x+1\right)=8{x}^{2}+34x+15$ ✓

As we can see above, this method can be time consuming and tedious, especially when it involves non-prime numbers with a lot of factors.

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