See also: Quadratic Functions and Quadratic Equations, Quadratic Equation Formula, Quadratic Equation Completing the Square, GCD and LCM, numbers
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:
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:
$$(x-3)(x-1)=0$$This implies that either $(x-3)=0$ or $(x-1)=0$
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: $${x}_{1}=3$$ $${x}_{2}=1$$
Find the roots of the quadratic equation
$${x}^{2}+5x=0$$We convert the left hand side (LHS) of the equation into factor form:
$$x(x+5)=0$$This implies that either $x=0$ or $(x+5)=0$
Hence, the roots are: $${x}_{1}=0$$ $${x}_{2}=-5$$
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:
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:
This technique uses the binomial expansion of perfect squares: $${(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$$ $${(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$$
${(a+b)}^{2}$ and ${(a-b)}^{2}$ are perfect squares.
So, we can have the following factorisations: $${x}^{2}+2ax+{a}^{2}={(x+a)}^{2}$$ $${x}^{2}-2ax+{a}^{2}={(x-a)}^{2}$$
Examples:
This technique uses the fact that: $${a}^{2}-{b}^{2}=(a+b)(a-b)$$
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:
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: $$(a{x}^{2}+px)+(qx+c)$$
Next, we factorise each group, so that we get this form: $$kx(mx+n)+l(mx+n)$$
And finally, we take out common factor from each group, to get the final factor form: $$(kx+l)(mx+n)$$
Examples:
$=3{x}^{2}+12x+1x+4$ | (split the $x$-term) |
$=(3{x}^{2}+12x)+(1x+4)$ | (group the terms in pairs) |
$=3x(x+4)+1(x+4)$ | (factorise each group) |
$=(x+4)(3x+1)$ | (take out common factor from each group) |
$=2{x}^{2}+14x-3x-21$ | (split the $x$-term) |
$=(2{x}^{2}+14x)-(3x+21)$ | (group the terms in pairs) |
$=2x(x+7)-3(x+7)$ | (factorise each group) |
$=(x+7)(2x-3)$ | (take out common factor from each group) |
A quadratic expression in the form of $$a{x}^{2}+bx+c$$ can be factorised into this form: $$(px+q)(rx+s)$$
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:
$(2x+1)(x+5)=2{x}^{2}+11x+5$ | ✗ |
$(2x+5)(x+1)=2{x}^{2}+7x+5$ | ✓ |
$(3x-1)(x-10)=3{x}^{2}-31x+10$ | ✗ |
$(3x-2)(x-5)=3{x}^{2}-17x+10$ | ✓ |
$(3x-5)(x-2)=3{x}^{2}-11x+10$ | ✗ |
$(3x-10)(x-1)=3{x}^{2}-13x+10$ | ✗ |
$(8x+3)(x+5)=8{x}^{2}+43x+15$ | ✗ | $(4x+3)(2x+5)=8{x}^{2}+26x+15$ | ✗ |
$(8x+5)(x+3)=8{x}^{2}+29x+15$ | ✗ | $(4x+5)(2x+3)=8{x}^{2}+22x+15$ | ✗ |
$(8x+1)(x+15)=8{x}^{2}+121x+15$ | ✗ | $(4x+1)(2x+15)=8{x}^{2}+62x+15$ | ✗ |
$(8x+15)(x+1)=8{x}^{2}+23x+15$ | ✗ | $(4x+15)(2x+1)=8{x}^{2}+34x+15$ | ✓ |
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.
Confused and have questions? We’ve got answers. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field.
See also: Quadratic Functions and Quadratic Equations, Quadratic Equation Formula, Quadratic Equation Completing the Square, GCD and LCM, numbers