Quadratic equations
A quadratic is a polynomial of degree 2. Some examples of quadratic equations and expressions in one variable are:
| $2x^2-3x+1=4$2x2−3x+1=4 | , | $\left(x+2\right)\left(x-9\right)$(x+2)(x−9) | , | $x^2=5$x2=5 |
Quadratics can be monic, where the coefficient of $x^2$x2 is 1, or non-monic if the coefficient of $x^2$x2 is not equal to 1.
As for linear equations, quadratic equations can be solved for an unknown by rearranging to isolate the variable. Often, isolating the variable can be made easier by first factorising the equation.
Factorisation techniques
Before solving quadratic equations, let's review common factorisation techniques. Looking for factorisation first when solving quadratics is a quick and effective strategy before moving to the quadratic formula. Here's a list:
- Highest Common Factor (HCF) factorisation: $AB+AC+\dots=A\left(B+C+\dots\right)$AB+AC+…=A(B+C+…), can be factorised by identifying the HCF for all terms. In this example, the HCF is $A$A.
- Difference of two squares: $A^2-B^2=\left(A+B\right)\left(A-B\right)$A2−B2=(A+B)(A−B). Look for the difference of two terms which are both perfect squares.
- Grouping in pairs: Look for four terms that can be split up into two pairs and factorised separately. Then, factorise using the HCF factorisation method.
- Perfect squares: $A^2+2AB+B^2=\left(A+B\right)^2$A2+2AB+B2=(A+B)2, $A^2-2AB+B^2=\left(A-B\right)^2$A2−2AB+B2=(A−B)2. Look for three terms where the first and third terms are perfect squares, and the middle term is twice the product of their square roots.
- Sum and product for monic quadratics: Look for three terms of the form $x^2+Px+Q$x2+Px+Q, where $P$P and $Q$Q are any numbers (and $x$x could be any variable). Check if the quadratic can be solved using the perfect square method first. Otherwise, find two numbers $A$A and $B$B that have a sum of $P$P and a product of $Q$Q, to give the factorised form $\left(x+A\right)\left(x+B\right)$(x+A)(x+B).
The key when factorising quadratics is to figure out which technique to use and when. Remember to always check if your answer can be further factorised before finalising the solution.
Worked examples
example 1
Factorise $xy-5y-2x+10$xy−5y−2x+10 completely.
Think about whether to take out positive or negative factors.
Do: Since these four terms have no common factors, try grouping in pairs.
$y$y goes into the first two pairs and $2$2 goes into the last two. So $xy-5y=y\left(x-5\right)$xy−5y=y(x−5) but $-2x+10=2\left(-x+5\right)$−2x+10=2(−x+5)
Therefore taking $2$2 out does not help us factorise further. Try $-2$−2 instead:
| $xy-5y-2x+10$xy−5y−2x+10 | $=$= | $y\left(x-5\right)-2\left(x-5\right)$y(x−5)−2(x−5) | |
| $=$= | $\left(x-5\right)\left(y-2\right)$(x−5)(y−2) | using HCF factorisation where $x-5$x−5 is the highest common factor |
example 2
Factorise $3x^2-3x-90$3x2−3x−90 completely.
Think about which 3 term method to use here, and whether to use another method first.
Do: This is a quadratic but non-monic, that is, the coefficient of $x^2$x2 is not equal to 1.
We cannot use the perfect squares technique here, but the three terms do have an HCF of $3$3, so factor that out first:
$3x^2-3x-90=3\left(x^2-x-30\right)$3x2−3x−90=3(x2−x−30)
The expression in the brackets is now a monic quadratic but is not a perfect square. Try using factor pairs.
Factor pairs of $-30$−30 are: $1$1 & $-30$−30, $-1$−1 & $30$30, $2$2 & $-15$−15, $-2$−2 & $15$15, $3$3 & $-10$−10, $-3$−3 & $10$10, $5$5 & $-6$−6, and $-5$−5 & $6$6
The only pair with a sum of $-1$−1 is $5$5 & $-6$−6.
Therefore:
$3\left(x^2-x-30\right)=3\left(x+5\right)\left(x-6\right)$3(x2−x−30)=3(x+5)(x−6)
Practice questions
Question 1
Factorise $k^2-81$k2−81.
Question 2
Factorise $x^2+12x+36$x2+12x+36.
Factorisation for non-monic quadratics
Most of the quadratics seen so far are monic, meaning their $x^2$x2 term only has a coefficient of $1$1. If the coefficient was not $1$1, then one way of factorising is to factor out that coefficient from the whole quadratic.
e.g. $2x^2-4x+6=2\left(x^2-2x+3\right)$2x2−4x+6=2(x2−2x+3).
Sometimes this won't work, specifically when the other two terms don't share the highest common factor. For example, looking at the quadratic expression $2x^2-x-1$2x2−x−1, only the $x^2$x2 term has a factor of 2, the other terms do not. So how can quadratics like this be factorised?
First, have a look at how a non-monic quadratic is composed:
Below are three different methods for factorising these more complicated forms of quadratics.
Cross method
The cross method has already been encountered once before with monic quadratics, and it's easy to see how this extends to non-monic quadratics.
For example, consider $5x^2+11x-12$5x2+11x−12. Draw a cross with a possible pair of factors of $5x^2$5x2 on one side and another possible factor pair of $-12$−12 on the other side.
Start with the factor pairs of $5x$5x & $x$x on the left, and $-6$−6 & $2$2 on the other:
$5x\times2+x\times\left(-6\right)=4x$5x×2+x×(−6)=4x, which is incorrect, so try again with another two pairs:
$5x\times3+x\times\left(-4\right)=11x$5x×3+x×(−4)=11x which is the correct answer. By reading across in the two circles, the quadratic must then factorise to $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3).
Sometimes, an unknown in an equation may need to be identified by considering the factor pairs in a non-monic quadratic. To do this, write all the pairs of factors, both positive and negative, and identify the sums for each pair by combining them systematically.
Practice questions
QUESTION 3
Find all positive and negative integers $k$k such that $x^2+kx+14$x2+kx+14 is factorisable.
Write your answers as a set; separate each term with a comma.
PSF method
The PSF (Product, Sum, Factor) method uses a similar idea to monic quadratics involving sums and products, but with different steps.
For a quadratic in the form $ax^2+bx+c$ax2+bx+c:
1. Find two numbers, $m$m & $n$n, that have a SUM of $b$b and a PRODUCT of $ac$ac.
2. Rewrite the quadratic as $ax^2+mx+nx+c$ax2+mx+nx+c.
3. Use grouping in pairs to factorise the four-termed expression.
Worked example
example 3
Using the same example as above, factorise $5x^2+11x-12$5x2+11x−12 using the PSF method.
Think about possible sums and products of $m$m & $n$n.
Do
We want the sum of $m$m & $n$n to be $11$11, and the product to be $5\times\left(-12\right)=-60$5×(−12)=−60.
The two numbers work out to be $-4$−4 & $15$15, so:
| $5x^2+11x-12$5x2+11x−12 | $=$= | $5x^2-4x+15x-12$5x2−4x+15x−12 |
| $=$= | $x\left(5x-4\right)+3\left(5x-4\right)$x(5x−4)+3(5x−4) | |
| $=$= | $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3) |
This is the same answer as was found using the cross method.
PSF variation
The above two methods are the most often used. However, a slightly different method can also be used to factorise directly using a formula.
$ax^2+bx+c=\frac{\left(ax+m\right)\left(ax+n\right)}{a}$ax2+bx+c=(ax+m)(ax+n)a, where $m+n=b$m+n=b & $mn=ac$mn=ac
Worked example
example 4
Factorise $5x^2-36x+7$5x2−36x+7 completely.
Think about whether it is easier to consider the product or the sum of $m$m & $n$n first.
Do
| $m+n$m+n | $=$= | $b$b |
| $=$= | $-36$−36 | |
| $mn$mn | $=$= | $ac$ac |
| $=$= | $5\times7$5×7 | |
| $=$= | $35$35 |
It's much easier to look at the product first as there are fewer possible pairs that multiply to give $35$35 than those that add to give $-36$−36. We can see that $m$m & $n$n $=$= $-1$−1 & $-35$−35. Then:
| $5x^2-36x+7$5x2−36x+7 | $=$= | $\frac{\left(5x-1\right)\left(5x-35\right)}{5}$(5x−1)(5x−35)5 |
| $=$= | $\frac{\left(5x-1\right)\left(x-7\right)\times5}{5}$(5x−1)(x−7)×55 | |
| $=$= | $\left(5x-1\right)\left(x-7\right)$(5x−1)(x−7) |
Practice questions
Question 4
Factorise the following trinomial:
$6x^2+13x+6$6x2+13x+6
Question 5
Factorise $-12x^2-7x+12$−12x2−7x+12.
Modelling situations using quadratics
As for linear equations, quadratics can be used to model a situation. Identify the unknowns and any other information that will help you construct an equation or expression to represent the situation. Then use factorisation methods if required.
Practice questions
QUESTION 6
Use the figure below to answer the following questions.
Let the length $L=-56y+11$L=−56y+11 and the width $W=5y^2$W=5y2.
Write the perimeter of the figure in terms of $y$y. Express the perimeter in simplified expanded form.
Fully factor the polynomial.