## A3.4 Quadratics

Image: Pixabay

Quadratic expressions have the general form \[ax^2+bx+c\] where \(a\), \(b\) and \(c\) are real numbers and \(a\neq 0\). Quadratics frequently arise in mathematics, science and engineering. This module explains how to factorise a quadratic into two linear factors. For example \[x^2+5x+6 = \left(x+2\right)\left(x+3\right).\]

The general form of a quadratic expression is: \(ax^{2}+bx+c\), \(a\neq0\), where \(a\), \(b\) and \(c\) are real constants and \(x\) is the variable.

We will initially work with expressions that have \(a=1\) so the expression becomes \(x^{2}+bx+c\).

### Expansion

To expand an expression of the form \((a+b)(c+d)\), multiply each term in the first bracket by each term in the second bracket.

\[\begin{align*} (x+2)(x+3) & =x(x+3)+2(x+3)\\ & =x^{2}+3x+2x+6\\ & =x^{2}+5x+6. \end{align*}\]

### Factorisation

Factorisation is the reverse of expansion:

\(x^{2}+5x+6\) is expressed as the product of two factors, \((x+2)\) and \((x+3).\) That is \[\begin{align*} x^{2}+5x+6 & =\left(x+2\right)\left(x+3\right). \end{align*}\]

Note that:

- Multiplying the first term in each bracket gives the term \(x^{2}\) in the expression as above.
- Multiplying the last term in each bracket gives the constant term, \(+6\) in the expression.
- The coefficient of the \(x\) term is the sum of the last term in each bracket \((+2+3=+5)\).

Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on factorising quadratics. The general form of a quadratic expression is given by A X squared plus B X plus C. Here A, B and C are constants and X is the unknown. For instance here’s an expression, two X squared minus three X minus four, A is plus two, B is minus three and C is minus four. Note carefully the sign for each of the numbers.

Let’s go back a step first and look at revision of expansion, in other words removing brackets. Here’s an expression, X plus two X plus three and when you expand that you multiply each term in the first bracket by the terms in the second bracket to get an answer of X squared plus five X plus six. We’ve done this in a previous movie.

Now factorisation is the reverse of expansion, in other words X square plus five X plus six can be expressed to the product of two factors, X plus two and X plus three. In other words as you go from left to right in this equation we’re factorising, as we go from right to left we’re expanding. A few important things to notice about factorisation; first of all multiplying the first terms in each bracket gives the X squared in the expression, in other words X times X is X squared. Multiplying the last terms in each bracket gives the constant term plus six in the expression, in other words plus two times plus three is plus six. Finally the co-efficient of the X terms, in other words the middle term, is the sum of the last term in each bracket, in other words here plus two and plus three is plus five. Now we’ll be using these steps to work out some of these examples that follow.

So in general for the expression A squared plus B X plus C where A is equal to one multiply two numbers to get C, we add the same two numbers to get B, in other words we insert one number into each of the brackets and secondly the order is not important, in other words the second brackets could be the first and vice versa, it doesn’t matter which way around you place the brackets.

So here’s our first example, X squared plus nine X plus 14. So straight away we can place two brackets, as I’ve done, to the right of the equal signs. Using the rules we need two numbers that when multiplied together gives us the last term, 14, and those same two numbers that, when added, gives us the co-efficient of X, the middle term, which is nine. So we can look at 14 and one, seven and two and so on until we get those two numbers that when multiplied gives us 14 but also added gives us nine and those two numbers are seven and two, seven times two is 14, seven plus two is nine. Notice if I use 14 and one multiplying them is fine but adding them will not give us nine.

Let’s look at another one; Y squared minus seven Y plus 12. Here we have a negative sign in the middle term which makes it slightly more difficult but we still use the same idea as in the previous example, in other words two numbers that multiply gives us 12, the same two numbers that when added gives us minus seven, so we can look at 12 and one, six and two, four and three. Notice when we look at those different combinations, those factors of 12, there’s only one that will satisfy the two conditions and that is minus three and minus four. Notice minus three times minus four is plus 12, minus three plus minus four is minus seven, so our answer is Y minus three Y minus four.

Here’s another one. Notice the terms are separated by negative signs, P squared minus five P minus 14. Again, using the rules, multiplying the two you must get minus 14, adding the two numbers we must get minus five. Here it’s not so difficult because we only have 14 and one, that gives us 14, but 14 and one there’s no way in which, when adding those two numbers, whether they be negative or positive, will give us minus five, so the solution to this is P minus seven and P plus two. Be extra careful with the negative signs here, notice that minus seven times plus two gives us minus 14 and minus seven plus two is minus five.

And finally here’s another one; A squared plus six A minus seven. We need two numbers that multiply to give minus seven and two numbers that add to give plus six. So we don’t have much choice here, the only factors we have of seven are seven and one, so they must be seven and one, but the difficult problem here is to work out where the negative signs go and this usually comes just with trial and error, in other words A plus seven A minus one, seven times minus one is minus seven, plus seven minus one is plus six. If you tried minus seven and plus one you would notice you would not get plus six A, so try the different methods to make sure you’re correct.

Now try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.

The basic rule is:

To factorise \(x^{2}+bx+c\), find two numbers \(m\) and \(n\) such that \[\begin{align*} x^{2}+bx+c & =\left(x+m\right)\left(x+n\right) \end{align*}\] where \(m\times n=c\) and \(m+n=b.\)

Note that order of the factors does not matter. That is \[\begin{align*} x^{2}+bx+c & =\left(x+m\right)\left(x+n\right)\\ & =\left(x+n\right)\left(x+m\right). \end{align*}\]

#### Example 1

Factorise \(x^{2}+9x+14.\)

**Solution:**

We want to write \[\begin{align*} x^{2}+9x+14 & =\left(x+m\right)\left(x+n\right) \end{align*}\] where according to the rule above, \[\begin{align*} m\times n & =14\ \text{and }\\ m+n & =9. \end{align*}\]

The factors of \(14\) 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 \(3\) satisfy the requirement that \(m+n=9\). So \[\begin{align*} x^{2}+9x+14 & =\left(x+2\right)\left(x+7\right). \end{align*}\]

#### Example 2

Factorise \(y^{2}-7y+12\).

**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 \(12\) are

\[\begin{align*} 1.\ m & =3 & 2.\ m & =-3 & 3.\ m & =2 & 4.\ m & =-2 & 5.\ m & =12 & 6.\ m & =-12\\ n & =4 & n & =-4 & n & =6 & n & =-6 & n & =1 & n & =-1. \end{align*}\] Of these, only the factors in \(2\) satisfy the requirement that \(m+n=-7\). So \[\begin{align*} y^{2}-7x+12 & =\left(y-3\right)\left(y-4\right). \end{align*}\]

#### Example 3

Factorise \(p^{2}-5p-14\).

**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, \[\begin{align*} m\times n & =14\ \text{and }\\ m+n & =-5. \end{align*}\]

The factors of \(-14\) 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 \(4\) satisfy the requirement that \(m+n=-5\). So \[\begin{align*} p^{2}-5x-14 & =\left(p+2\right)\left(p-7\right). \end{align*}\]

#### Example 4

Factorise \(a^{2}+6a-7\).

**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, \[\begin{align*} m\times n & =-7\ \text{and }\\ m+n & =6. \end{align*}\]

The factors of \(-7\) are

\[\begin{align*} 1.\ m & =1 & 2.\ m & =-1\\ n & =-7 & n & =7 \end{align*}\]

Of these, only the factors in \(2\) satisfy the requirement that \(m+n=6\). So \[\begin{align*} a^{2}+6a-7 & =\left(a-1\right)\left(a+7\right). \end{align*}\]

#### Example 5 (No Real Factors)

Factorise \(a^{2}+3a+6\).

**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, \[\begin{align*} m\times n & =6\ \text{and }\\ m+n & =3. \end{align*}\]

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 \(m+n=3\). So it is not possible to factorise the expression \[\begin{align*} a^{2}+3a+6 & . \end{align*}\] In this case we say there are no real factors.1 Geometrically this means that the graph of \(y=a^{2}+3a+6\) does not touch or intersect the \(a-\text{axis$.$ }\) There are complex factors but these are not dealt with in this module. It is important to understand when this occurs and is discussed in a later section.

### Factorisation when \(a\neq1\)

In this section we deal with factorisation of expressions of the form \[\begin{align*} ax^{2}+bx+c \end{align*}\] where \(a\neq1.\)

Expressions of the type \(ax^{2}+bx+c\) can be factorised using a technique similar to that used for expressions of the type \(x^{2}+bx+c.\)

In this case the coefficient of \(x\), in at least one bracket, will not equal 1.

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 term in each bracket gives the \(x^{2}\) term. In this case \(6x^{2}.\)
- multiplying the last term in each bracket gives the constant term, in this case \(2.\)
- the coefficient of the \(x-\)term is the sum of the \(x\) terms in the expansion. In this case \(3x+4x=7x.\)

We can use these ideas to factorise expressions like \(ax^{2}+bx+c\).

#### Example 6

Factorise \(2x^{2}+7x+6\).

**Solution:**

The only factors of the coefficient of the \(x^{2}\) term are \(2\) and 1. So we are looking for a factorisation like \[\begin{align*} 2x^{2}+7x+6 & =\left(2x+m\right)\left(x+n\right)\text{}\\ & =2x^{2}+2nx+mx+nm\\ & =2x^{2}+\left(2n+m\right)x+nm \end{align*}\] where \(mn=6\) and \(2n+m=7.\) We have the following possibilities:2 Note that negative factors like \(m=-6,\,n=-1\) and \(m=-1,\,n=-6\) don’t need to be considered as they don’t satisfy condition \(2n+m=7\).

\(m=3,\ \:n=2\)

\(m=2,\ \ n=3\)

\(m=6,\ \:n=1\)

Of these possibilities, the only one that satisfies \(2n+m=7\) is number \(1.\) That is \(m=2\text{ and }n=2\), so \[\begin{align*} 2x^{2}+7x+6 & =\left(2x+3\right)\left(x+2\right). \end{align*}\]

#### Example 7

Factorise \(2x^{2}-10x+12.\)

**Solution:**

At first this looks like a case where \(a=2\) but a factor of \(2\) can be taken out to get: 3 You should always check if there is a number that divides into all terms of the quadratic. \[\begin{align*} 2x^{2}-10x+12 & =2\left(x^{2}-5x+6\right) \end{align*}\] Now we can use the methods in Examples \(1-4\) above to get

\[\begin{align*} 2x^{2}-10x+12 & =2\left(x+m\right)\left(x+n\right) \end{align*}\] where \(mn=6\) and \(m+n=-5.\) This implies \(m=-2\) and \(n=-3\) and so: \[\begin{align*} 2x^{2}-10x+12 & =2\left(x^{2}-5x+6\right)\\ & =2\left(x-2\right)\left(x-3\right). \end{align*}\]

#### Example 8

Factorise \(6x^{2}+13x-8\).

**Solution:**

In this case there are no numbers that divide into each term as we had in Example 7 above. The coefficient of the \(x^{2}\) term is \(6\) which has factors \[\begin{align*} 6 & =6\times1\\ & =2\times3. \end{align*}\] So we are looking for a factorisation such as: \[\begin{align*} 6x^{2}+13x-8 & =\left(6x+m\right)\left(x+n\right)\\ & =6x^{2}+6xn+mx-8\\ & =6x^{2}+\left(6n+m\right)x-8 & \left(8.1\right) \end{align*}\] or \[\begin{align*} 6x^{2}+13x-8 & =\left(3x+m\right)\left(2x+n\right)\\ & =6x^{2}+3xn+2mx-8\\ & =6x^{2}+\left(3n+2m\right)x-8 & \left(8.2\right) \end{align*}\] In both cases, \(mn=-8.\) 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 \(\left(8.1\right)\) we know that \(6n+m=13\). This is not satisfied by any of the \(m\) and \(n\) values above.

For eqn \(\left(8.2\right)\) we know that \(3n+2m=13\). This is satisfied by \(m=8\) and \(n=-1\) and so \[\begin{align*} 6x^{2}+13x-8 & =\left(3x+8\right)\left(2x-1\right). \end{align*}\]

#### Example 9

Factorise \(4x^{2}+4x+1\)

**Solution:**

In this case there are no numbers that divide into each term as we had in Example 7 above. The coefficient of the \(x^{2}\) term is \(4\) which has factors \[\begin{align*} 4 & =4\times1\\ & =2\times2. \end{align*}\]

So we are looking for a factorisation such as: \[\begin{align*} 4x^{2}+4x+1 & =\left(4x+m\right)\left(x+n\right)\\ & =4x^{2}+4xn+mx+mn\\ & =4x^{2}+\left(4n+m\right)x+1 & \left(9.1\right) \end{align*}\] or \[\begin{align*} 4x^{2}+4x+1 & =\left(2x+m\right)\left(2x+n\right)\\ & =4x^{2}+2xn+2mx+mn\\ & =4x^{2}+\left(2n+2m\right)x+1 & \left(9.2\right) \end{align*}\] where \(mn=1\). That means \[\begin{align*} m & =1 & n= & 1 & \left(9.3\right)\\ m & =-1 & n= & -1. & \left(9.4\right) \end{align*}\] Suppose eqn \(\left(9.1\right)\) is correct then \(\left(4n+m\right)=4.\) But this is not possible with the choices for \(m\) and \(n\) in \(\left(9.3\right)\) and \(\left(9.4\right)\). Hence the factorisation must be as in eqn \(\left(9.2\right)\) with \(2n+2m=4.\) The latter is achieved with \(\left(9.3\right)\) and so \[\begin{align*} 4x^{2}+4x+1 & =\left(2x+1\right)\left(2x+1\right). \end{align*}\]

The approach given in examples 6-9 is okay provided there are not too many factors for \(a\) and \(c\). If the number of factors is excessive, we can employ other methods.

### When Can you Get Real Linear Factors for a Quadratic?

In Example 5 above we found that we could not get real linear factors for \[\begin{align*} a^{2}+3a+6 & . \end{align*}\]

This raises the question of when real solutions to general quadratics may be found. The most general quadratic has the form 4 Note that the graph of \[\begin{align*} y & =ax^{2}+bx+c \end{align*}\] is a parabola. \[\begin{align*} ax^{2}+bx+c & . \end{align*}\]

To determine if there are real linear factors, we introduce the discriminant.

#### The Discriminant

The discriminant denoted \(\Delta,\) for the general quadratic \[\begin{align*} ax^{2}+bx+c \end{align*}\] is \[\begin{align*} \Delta & =b^{2}-4ac. \end{align*}\]

If \[\begin{align*} \Delta & =\begin{cases} 0\text{ there is one repeated linear factor}\\ >0\text{ there are two distinct linear factors}\\ <0\text{ there are no real linear factors.} \end{cases} \end{align*}\] The discriminant tells us how many real roots there are to the equation 5 Geometrically, the discriminant tells us how many times the graph of \[\begin{align*} y & =ax^{2}+bx+c \end{align*}\] intersects the \(x-\)axis. If \(\Delta=0\), the graph just touches the \(x-\)axis at one point. If \(\Delta>0,\) the graph intersects the \(x-\)axis at two points. If \(\Delta<0,\) the graph does not intersect the \(x-\)axis. \[\begin{align*} ax^{2}+bx+c=0 & . \end{align*}\]

#### Example 10

Factorise (if possible) \(x^{2}+6x+12\)

**Solution:**

In this case: \(a=1\) , \(b=+6\) , \(c=+12\) therefore

\[\begin{align*} \Delta & =\left(b^{2}-4ac\right)\\ & =36-4(1)(12)\\ & =36-48\\ & =-12\\ & <0. \end{align*}\] The discriminant is negative therefore \(x^{2}+6x+12\) has no real factors.

### Exercise 1

Factorise the following expressions (if possible):

\(\text{a)$\ x^{2}+10x+21\qquad\text{b)$\ z^{2}+11z+18\qquad\text{c)$\ x^{2}+5x-14$ }$ }$ }\)

\(\text{d)$\ m^{2}-m-72\qquad\text{e)$\ x^{2}+6x+9\qquad\text{f)$\ a^{2}-15a+44$ }$ }$ }\)

\(\text{g)$\ x^{2}-2x-24\qquad\text{h)$\ y^{2}-10y+16\qquad\text{i)$\ z^{2}+4z-60$ }$ }$ }\)

\(\text{j)$\ n^{2}+6n-16\qquad\text{k)$\ a^{2}+5a+10\qquad\text{l)$\ s^{2}+2s-48$ }$ }$ }\)

\(\text{m)$\ y^{2}+7y+19\qquad\text{n)$\ x^{2}+16x+39\qquad\text{o)$\ x^{2}-14x+45$ }$ }$ }\)

Note that order of factors does not matter.

\(\text{a)$\ \left(x+7\right)\left(x+3\right)\qquad\text{b)$\ \left(z+9\right)\left(z+2\right)\qquad\text{c)$\ \left(x+7\right)\left(x-2\right)$ }$ }$ }\)

\(\text{d)$\ \left(m-9\right)\left(m+8\right)\qquad\text{e)$\ \left(x+3\right)\left(x+3\right)\qquad\text{f)$\ \left(a-11\right)\left(a-4\right)$ }$ }$ }\)

\(\text{g)$\ \left(x-6\right)\left(x+4\right)\qquad\text{h)$\ \left(y-2\right)\left(y-8\right)\qquad\text{i)$\ \left(z-6\right)\left(z+10\right)$ }$ }$ }\)

\(\text{j)$\ \left(n-2\right)\left(n+8\right)\qquad\text{k) no real factors$\qquad\text{l)$\ \left(s+8\right)\left(s-6\right)$ }$ }$ }\)

\(\text{m) no real factors$\qquad\text{n)$\ \left(x+13\right)\left(x+3\right)\qquad\text{o)$\ \left(x-9\right)\left(x-5\right).$ }$ }$ }\)

### Exercise 2

Factorise the following if possible.

\(\text{a)$\ 5x^{2}+13x+6\qquad\text{b)$\ 2x^{2}+x-15\qquad\text{c)$\ 3m^{2}-m-2$ }$ }$ }\)

\(\text{d)$\ 3y^{2}-10y+8\qquad\text{e)$\ 2a^{2}+11a+12\qquad\text{f)$\ 6x^{2}-11x+5$ }$ }$ }\)

Note that order of factors does not matter.

\(\text{a)$\ \left(5x+3\right)\left(x+2\right)\qquad\text{b)$\ \left(2x-5\right)\left(x+3\right)\qquad\text{c)$\ \left(3m+2\right)\left(m-1\right)$ }$ }$ }\)

\(\text{d)$\ \left(3y-4\right)\left(y-2\right)\qquad\text{e)$\ \left(2a+3\right)\left(a+4\right)\qquad\text{f)$\ \left(6x-5\right)\left(x-1\right).$ }$ }$ }\)

Download this page, A3.4 Factorisation: Quadratics (PDF 258 KB)

What's next... Completing the square