Factorise the expression $x^2+8x+15$

### Quadratic factorising

This is one of those iconic maths topics which people talk about, when they want to name-check a **difficult** pre-16 school level idea. The post-16 equivalent is Calculus.

When you first look at the quadratic expression $x^2+3x+2$, you are likely to think that it cannot be factorised; that it is not the result of a multiplication, but it is fairly straightforward to demonstrate that this is not the case.

$$ \begin{aligned} &\big(x+1 \big)\big(x+2 \big)\\ & = x^2+2x+1x+2 \\ &= x^2+3x+2 \end{aligned} $$From the calculation above it is clear that $(x+1)$ and $(x+2)$ are both factors of $x^2+3x+2$.

For more details please work through the worked examples on the next tab.

### Worked example 1: Factorising quadratics #1 $x^2+ax+b$

- Notes
In this simple situation, all of the signs in the quadratic are positive. This tells us that the signs in the brackets will also both be positive.

$\begin{aligned} &1 \times 15\\ &3 \times 5 \end{aligned}$

$\begin{aligned}&x^2+8x+15\\ &=(x+3)(x+5)\end{aligned}$

$(x+a)(x+b) = x^2+(a+b)x+ab$

When multiplying out two brackets to make a quadratic, as above we observe that the coefficient of $x^2$ is $1$, the coefficient of $x$ is the sum of $a$ and $b$, $(a+b)$ and the constant is the product of $a$ and $b$, which is $ab$.

This means that when we want to reverse the process, to find the values of $a$ and $b$, we are looking for a pair of numbers which add together to make the coefficient of $x$ and which multiply together to make the constant. Here we want a pair of numbers which add together to make $8$ and which multiply together to make $15$.

The simplest way of doing this is to start by writing down the factor pairs of $15$. From here it is simple to see that only one of these factor pairs has a sum of $8$, which is $3 \times 5$. This gives us the answer.

- $(x+3)(x+5)$

### Worked example 2: Factorising quadratics #2 $x^2-ax+b$

- Notes
Factorise the expression $x^2-8x+15$

$\begin{aligned}&x^2-8x+15\\ &=(x-3)(x-5)\end{aligned}$

This problem is only different from the first example in one regard: that is, the $+$ sign before the coefficient of $x$ is negative.

So on this occasion we are looking for a pair of numbers which multiply together to give $+15$, but this time, they will add together to give $-8$.

Now, the only way to get a positive result from a multiplication is for both of the numbers to be either both positive or both negative. Clearly, if they are going to add together to make a negative then the signs must be both negative.

- $(x-3)(x-5)$

### Worked example 3: Factorising quadratics #3 $x^2 \pm ax-b$

- Notes
- Factorise the quadratic expression $b^2 + 3b - 10$
Before you start on this problem, please have a go at the first couple of exercises on the types in examples 1 & 2. Being comfortable with the method in these more straightforward types will prepare you well for the next step up.

$\begin{aligned}&b^2+3b-10\\ &=\big(b-2\big)\big(b+5\big)\end{aligned}$

These are a little bit more tricky and require extra thought, though the process is very similar to what we have done previously.

Examine the quadratic and you will notice that the constant (the result of the product of the two numbers we are looking for) is negative and the only way that can occur is for the signs in the brackets to be one of each type.

We can begin in the same way as before, by writing out the factor pairs of $10$, which are $1 \times 10$ and $2 \times 5$.

With these, we know that one of the brackets will have a plus sign and the other will have a negative sign, so rather than asking for two numbers which add to give $+3$, we are looking for a pair of numbers with a difference of $+3$. The only way to get $+3$ is to use the factor pair $-2 \times +5$ or $+2 \times -5$. Now the first of these will give $+3$ and the 2nd will give $-3$, which we don't want.

Hence we want $-2 \times +5$.

- $\big(b-2 \big)\big(b+5 \big)$

### Worked example 4: Factorising a difference of two squares $x^2 - a^2$

- Notes
- $x^2 - 36$
A difference of two squares is a special version of the one of each signs type of quadratic, but because the coefficient of $x$ is zero, we are looking for a pair of numbers which add together to make $0$.

- $\begin{aligned} &x^2 - 36\\ &=\big(x-6 \big)\big(x+6 \big) \end{aligned}$
The factor pairs of $36$ are

$\begin{aligned} &1 \times 36\\ &2 \times 18\\ &3 \times 12\\ &4 \times 9\\ &6 \times 6 \end{aligned}$We know that one will be positive and the other negative and the only way we get a sum of $0$, is to use $\pm6$

You should notice that both $x^2$ and $36$ are perfect squares. These will always factorise using this form, hence we get this general result. Learn to notice them as they pop up all over the place and the ability to recognise them will help you greatly.

$a^2-b^2=(a-b)(a+b)$

- $\big(x-6 \big)\big(x+6 \big)$

### Exercise 1: Factorising quadratics #1 ($x^2+ax+b$)

**Instructions:**Factorise the following quadratic expressions.

- Q1$v^2 +16v + 15$\[(v+1)(v+15)\]
- Q2$a^2 + 4a +3 $\[(a+1)(a+3)\]
- Q3$m^2 +5m +4 $\[(m+4)(m+1)\]
- Q4$s^2 + 6s + 9$\[(s+3)(s+3)\]
- Q5$u^2 + 13u + 12$\[(u+1)(u+12)\]
- Q6$x^2 + 3x + 2$\[(x+1)(x+2)\]
- Q7$q^2 + 12q + 20$\[(q+2)(q+10)\]
- Q8$a^2 + 11a + 18$\[(a+2)(a+9)\]
- Q9$h^2 + 5h + 6$\[(h+2)(h+3)\]
- Q10$w^2 + 8w + 15$\[(w+3)(w+5)\]

### Exercise 2: Factorising quadratics #2 ($x^2-ax+b$)

**Instructions:**Factorise the following quadratic expressions.

- Q1$b^2 - 4b + 4$\[(b-2 )(b-2 )\]
- Q2$x^2 - 2x + 1$\[(x-1)(x-1)\]
- Q3$s^2 - 3s + 2$\[(s-1)(s-2)\]
- Q4$q^2 - 8q + 16$\[(q-4)(q-4)\]
- Q5$c^2 - 5c + 6$\[(c-2)(c-3)\]
- Q6$p^2 - 6p + 5$\[(p-1)(p-5)\]
- Q7$k^2 - 6k + 8$\[(k-2)(k-4)\]
- Q8$m^2 - 5m + 4$\[(m-1)(m-4)\]
- Q9$a^2 - 4a + 3$\[(a-1)(a-3)\]
- Q10$w^2 - 7w + 12$\[(w-3)(w-4)\]

### Exercise 3: Factorising quadratics #3 ($x^2 \pm ax - b$)

**Instructions:**Fully factorise the following quadratic expressions.

- Q1$d^2 +d - 2$\[(d-1)(d+2)\]
- Q2$k^2 -k - 2$\[(k-2)(k+1)\]
- Q3$r^2 -r - 6$\[(r-3)(r+2)\]
- Q4$x^2 + x - 6 $\[(x-2)(x+3)\]
- Q5$e^2 +5e - 6$\[(e-1)(e+6)\]
- Q6$m^2 -5m - 6$\[(m-6)(m+1)\]
- Q7$s^2 - 4$\[(s-2)(s+2)\]
- Q8$w^2 -4w - 5$\[(w-5)(w+1)\]
- Q9$f^2 +2f - 8$\[(f-2)(f+4)\]
- Q10$z^2 -2z - 8$\[(z-4)(z+2)\]

### Exercise 4: Mixed quadratic types

**Instructions:**Fully factorise the following quadratic expressions.

- Q1$d^2 -2d -24$\[(d-6 )(d+4 )\]
- Q2$j^2 +11j +30$\[(j+5 )(j+6 )\]
- Q3$q^2 -7q - 18$\[(q-9 )(q+2 )\]
- Q4$u^2 +5u -14$\[(u-2 )(u+7 )\]
- Q5$w^2 +5w - 24$\[(w-3 )(w+8 )\]
- Q6$e^2 -5e - 24$\[(e-8 )(e+3 )\]
- Q7$r^2 - 15r + 14$\[(r-1 )(r-14 )\]
- Q8$z^2 + 20z + 36$\[(z+2 )(z+18 )\]
- Q9$k^2 - 6k - 27$\[(k-9 )(k+3 )\]
- Q10$x^2 + 7x + 10$\[(x+2 )(x+5 )\]