# Solving Quadratics Through Factorising

## Solving Quadratics Through Factorising Revision

**Solving Quadratic Equations by Factorising**

**Quadratics** are **algebraic expressions** that include the term, x^2, in the general form,

ax^2 + bx + c

If you are on the foundation course, any quadratic equation you’re expected to solve will always have a=1, with all terms on one side and a **zero on the other**. If you are on the higher course, you may have to do some **rearranging** in order to get all the terms on one side. Make sure you are happy with the following topics before continuing.

**Solving through Factorising (a=1)**

We can solve quadratics through factorising by following these **4** easy steps.

**Example:** Solve the quadratic equation, \textcolor{blue}{ x^2-3x+2=0} by factorisation.

**Step 1:**Factorise the quadratic \textcolor{blue}{(x -2)(x-1)=0}**Step 2:**Form two linear equations

Since the right-hand side of the equation is zero, the result of multiplying the two brackets together on the left-hand side **must be zero**. Therefore, at least one of the brackets must be equal to zero. So, in summary,

If (x-2)(x-1)=0, then either (x-2)=0 or (x-1)=0.

**Step 3:**Solve the equations to find the roots of the equation

Equation 1:

\begin{aligned}(+2) \,\,\,\,\,\,\,\,\, x - 2 &= 0 \\ x &= 2\end{aligned}

Equation 2:

\begin{aligned}(+1) \,\,\,\,\,\,\,\,\, x - 1 &= 0 \\ x &= 1\end{aligned}

The final roots are:

x = 2 \,\,\,\,\text{and}\,\,\,\, x=1

**Note:** The solutions to a quadratic equation are also called **roots**, because they correspond to where a quadratic graph crosses the x-axis.

**Solving through Factorising (a>1)**

Solve the following quadratic equation through factorising 2x^2-3x-9=0

**Step 1:** Rearrange the given quadratic so that is it **equal to zero** (=0)

This quadratic is already equal to 0 so there is nothing more to do.

**Step 2:** Factorise the quadratic,

(2x+3)(x-3)=0

**Step 3:** Form two linear equations

2x+3=0 \,\,\text{and}\,\, x-3=0

**Step 4:** Solve the equations to find the roots of the equation

Equation 1:

\begin{aligned}(-3)\,\,\,\,\,\,\,2x+3&=0 \\ (\div2)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x&=-3 \\ \,\,\,\,\,\,\,\,\, x& = -\dfrac{3}{2}\end{aligned}

Equation 2:

\begin{aligned}(+3)\,\,\,\,\,\,\,\,\,x-3&=0 \\ x &=3 \end{aligned}

So, the 2 solutions to the equation are x=3 and x=-\frac{3}{2}.

**Quadratic Equations and Sketching Graphs**

It is possible to use factorisation to allow you to sketch a quadratic graph.

**Example:** Use factorisation to find the roots of x^2 -2x -3=0 and hence sketch the quadratic.

Like always, the equation needs to be =0

**Step 1:** First we need to factorise the left hand side of the equation

(x-3)(x+1) = 0

**Step 2:** Solve the quadratic as show in the above examples

\begin{aligned}(+3)\,\,\,\,\,\,\,\,\,x-3&=0 \\ x&=3\end{aligned}

\begin{aligned}(-1) \,\,\,\,\,\,\,\,\,x+1&=0 \\ \,\,\,\,\,\,\,\,\, x&=-1\end{aligned}

**Step 3:** Find the coordinates of the roots. We know this equation has the solutions x= 3 and x=-1

When we set the equation to 0, y=0. This means we can form the two coordinates

(3,0) and (-1,0)

**Step 4:** Identify the y-intercept. This is the point at which the curve crosses the y-axis and is given by the value of c in the general quadratic ax^2 + bx +c=0, which in this case is -3.

Once you have found these values, you can sketch the graph, see below.

**Note:** This graph was done by a computer, but a “sketch” doesn’t have to be perfect, it just has to be the right shape and cross the axes at the right points.

## Solving Quadratics Through Factorising Example Questions

**Question 1:** Use factorisation to solve the equation p^2-3p-10=0

**[3 marks]**

The quadratic on the left hand side of the equation factorises so that,

p^2-3p-10=(p+2)(p-5)

Therefore, we can rewrite the equation as

(p+2)(p-5)=0

For the left-hand side to be zero we require one of the brackets to be zero, hence, the two solutions to this quadratic equation are,

p=-2 and p=5

**Question 2:** Use factorisation to solve the equation x^2-8x+15=0

**[3 marks]**

The quadratic on the left hand side of the equation factorises so that,

x^2-8x+15=(x-5)(x-3)

Therefore, we can rewrite the equation as

(x-5)(x-3)=0

For the left-hand side to be zero we require one of the brackets to be zero, hence, the two solutions to this quadratic equation are,

x=5 and x=3

**Question 3: **Use factorisation to find the roots of the quadratic x^2-6x+8

**[3 marks]**

To find the roots of a quadratic, we must set it equal to zero and find the solutions of that equation,

x^2-6x+8=0

Now, we must factorise. Observing that (-2)\times(-4)=8 and -2+(-4)=-6, we get that this quadratic factorises to

(x-2)(x-4)=0

So, for the left-hand side to be zero we require one of the brackets to be zero.

Therefore, the two roots of this quadratic equation are

x=2 and x=4

**Question 4:** Use factorisation to solve the equation 2x^2+13x+9=-6

**[4 marks]**

First rearrange the equation so that it equals 0.

2x^2+13x+15=0

The quadratic on the left hand side of the equation factorises so that,

2x^2+13x+15=(2x+3)(x+5)=0

For the left-hand side to be zero we require one of the brackets to be zero, hence, the two solutions to this quadratic equation are,

x=-\dfrac{3}{2} and x=-5

**Question 5:** Use factorisation to solve the equation 3k^2+17k-6=0

**[4 marks]**

The quadratic on the left hand side of the equation factorises so that,

(3k-1)(k+6)=0

So, for the left-hand side to be zero we require one of the brackets to be zero. If the first bracket is zero, then we get

3k-1=0

If we add 1 to both sides and then divide by 3, we get the solution

k=\dfrac{1}{3}

If the second bracket is zero, then we get

k=-6

Therefore, the two solutions to this quadratic equation are,

k=\frac{1}{3} and k=-6