# Factorising Quadratic Equations Support Page

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Welcome to the Math Salamanders' Factorising Quadratic Equations Support Page.

Here you will find help and support to enable you to learn to factorise a range of different quadratic equations of the form ax2 + bx + c = 0.

All the quadratic equations in this section factorise into integer solutions inside each bracket.

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## Definition of a Quadratic Equation

A quadratic equation is an equation involving the square of a number which is the highest power in the equation.

It is usually in the form:

ax2 + bx + c = 0

where a, b and c are numbers, and x is an unknown.

It is also sometimes called a second degree polynomial equation.

Examples

The following equations are quadratic expressions:

x2 + 7x + 4

2y2 - 28

4a 2 - ½ a + 15

The following equations are NOT quadratic expressions:

x3 + 7x + 4

2y - 28

4a 5 - ½ a + 15

How to solve a quadratic equation

Please note that this method will only work for specific cases of a quadratic equation with integer/fraction answers.

Take a general quadratic equation ax2 + bx + c = 0

Case 1) Is c=0?

If your quadratic equation just has x2 and x values, then we can simply use x as a common factor.

Example 1: 6x2 + 4x = 0

x is a common factor, so we can factorise this equation.

x(6x + 4) = 0

Now, we know that if two numbers multiply to give zero, then one of the numbers has to be equal to zero.

This means that either x = 0 or 6x + 4 = 0, in which case 6x=-4 so x=-4/6 or -2/3

Solutions: x=0 or x=-2/3

Example 2: 3x2 - 7x = 0

x is a common factor, so we can factorise this equation.

x(3x - 7) = 0

This means that either x=0 or 3x - 7 = 0

Solutions: x=0 or x=7/3

Case 2) Is a=1?

In simpler cases, the value of a=1 and you have x2 + bx + c = 0

Find two integers which multiply together to make c, but add together to give b.

This gives you the solutions for your two brackets, and hence the answer to the equation.

Example 1: x2 + 6x + 5= 0

In this example, we only have one x2 so a=1.

We can see that the c=5 and the factors of 5 are only 1 and 5 (as it is prime).

We also know that 1+5=6 so we have found our two numbers to go inside the brackets.

We end up with (x+1)(x+6)=0

This means that either x+1=0 (so x=-1) or x+6=0 (so x=-6).

Solutions: x=-1 or -6

Example 2: x2 -2x - 8 = 0

We can see in this example that a=1, b=-2 and c=-8

The factors of -8 are:

• 1 and -8
• -1 and 8
• 2 and -4
• -2 and 4

Looking at these factors, the pair of factors that add up to give b (which is -2) are 2 and -4.

We end up with (x + 2)(x - 4) = 0

This means that either x+2=0 (so x=-2) or x-4=0 (so x=4)

Solutions: x=-2 or 4

Example 3: x2 -7x + 10 = 0

We can see in this example that a=1, b=-7 and c=10

As c is positive, but b is negative, there must be 2 negative factors involved.

The (negative) factors of 10 are:

• -1 and -10
• -2 and -5

Looking at these factors, the pair of factors that add up to give b (which is -7) are -2 and -5.

We end up with (x-2)(x-5)=0

Solutions: x=2 or 5

Case 3) a is an integer >1.

This is quite tricky, so it is best to go through it with a worked example!

Example 1:

Find the solution of 2x2 - x - 6 = 0

Step i) Multiply a and c together.

This gives us 2x(-6)=-12

Step ii) Write down pairs of factors of a x c

Pairs of factors of -12 are:

• 1 and -12
• -1 and 12
• 2 and -6
• -2 and 6
• 3 and -4
• -3 and 4

Step iii) Look for a pair of factors which add up to make b.

The only pair of factors that add up to make -1 is 3 and -4.

Step iv) Rewrite the equation using this information to split up the x values

2x2 -x -6 = 2x2 + 3x - 4x - 6

Step v) Now factorise this expression in 2 parts with common factors

2x2 + 3x - 4x - 6 = x(2x + 3) - 2(2x + 3)

This gives us: (x-2) (2x + 3) or (2x + 3)(x - 2)

Solutions x=-3/2 or 2

Example 2:

Let's try 3x2 - 10x + 8 = 0

i) Multiply a and c together. This gives 3 x 8 = 24

ii) Write down pairs of factors of 24.

As b is negative, we are looking for 2 negative factors.

• -1 and -24
• -2 and -12
• -3 and -8
• -4 and -6

iii) Look for pairs of factors that add up to make b

The only two factors that add up to make -10 are -4 and -6.

iv) Rewrite the equation

3x2 - 10x + 8 = 3x2 - 4x - 6x + 8

v) Factorise this expression with common factors

3x2 - 4x - 6x + 8 = x(3x - 4) - 2(3x - 4)

This gives us: (x - 2)(3x - 4) or (3x - 4)(x - 2)

Solution: x = 4/3 or 2

Example 3:

Find the solutions to 2x2 - 9x - 5 = 0

i) Multiply a and c together. This gives us 2 x -5 = -10

ii) Write down the pairs of factors of -10.

Factors of negative 10 are:

• 1 and -10
• -1 and 10
• 2 and -5
• -2 and 5

iii) Look for the pair of factors that add up to make b.

The only two factors that add up to make b (which is -9) are 1 and -10.

iv) Rewrite the equation

2x2 - 9x - 5 = 2x2 -10x + x - 5

v) Factorise this expression

2x2 -10x + x - 5 = 2x(x - 5) + (x - 5)

This gives us (2x + 1)(x - 5)

Solutions: x = -1/2 or x = 5

Checking a Solution

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