# how do you do factorise quadratic equations ?

## 2 Answers

- 2 votes

Take the equation

x(2) + 7x + 12 = 0 (2) squared

You need to find two numbers that add to make 7 and multiply to make twelve.

3 and 4. 3 * 4 = 12, 3 + 4 =7.

(x + 3)(x + 4) = 0

To solve it you then switch the signs so: x = -3 and x = -4

Substitute them into the equation to check that they work:

-3(2) + 7*-3 + 12 = 0

-4(2) + 7*-4 + 12 = 0

They both work :)

Try another:

x(2) + 9x + 20 = 0

4 + 5 = 9, 4 * 5 = 20

(x + 4)(x + 5) = 0

switch the signs: x = -4, x = -5

If you are unable to find numbers that work however you have to do the quadratic equation:

where ax(2) + bx + c =0 you just sub in the numbers to the equation below.

Hope I have been of assistance to you :)

- 0 votes

If there is a coefficient in front of the x^{2 }: e.g. **2x ^{2} + 7x + 3**

You multiply the number in front of the x^{2} and the constant at the end: **2 x 3 = 6**

You then need to find two numbers that **multiply to make 6**, and **add to make 7** *(the number in front of x)*.

**1 x 6 = 6** , **1 + 6 = 7**

Instead of writing **2x ^{2} + 7x + 3 ** , you write

**2x**

^{2}+ 1x + 6x + 3(Exchanging the **7x** for **1x + 6x)**

**2x ^{2} + 1x + 6x + 3**

See what is common to the first two terms **2x ^{2 }**

**+ 1x**....... it's x

Put **x** outside the bracket: **x(2x + 1)** x goes into **2x**** ^{2 }**, 2x times & x goes into 1x, 1 time

Do the same for the other two terms **6x + 3 ** : 3 is common to both terms

Put** 3** outside the bracket: **3(2x+1)**

Add **x(2x + 1) + 3(2x + 1) ** (note that the two brackets should work out to be the same)

Take the values outside the brackets **(x + 3) ** and multiply by the value inside the brackets **(2x + 1)**

Final answer = **(x + 3)(2x + 1)**