**JSS 3 Mathematics Second Term Week 1**

**Topic: FACTORISATION OF QUADRATIC EXPRESSIONS**

You can also factorise quadratic expressions. Remember that factorizing an expression simplifies it in some way. Factorizing is the reverse of expanding brackets.

**Factorizing Quadratic Expressions**

Multiply these brackets to remind yourself how to factorise.

( x + 2 ) ( x + 5 ) = x^{2} + 7x + 10

( x + 2 ) ( x + 3 ) = x^{2} + 5x + 6

( x – 3 ) ( x – 5 ) = x ^{2} – 8x + 15

( x + 6 ) ( x – 5 ) = x^{2} + x – 30

( x – 6 ) ( x + 5 ) = x^{2} – x – 30

**Factorising**

To factorise an expression such x^{2} + 5x + 6, you need to look for two numbers that add up to make 5 and multiply to give 6.

The factor pairs of 6 are:

1 and 6

2 and 3

2 and 3 add up to 5. So: (x +2) (x+3) = x^{2} + 5x + 6

Factorising expressions gets trickier with negative numbers.

**Example**

Factorise the expression: c^{2}– 3c – 10

Write down the expression: c^{2}– 3c – 10

Remember that to factorise an expression we need to look for common factor pairs. In this example we are looking for two numbers that:

multiply to give -10

add to give -3

Think of all the factor pairs of -10:

1 and -10

-1 and 10

2 and -5

-2 and 5

Which of these factor pairs can be added to get -3?

Only 2 + (-5) = -3

So the answer is:

c^{2} – 3c – 10 = (c + 2)(c – 5)

**Factorising the difference of two squares**

Some quadratic expressions have only a term in x^{2} and a number such as x^{2} – 25.

These quadratic expressions have no x term.

Using our method to factorise quadratics means we look for two numbers that multiply to make -25 and add to make 0.

The only factor pair that will work are 5 and -5. So:

(x + 5)(x – 5) = x² – 25

Not all quadratic expressions without an x term can be factorised.

**Examples**

Factorise:

x^{2} – 4 = (x + 2)(x – 2)

x^{2} – 81 = (x + 9)(x – 9)

x^{2} – 9 = (x + 3)(x – 3)

**Solving Quadratic Equations by Factorising**

To solve a quadratic equation, the first step is to write it in the form: ax^{2} + bx + c = 0. Then factorise the equation as you have revised in the previous section.

If we have two numbers, A and B, and we know that A × B = 0, then it must follow that either A = 0, or B = 0 (or both). When we multiply any number by 0, we get 0.

**Example**

Solve the equation: x^{2} – 9x + 20 = 0

### Solution

First, factorise the quadratic equation x^{2}– 9x + 20 = 0

Find two numbers which add up to 9 and multiply to give 20. These numbers are 4 and 5.

(x – 4) (x – 5) = 0

Now find the value x so that when these brackets are multiplied together the answer is 0.

This means either (x – 4) = 0 or (x – 5) = 0

So x = 4 or x = 5.

You can check these answers by substituting 4 and 5 in to the equation:

x^{2}– 9x + 20

Substituting 4 gives:

4^{2} – 9 × 4 + 20 = 16 – 36 + 20 = 0

Substituting 5 gives:

5^{2} – 9 × 5 + 20 = 25 – 45 + 20 = 0

Remember these 3 simple steps and you will be able to solve quadratic equations.

Now try this question.

**Completing the square**

This is another way to solve a quadratic equation if the equation will not factorise.

It is often convenient to write an algebraic expression as a square plus another term. The other term is found by dividing the coefficient of x by 2, and squaring it.

Any quadratic equation can be rearranged so that it can be solved in this way.

Have a look at this example.

**Example **

Rewrite x^{2} + 6x as a square plus another term.

The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9.

x^{2} + 6x = (x^{2} + 6x + 9) – 9

= (x + 3)^{2} – 9

**Example **

We have seen in the previous example that x^{2} + 6x = (x + 3)^{2} – 9

So work out x^{2} + 6x – 2

x^{2} + 6x – 2 = ( x^{2} + 6x + 9 ) – 9 – 2 = (x + 3)^{2} – 11

Now try one for yourself.

**Example**

Solve x^{2} + 6x – 2 = 0

From the previous examples, we know that x^{2} + 6x – 2 = 0 can be written as (x + 3)^{2} – 11 = 0

So, to solve the equation, take the square root of both sides. So (x + 3)^{2} = 11

x + 3 = + √11

or x + 3 = – √11

x = – 3 + √11

or x = – 3 – √11

x = – 3 + 3.317 or x = – 3 – 3.317 (√11 is 3.317)

**x = 0.317** (3 s.f) or **x = – 6.317** (3 s.f)

**Example **

Rewrite 2x^{2} + 20x + 3

Rewrite to get x^{2} on its own.

2( x^{2} + 10x ) + 3

The coefficient of x is 10. Divide 10 by 2, and square to get 25.

= 2 ( ( x + 5)^{2} – 25) + 3

= 2 (x + 5)^{2} – 50 + 3

= 2 (x + 5)^{2} – 47

Now use the previous example to solve 2x^{2} + 20x + 3 = 0

From the previous example, we know that 2x^{2} + 20x + 3 can be rewritten as:

2 (x + 5)^{2} – 47

Therefore, we can rewrite the equation as:

2(x + 5 )^{2} – 47 = 0

2(x + 5 )^{2} = 47

(x + 5 )^{2} = 23.5 (dividing both sides by 2)

Take the square root of both sides.

x + 5 = √23.5

or x + 5 = – √23.5

x = – 5 + √23.5

or x = – 5 – √23.5

x = – 5 + √23.5

or x = – 5 – √23.5

**x = – 0.152** (3 s.f) or **x = – 9.85** (3 s.f)

**Exercise**

1. x^{2 }+ 4x – 5

2. x^{2 }+ 6x – 7

3. (a + 4)^{2}

4. (3x + y)^{2 }

5. 49m^{2 }– n^{2 }

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