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In today’s Mathematics class, We will be learning about how to Factorize Quadratic Equations. We hope you enjoy the class!
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 factorized.
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 factorizing
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.
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
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)
EVALUATION
Factorise the following

 x^{2 }+ 4x – 5
 x^{2 }+ 6x – 7
 (a + 4)^{2}
 (3x + y)^{2}
 49m^{2 }– n^{2 }
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
We have come to the end of this term and hence JSS3. It’s been a remarkable journey and we are glad that you have made it this far. For making it this far, we commend you for being resilient, you have taken charge of your education and future.
We wish you success in your Junior WAEC exams and we hope to see you in SS1.
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