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FACTORISATION: COMMON FACTORS
FACTORISATION OF QUADRATIC EXPRESSIONS
SIMILARITY AND ENLARGEMENT
SIMULTANEOUS LINEAR EQUATION
TRIGONOMETRY – SINE AND COSINE OF ANGLES
AREA OF PLANE SHAPES
AREAS AND VOLUMES OF SIMILAR SHAPES
Topic: FACTORIZATION: COMMON FACTORS
a(b + c) = ab + ac
The reverse process, ab + ac = a(b + c), is called taking out the common factor.
Consider the factorisation of the expression 5x + 15.
Clearly, 5x = 5 X x
15 = 3 X 5
∴ HCF = 5
Thus 5x + 15 = 5 X x + 3 X 5
= 5(x + 3)
Note that the common factor 5 has been taken out and placed in front of the brackets. The expression inside the brackets is obtained by dividing each term by 5.
In general:
To factorise an algebraic expression, take out the highest common factor and place it in front of the brackets. Then the expression inside the brackets is obtained by dividing each term by the highest common factor.
Example
Factorise the following
a. 15x – 20
b. 3px + 12qx
c. x2 + x
d. 8x2y + 6xy2
Solution:
a. 15x – 20 {HCF = 5}
= 5(3x – 4)
b. 3px + 12qx {HCF = 3x}
= 3x(p + 4q)
c. x2 + x {HCF = x}
= x(x + 1)
d. 8x2y + 6xy2 {HCF = 2xy}
= 2xy(4x + 3y)
Note:
The process of taking out a common factor is of great importance in algebra. With practice you will be able to find the highest common factor (HCF) readily and hence factorise the given expression.
Example
Fcartorise the following:
a. -8x + 12
b. –x2 – x + ax
c. -6×2 + 4x – 2px
Solution:
a. -8x
-8x + 12 {HCF = 4}
= 4(-2x + 3)
It is customary to have the first term in the brackets positive. So, we take out -4 as the common factor rather than 4.
So, -8x + 12 = -4(2x – 3)
–x2 – x + ax {Take out –x as the common factor}
= -x(x + 1 – a)
C. -6×2 + 4x – 2px {Take out the common factor -2x}
= -2x(3x – 2 + p)
Simplifying calculations by factorization
Example
By factorising, simplify 79 X 37 + 21 X 37.
37 is a common factor of 79 X 37 and 21 X 37.
79 X 37 + 21 X 37 = 37(79 + 21)
= 37 X 100
= 3 700
Factorise the expression πr2 + 2πrh. Hence, find the value of πr2 + 2πrh when π = 22/7, r = 14 and h = 43.
π r2 +2πrh = πr(r + 2h)
when π = 22/7, r = 14 and h = 43,
πr2 + 2πrh = πr(r + 2h)
= 22/7 X 14(14 + 2 X 43)
= 22 X 2(14 + 86)
= 44 X 100
= 4400
Factorization by grouping
This lesson introduces the technique of factoring by grouping, as well as factoring the sum and difference of cubes. Factoring by grouping builds on the ideas that were presented in the section on factoring the common factor. While there are many different types of grouping, as you will learn in higher algebra courses, all of the grouping problems in this book involve four terms, and they work by grouping the first two terms and the second two terms together. If you do it right, a common factor will always emerge! Remember, the method of grouping is one of trial and error. As always, there is no substitute for practice and experience. Remember, if you can learn this topic now, it will help you later.
Example
Factorise x3 + 2×2 + 8x + 16
Solution: There are no common factors to all four terms. It is not a trinomial, and nothing discussed so far works to factor this. So, try grouping the first two terms together, and the last two terms together, and factor out the common factor within each grouping as follows:
(x3 + 2×2) + (8x + 16) = x2 (x + 2)+ 8(x + 2)
Notice that there is a common factor of (x+2) that can be factored out:
= (x + 2)( x2 + 8)
Example
Factorise xy − 4y + 3x − 12
Solution: Again, there are no common factors, and this is not a trinomial. Group the first two and the last two terms together, and factor out the common factor from each grouping:
(xy − 4y) + (3x − 12) = y(x − 4) + 3(x − 4)
Now, take out the common factor, which is (x − 4):
= (x − 4)(y + 3)
Example
Factorise xy − 4y − 3x + 12
Solution: Group the first two and the last two terms together, and factor out the common factor from each grouping:
(xy − 4y) + (−3x + 12) = y(x − 4) + 3(−x + 4)
This time there is no common factor. Try again, this time factoring a −3 from the last grouping. This works!
xy − 4y − 3x + 12 = y(x − 4) − 3(x − 4)
= (x − 4)(y − 3)
Example
Factorise xy − 4y + 3x + 12
Solution: Group the first two and the last two terms:
xy − 4y + 3x + 12 = y(x − 4) + 3(x + 4)
At this point, it is important to realize that no common factor resulted. Do not try to factor out something that is not common to both groupings. In fact, there is no way to group this problem to get a common factor. This one cannot be factored by grouping. In fact, it can not be factored by any method. Remember, not all problems can be factored. Remember that the entire process of grouping is one of trial and error, and, as you will see later, there are different types of grouping.
Assessment
Factorise the following: 4xy + 6zy, kx – 6k – 9x + 24
Factorise 12mn + 8kn and solve for k if m=5 and n=2
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 ) = x2 + 7x + 10
( x + 2 ) ( x + 3 ) = x2 + 5x + 6
( x – 3 ) ( x – 5 ) = x 2 – 8x + 15
( x + 6 ) ( x – 5 ) = x2 + x – 30
( x – 6 ) ( x + 5 ) = x2 – x – 30
Factorising
To factorise an expression such x2 + 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) = x2 + 5x + 6
Factorising expressions gets trickier with negative numbers.
Example
Factorise the expression: c2– 3c – 10
Write down the expression: c2– 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:
c2 – 3c – 10 = (c + 2)(c – 5)
Factorising the difference of two squares
Some quadratic expressions have only a term in x2 and a number such as x2 – 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:
x2 – 4 = (x + 2)(x – 2)
x2 – 81 = (x + 9)(x – 9)
x2 – 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: ax2 + 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: x2 – 9x + 20 = 0
Solution
First, factorise the quadratic equation x2– 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:
x2– 9x + 20
Substituting 4 gives:
42 – 9 × 4 + 20 = 16 – 36 + 20 = 0
Substituting 5 gives:
52 – 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 x2 + 6x as a square plus another term.
The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9.
x2 + 6x = (x2 + 6x + 9) – 9
= (x + 3)2 – 9
Example
We have seen in the previous example that x2 + 6x = (x + 3)2 – 9
So work out x2 + 6x – 2
x2 + 6x – 2 = ( x2 + 6x + 9 ) – 9 – 2 = (x + 3)2 – 11
Now try one for yourself.
Example
Solve x2 + 6x – 2 = 0
From the previous examples, we know that x2 + 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 2×2 + 20x + 3
Rewrite to get x2 on its own.
2( x2 + 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 2×2 + 20x + 3 = 0
From the previous example, we know that 2×2 + 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. x2 + 4x – 5
2. x2 + 6x – 7
3. (a + 4)2
4. (3x + y)2
5. 49m2 – n2
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 ) = x2 + 7x + 10
( x + 2 ) ( x + 3 ) = x2 + 5x + 6
( x – 3 ) ( x – 5 ) = x 2 – 8x + 15
( x + 6 ) ( x – 5 ) = x2 + x – 30
( x – 6 ) ( x + 5 ) = x2 – x – 30
Factorising
To factorise an expression such x2 + 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) = x2 + 5x + 6
Factorising expressions gets trickier with negative numbers.
Example
Factorise the expression: c2– 3c – 10
Write down the expression: c2– 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:
c2 – 3c – 10 = (c + 2)(c – 5)
Factorising the difference of two squares
Some quadratic expressions have only a term in x2 and a number such as x2 – 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:
x2 – 4 = (x + 2)(x – 2)
x2 – 81 = (x + 9)(x – 9)
x2 – 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: ax2 + 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: x2 – 9x + 20 = 0
Solution
First, factorise the quadratic equation x2– 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:
x2– 9x + 20
Substituting 4 gives:
42 – 9 × 4 + 20 = 16 – 36 + 20 = 0
Substituting 5 gives:
52 – 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 x2 + 6x as a square plus another term.
The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9.
x2 + 6x = (x2 + 6x + 9) – 9
= (x + 3)2 – 9
Example
We have seen in the previous example that x2 + 6x = (x + 3)2 – 9
So work out x2 + 6x – 2
x2 + 6x – 2 = ( x2 + 6x + 9 ) – 9 – 2 = (x + 3)2 – 11
Now try one for yourself.
Example
Solve x2 + 6x – 2 = 0
From the previous examples, we know that x2 + 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 2×2 + 20x + 3
Rewrite to get x2 on its own.
2( x2 + 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 2×2 + 20x + 3 = 0
From the previous example, we know that 2×2 + 20x + 3 can be rewritten as:
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