Factorisation
THIS WEEK, we will complete the review of Factorisation by going through the homework from last week’s lesson. Please be reminded that: Factorisation converts an algebraic expression to the product of the factors.
It is by studying the various methods that you will be able to select the appropriate one.
The CXC mathematics reports indicate that most difficulty is experienced in the grouping and quadratic factors methods.
More time should be spent preparing these. Be guided by this, and be sure to express the factors in the simplest form, for example, 2(x + 3) instead of (2x + 6).
The following are the solutions:
(a) Factorise: x² + 9x + 14
In this case, we find two numbers, a and b, such that a + b = 9 and a x b = 14.
The numbers are 2 and 7. ■ x² + 9x + 14 = (x + 2) (x + 7).
(b) Factorise : x² – 6x – 16 = Using a similar method.
x² – 6x – 16 = (x – 8)( x + 2)
(c) Factorise: 3x² – 7x – 6
Given 3x² – 7x – 6 and using the method similar to that used above
3x² – 7x – 6 = (3x + 2) (x – 3).
Please clear the brackets to verify your answer as follows,
(3x + 2) (x – 3) = 3x2 + 2x – 9x – 6 = 3x² – 7x –6.
(d) Factorise: 3x –8y – 4xy + 6
Using the grouping method:
The terms are rearranged such that both terms have the same common factor.
3x + 6 –8y – 4xy = (3x + 6) – (8y + 4xy)
3(x + 2) – 4y(2 + x) NB: ( x + 2 ) is common to both.
= (x + 2) (3 – 4y)
You may expand the answer to ensure that the answer is the same as the given expression. Indeed, this is required if time allows.
(e) Factorise: 25a² – 9 b²
The difference of two squares method is used.
As a reminder, we must find the square root of 25a2 and 9b2.
Since the square roots are 5a and 3b, respectively, then: 25a² – 9b² = (5a – 3b)(5a + 3b)
I am sure that you realise that (5a + 3b)(5a – 3b) is
also correct.
Finding square root provides a challenge for some. Please practise and check your results.
(f) Factorise: 3x² – 3y² – 4x + 4y
Using the grouping method:
3x² – 3y² – 4x + 4y = (3x² – 3y²) – (4x – 4y)
Factorising:
= 3(x² – y²) – 4(x – y)
= 3(x – y)(x + y) – 4(x – y) = 3(x + y) (x – y) – 4(x – y) = ( x – y) ( 3x + 3y – 4)
Answer: (x – y) (3x + 3y – 4)
(g) Factorise: 1 – 64x2
By using the difference of two squares method, the square roots of 1 and 64x2 are 1 and 8x, respectively, 1 – 64x2 = ( 1 – 8x)( 1 + 8x)
= ( 1 – 8x)( 1 + 8x)
On your own, please factorize:
1. x2 + 4x – 32
2. 2x2 + 5x – 12
3. 64t2 – 9
4. 4a2 – 16
5. 4x2 – 12xy + 9y2
6. pq – 2p – 3q + 6
You must ensure that you are familiar with the four methods – common factor, grouping, quadratic factors and difference of two squares – demonstrated above and know when to use each. You cannot afford to not earn the marks available from these questions.
Now let us review another topic, Simultaneous Linear Equations. The specific objective was presented previously. SIMULTANEOUS LINEAR EQUATIONS
■ The solution of the simultaneous equations is the pair of x and y values which satisfy both equations.
■ If both equations are plotted on a graph, it is the point of intersection of both lines.
■ You may use the elimination or substitution method. Both methods should be studied. You should also know the appropriate circumstances to use each.
■ It would be appropriate to review directed numbers and the simplification of algebraic expressions at the outset.
■ The method using matrices will be presented in future lessons.
The elimination method is illustrated as follows: EXAMPLE 1
Solve the simultaneous equations:
1. 2x + y = 3
2. 5x – 2y = 12 Multiply equation (1) by 2.
3. 4x + 2y = 6
Add equations (2) and (3)
This is done as the coefficients of y are plus and minus 2. 9x = 18 y is eliminated.
■ x = 18 = 2
9
Substituting x = 2 into (1)
2 x 2+ y = 3
■ 4 + y = 3
■ y = 3 – 4 = –1
■ y = –1
Answer is x = 2, y = – 1
You may check your answer by substituting the values x = 2 and y = – 1 into both equations to show that they satisfy the equations.
Do you realise that in EXAMPLE 1, since the coefficients of y are – 2 and 2, respectively, in both equations, you eliminate y by adding? If the coefficients are the same, that is, if the coefficients of y are both 2, then you eliminate by subtracting. EXAMPLE 2
Solve the simultaneous equations:
1. 5x – 2y = 12
2. 2x + y = 3
Multiply equation (2) by 2 to form equation (3) and then subtract equation (1) from equation (3).
3. 4x + 2y = 6
5x – 2y = 12.
(Both equations have the same coefficient of y)
■ 9 x = 18
■ x = 2
Substituting x = 2 in (2)
■ 4 + y= 3
■ y = 3 – 4 = – 1 Answer: x = 2 and y = – 1.
The following is an example of the substitution method: EXAMPLE 3
Let us try another example.
1. 3x – 2y = 10
2. 2x + 5y = 13
The elimination method is the appropriate one to be used here.
Multiply equation (1) x 2 to form equation (3)
3. 6x – 4y = 20 And multiply equation (2) x 3 to form equation (4) CONTINUED ON PAGE 24