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Factorising
quadratic expressions
Clement Radcliffe,Contributor
As
we continue to review algebra, I wish
to remind you of the following:
- The
concepts included in algebra are
fairly routine and with effort you
all should be able to do them well.
- Many
areas were done in the lower forms
and must be effectively revised.
- Algebra
should be selected as one of the
compulsory topics in section 2.
We
will now review last week's homework.
1.
Solve : x⁄4
+ 16 = 2x
Solution
The
appropriate method is to multiply
both sides by 4.
4
x x⁄4
+ 4 x 16 = 4 x 2x
x
+ 64 = 8x
7x
= 64
x
= 64⁄7
Note
You
may also simplify the left-hand side
prior to equating both sides. You
may wish to try this approach on your
own.
2.
Solve : ( 2x - 3 )/2 - ( x + 4 )/4
= 1
Solution
In
this case, the method recommended
above may also be used.
As
the LCM of 2 and 4 is 4, simplify
the left-hand side :
( 2x - 3 )/2 - ( x + 4 )/4
(
2(2x - 3) - (x + 4) )/4 = ( 4x - 6
- x - 4 )/4 = ( 3x - 10 )/4
Equating
both sides:
(
3x - 10 )/4 = 1
3x
- 10 = 4
3x
= 14
or
x
= 14⁄3
3.
Factorise: (a) 7x2 - 21x
Solution
As
the common factor is 7x
7x2
- 21x = 7x (x - 3)
(b)
axy - a2y
Solution
axy
- a2y = ay(x - a)
4.
Factorise: 3x - 8y - 4xy + 6
Solution
3x
- 8y - 4xy + 6 Re-organising
3x
+ 6 - 8y - 4xy
Using
grouping, that is, repeated common
factor method:
3x
+ 6 - 8y - 4xy = 3(x + 2) - 4y(2 +
x)
=
(x + 2)(3- 4y)
Continuing
the review of factorisation, we will
proceed with factorising quadatic
expressions.
EXAMPLES
1.
Factorise x2 + 8x + 15
This
method is based on the principle that
(x + b)(x + c) = x? + (b + c) x +
bc. Do you see the relationship between
(b + c) which is the coefficient of
x, bc which is the constant term,
and b and c which are the values in
the brackets on the left-hand side?
This relationship and the 'trial and
error' plays an important role in
this method.
Using
the above:
x2
+ 8x + 15 = (x + 5)(x +3)
If
you have not realised the relationship
mentioned above, then please note
that:
- 5
+ 3 = 8 (coefficient of x)
- 5
x 3 = 15 (The constant term)
You
may use 'trial and error' to identify
5 and 3, the values which satisfy
the relationship.
2.
Factorise: 2x2 + 5x -12
Despite
the coefficient of x? being 2, a method
similar to that of example 5 above
is used.
2x2
+ 5x - 12 = (2x - 3)(x + 4)
EXAMPLES
OF METHOD OF DIFFERENCE OF TWO SQUARES
3.
Factorise: 25 - x2
This
is based on the fact that a? - b?
= (a - b)(a + b). The critical problem
is, therefore, to find the square
root of each term.
As
square root of 25 = 5 and square root
of x2 = x
25
- x2 = (5 - x)(5 + x).
We
will try another example.
4.
Factorise: 9x2 - 16
By
using the method of difference of
two squares you can show that since
square
root of 9x2 = 3x and square
root of 16 = 4, then
9x2
- 16 = (3x - 4)(3x + 4).
Remember
to check your answers by expanding
the factors.
Now,
please attempt the following for homework.
Factorise:
(a)
x2 + 7x + 12
(b)
x2 - 4x - 21
(c)
3x2 - 7x -6
(d)
2x2 + 5x -12
(e)
3x -8y - 4xy + 6
(f)
x2 - y2 - 4x
+ 4y
(g)
9a2 - b2
(h)
36x2 -1
Clement
Radcliffe is an independent contributor.
Send questions and comments to kerry-ann.hepburn@gleanerjm.com
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