Download Revision Algebra II - Warilla High School

Revision Algebra II
Factorisation of Binomials,
Trinomials, Sum & Difference of
Two Cubics
By I Porter
Factors
A factor of a term or number divides into that term or number, without a remainder.
For examples, the factors of
• 18 are 1, 2, 3, 6, 9 and 18.
• 6x2 are 1, 2, 3, 6, x, 2x, 3x, 6x, x2, 2x2, 3x2 and 6x2.
• 4xy are 1, 2, 4, x, y, 2x, 2y, 4x, 4y, xy, 2xy and 4xy.
• The highest common factor (HCF) of 4xy and 6x2 is 2x.
The product of a(b + c) is ab + ac and the product of (a + b)(c + d) is
ac + ad + bc + bd, we say, these two results are written in expanded form.
When they are written in the form a(b + c) and (a + b)(c + d), we say they are in
factored form.
Factorisation is the process whereby an expression in expanded form, such as ab + ac,
is changed to factored form, such as a(b + c).
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Factorising by Grouping
Algebraic expressions with 4 terms do not have a factor common to every term.
Such expressions can often be factorised by first grouping the terms in pairs.
ax+ bx + ay +by = x(a + b) + y(a+ b) = (a + b)(x + y)
Examples: Fully factorise the following
a) a(b - 3) + 6(b - 3)
b) ac + ad + bc + bd
c) 4x2 - 4xy + x - y
= a(b - 3) + 6(b - 3)
= ac + ad + bc + bd
= 4x2 - 4xy + x - y
= (b - 3) (a + 6)
= a(c + d) + b(c + d)
= 4x(x - y) + 1(x - y)
= (c + d) (a + b)
= (x - y) (4x + 1)
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More Examples:
[negative signs as a factor]
d) mn + 4n - m - 4
e) 6x2 - 4x - 3xy + 2y
= mn + 4n - (m + 4)
= 2 x 3x2 - 2 x 2x - 3xy + 2y
= n(m + 4) - 1(m + 4)
= 2x(3x - 2) - y(3x - 2)
= (m + 4) (n - 1)
= (3x - 2)(2x - y)
Exercise: Fully factorise the following
a) ab + 5a + 2b + 10
= (a + 2)(b + 5)
b) xy - 3x + 2y - 6
= (x + 2)(y - 3)
c) 3c2 - 2cd + 3cd2 - 2d3
= (c + d2)(3c - 2d)
d) ab + 2b - 3a - 6
e) 2x2 - 6x - xy + 3y
f) 4c2 + 12c - cd - 3d
= (2x - y)(x - 3)
= (4c - d)(c + 3)
= (a + 2)(b - 3)
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Factorising the difference of 2 squares: a2 - b2 = (a - b)(a + b)
Examples: Factorise.
a) n2 - p2
= (n - p) (n + p)
d) 4x2 - y2
b) x2 - 9
c) 9x2 - 25
= (x)2 - (3)2
= (3x)2 - (5)2
= (x - 3) (x + 3)
= (3x - 5) (3x + 5)
e) 5x2 - 45
f) 32x2 - 18y2
= (2x)2 - (y)2
= 5(x2 - 9)
= 2(16x2 - 9y2)
= (2x - y) (2x + y)
= 5[(x)2 - (3)2]
= 2[(4x)2 - (3y)2]
= 5(x - 3) (x + 3)
= 2(4x - 3y) (4x + 3y)
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Exercise: Factorise
a) x2 - 36
= (x - 6)(x + 6)
b) 4x2 - 49
= (2x - 7)(2x + 7)
d) 8x2 - 50
e) 18x2 - 72y2
= 2(2x - 5)(2x + 5)
= 18(x - 2y)(x + 2y)
Factorisation of CUBICS: a3 - b3 and a3 + b3
a3 - b3 = (a - b)(a2 + ab + b2)
c) 9x2 - 25y2
= (3x - 5y)(3x + 5y)
Must know for HSC
a3 + b3 = (a + b)(a2 - ab + b2)
Examples: Factorise
a) x3 - 8 = x3 - 23
b) 8x3 + 27 = (2x)3 + (3)3
= (x - 2)(x2 + 2x + 22)
= (2x + 3)(4x2 - 6x + 32)
= (x - 2)(x2 + 2x + 4)
= (2x + 3)(4x2 - 6x + 9)
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Exercise: Factorise
a) x3 - 125
= (x - 5)(x2 + 5x + 25)
b) 8x3 + y3
= (2x + y)(4x2 - 2xy + y2)
c) 9x3 - 72y3
= 9(x - 2y)(x2 + 2xy + 2y2)
Factoring quadratic trinomials: ax2 ± bx ± c
• A trinomial has 3 terms.
• A quadratic trinomial is an expression with 3 terms where the highest power of
a term is 2.
Sum and Product Method.
To factorise a trinomial of the form x2 ± bx ± c (where b and c are non-zero
integers) look for two numbers P and Q such that:
* Their sum is b : (P + Q = b)
* Their product is c : (PQ = c)
[ The first step to start may be to write out the factors of c in a systematic process.]
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Examples: Factorise the following:
a) x2 + 7x +12
c) x2 + 2x - 24
Sum = +7
Product = +12
Factors of +12
12 = 1 x 12
=2x6
=3x4
Factors of -24
-24 = -1 x 24
= -2 x 12
= -3 x 8
= -4 x 6
Factor sums
1 + 12 = +13
2 + 6 = +8
3 + 4 = +7
Why not the (-) factors?
Sum and Product are both (+)
So, x2 + 7x +12 = (x + 3)(x + 4)
b) x2 - 8x +12
Sum = -8
Product = +12
Factors of +12
12 = -1 x -12
= -2 x -6
= -3 x -4
Why not the (+) factors?
Sum (-) and Product (+) both are (-)
So, x2 - 8x +12 = (x - 2)(x - 6)
Factor sums
-1 + 24 = +23
-2 + 12 = +10
-3 + 8 = +5
-4 + 6 = +2
Why do the factors have different signs?
Sum (+) and Product (-), largest number is (+)
So, x2 + 2x - 24 = (x - 4)(x + 6)
d) x2 - 3x - 40
Factor sums
-1 + -12 = -13
-2 + -6 = -8
-3 + -4 = -7
Sum = +2
Product = -24
Sum = -3
Product = -40
Factors of -40
-40 = 1 x -40
= 2 x -20
= 4 x -10
= 5 x -8
Factor sums
1 + -40 = -39
2 + -20 = -18
4 + -10 = -6
5 + -8 = -3
Why do the factors have different signs?
Sum (-) and Product (-), largest number is (-)
So, x2 - 3x - 40 = (x + 5)(x - 8)
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Exercise: Factorise
a) x2 - 9x + 20
b) x2 + 10x + 24
= (x - 4)(x - 5)
c) x2 - 8x - 48
= (x + 4)(x + 6)
d) x2 + 2x - 48
= (x + 8)(x - 6)
= (x + 4)(x - 12)
e) x2 - 15x + 54
e) x2 + 12x - 64
= (x - 9)(x - 6)
= (x + 16)(x - 4)
Examples: Factorise
a) 2x2 - 10x - 28
= 2(x2 - 5x - 14)
b) 4x2 - 20x + 24
Sum = -5
Product = -14
Factors of -14
-14 = 1 x -14
= 2 x -7
= 4(x2 - 5x + 6)
Factor sums
1 + -14 = -13
2 + -7 = -5
So, 2x2 - 10x - 28 = 2(x + 2)(x - 7)
Sum = -5
Product = +6
Factors of +6
+6 = -1 x -6
= -2 x -3
Factor sums
-1 + -6 = -7
-2 + -3 = -5
So, 4x2 - 20x - 28 = 4(x - 2)(x - 3)
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Exercise: Factorise
a) 7x2 - 63x + 140
b) 12x2 - 12x - 72
= 7(x - 4)(x - 5)
c) 9x2 + 36x + 27
= 12(x + 2)(x - 3)
Harder factorisation of quadratic trinomials: ax2 + bx + c
= 9(x + 1)(x + 3)
[ Variation of sum + product method]
Examples: Factorise
a) 3x2 - 2x - 8
3x2 - 2x - 8
Step 1: multipy first and last
term. 3 x -8 = -24
Step 2: Look for two numbers
that have a sum of -2 and product
of -24. (I.e. -6 and +4)
Step 3: Split the middle term
using the number found in step 2
3x2 - 6x + 4x - 8
Step 4: Factorise.
= 3x2 - 6x + 4x - 8
b) 6x2 + 13x - 5
Step 1: multipy first and last
term. 6 x -5 = -30
Step 2: Look for two numbers
that have a sum of +13 and
product of -30. (I.e. -2 and +15)
Step 3: Split the middle term
using the number found in step 2
6x2 - 2x + 15x - 5
Step 4: Factorise.
6x2 + 13x - 5 = 6x2 - 2x + 15x - 5
= 3x(x - 2) + 4(x - 2)
= 2x(3x - 1) + 5(3x - 1)
= (x - 2) (3x + 4)
= (3x - 1) (2x + 5)
If you can use the CROSS method successfully, KEEP USING IT!
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Exercise: Factorise
a) 2x2 + 5x + 2
(2x + 1)(x + 2)
b) 5x2 - 17x - 12
(5x + 3)(x - 4)
c) 6x2 - 7x + 1
(6x - 1)(x - 1)
d) 4x2 + 7x - 2
(4x - 1)(x + 2)
e) 6x2 - 7x - 20
(2x - 5)(3x + 4)
f) 8 + 10x - 3x2
(2 + 3x)(4 - x)
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