2.6 Factorisation of Trinomials:
This concept is important when we need to simplify
algebraic expressions.
Do you know that 5-(3a2-2a)( 6-3a2+2a) = (3a+1)(a-1) (3a-5)(a+1) ? check the
correctness by substituting a=-1 and a=1. How do we prove that this equation holds
good for all values of a?
We have seen earlier that HCF is useful in
simplifying the algebraic expressions by taking this highest common factor out
side of of algebraic expressions.
The HCF of 4x2y, 8x3and 12xy is 4x
4x2y+8x3+12xy
can be written as 4x (xy+2x2+3y)
The process of writing an algebraic expression as the
product of two or more expressions (called factors)
is called ‘factorisation’.
How do we factorize
trinomials of type x2+mx +c?
Example : Let us take the expression x2+x(a+b)+ab
x2+x(a+b)+ab
= (x2+xa)+(xb+ab) ( By rearranging
the terms)
= x(x+a)+b(x+a) ( x is common factor
of x2 and xa and b is common factor of xb
and ab)
= (x+a)(x+b)
We say x+a and x+b are factors of the expression x2+x(a+b)+ab
In other terms we say x2+x(a+b)+ab
can be split in to product of x+a and x+b.
Example :
x2+5x+6 can be rewritten as
=x2+3x+2x+6
=x(x+3)+2(x+3)
=(x+3)*(x+2)
Thus x+3 and x+2 are factors of x2+5x+6.
They are of the form x+a and x+b.
These factors x+a and x+b
of x2+5x+6 are such that a+b= 5 and ab=6. By trial and error method we find that a=3 and b=2 satisfy
the condition a+b=5 and ab=6.
This is the reason why we split 5x as 3x+2x and not
as x+4x or anything else in the above example.
It is to be noted that not all trinomials of type x2+mx
+c always have factors.
In later sections you will learn the formula to
find the factors for algebraic expression of type x2+mx +c.
x2+5x+6 is of the form x2+mx
+c with
m = 5 and c=6
2.6 Problem 1: Factorise x2+27x+176
Solution:
We need to find a and b
such that a+b=27 and ab=176
The pairs of factors of 176 are (2, 88), (4, 44),
(8, 22), (16, 11)
The –ve pairs of factors (Ex (-2, -88)) are
neglected as their sum cant be a +ve number.
Of these pairs, we notice that the pair (16, 11)
satisfies the desired condition with a= 16 and b=11
So x2+27x+176 can be rewritten as
x2+16x+11x+ 176
=x(x+16) +11(x+16)
=(x+16) (x+11)
Thus (x+16) and (x+11)
are factors of x2+27x+176
Verification:
(x+16)(x+11) is of the form (x+a)*(x+b) with a=16 and b=11
(x+16)*(x+11) = x2+ x(16+11)+
16*11= x2+27x+176 which is the given algebraic expression
2.6 Problem 2 : Factorise x2-6x-135
Solution:
We need to find a and b
such that a+b= -6 and ab=
-135
The pairs of factors of -135 are (3,-45), (-3,
+45), (5,-27), (-5, +27), (9,-15), (-9, +15)
Of these pairs we notice that 9-15 = -6 and 9*-15 =
-135 satisfy the desired condition with a= 9 and b= -15
x2-15x+9x
-135
=x(x-15)+9(x-15)
=(x-15)(x+9)
Thus (x-15) and (x+9) are factors of x2-6x-135
Verification:
(x-15)(x+9) is of the form
(x+a)*(x+b) with a=-15, b=9
(x-15)*(x+9) = x2+ x(-15+9)+ (-15*9)= x2-6x-135 which is the given
algebraic expression
2.6 Problem 3: Factorise m2+4m-96
Solution:
We need to find a and b
such that a+b= 4 and ab= -96
The pairs of factors of -96 are (2,-48), (-2, 48),
(3,-32), (-3, +32), (4,-24), (-4, +24), (6,-16), (-6,16),
(8,-12), (-8,12)
Of these pairs we notice that -8+12 = 4 and -8*12 =
-96 satisfy the desired condition with a= -8 and b=12
m2-8m+12m -96
=m(m-8)+12(m-8)
=(m-8)(m+12)
Thus (m-8) and (m+12) are factors of m2+4m-96
Verification:
(m-8)(m+12) is of the form
(m+a)*(m+b) with a=-8, b=12
(m-8)*(m+12) = m2+ m(-8+12)+ -8*12= m2+4m-96 which is the given
algebraic expression
Let us try to factorize the expression of type px2+mx
+c (note the co-efficient of x2, is p and not 1)
We need to find a and b such that a+b=m and ab=pc
2.6 Problem 4 : Factorize 24x2-65x+21
Solution:
We need to find a and b
such that a+b= -65 and ab=
24*21 =504
The pairs of factors of 24*21 are(2,252), (-2,-252), (3, 138 ), (-3,-138),
(4,126), (-4,-126), (6,83), (-6,-83),
(8,63), (-8,-63), (9,56), (-9,-56), (12,42), (-12,-42)
Of
these pairs we notice that -9-56 =
-65 and -9*(-56) = 504=24*21 satisfies the desired condition with a= -9 and b= -56
24x2-65x+21
=24x2-9x -56x+21 ( -65x
is rewritten as -9x -56x)
=3x(8x-3) -7(8x-3) (3x is
common factor of 24x2 and 9x. -7 is common factor of -56x and 21)
= (8x-3)(3x-7) ( 8x-3 is
common factor )
Thus 8x-3 and 3x-7 are factors of 24x2-65x+21
Verification:
(8x-3)(3x-7)
=8x(3x-7)-3(3x-7) (
Multiply each of the terms)
=24x2-56x
-9x+21 (simplification)
=24x2-65x+21 which is the term given in
the problem
2.6 Problem 5: Factorize 6p2+11pq -10q2
Solution:
We need to find a and b
such that a+b= 11 and ab=
6*(-10) =-60
The pairs of factors of -60
are(2,-30), (-2,30),(3, -20 ),(-3,20) (4,-15), (-4,15),
(5,-12),(-5,12),(6,-10),(-6,10)
Of
these pairs we notice that -4+15 =
11 and -4*15 = -60
which satisfies our need of
finding a and b
6p2+11pq -10q2
=6p2+15pq -4pq-10q2(
11pq = 15pq-4pq)
=3p(2p+5q) -2q(2p+5q)
=(2p+5q)(3p-2q)
Thus 2p+5q and 3p-2q are factors of 6p2+11pq -10q2
Verification:
(2p+5q)(3p-2q)
=2p(3p-2q)+5q(3p-2q) (
Multiply each of the terms)
=6p2-4pq +15qp-10q2 (simplification)
= 6p2+11pq -10q2 which is the
term given in the problem
2.6 Problem 6: Factorize 5-(3a2-2a) (6-3a2+2a)
Solution:
For easy working let x =3a2-2a
Thus we need to factorize 5-x(
6-x)
5-x( 6-x)
= 5 -6x + x2
= x2 -6x +5 = x2 -5x -x+5
= x(x-5)-1(x-5)
= (x-1)(x-5)
By substituting value for x we get
5-(3a2-2a)(
6-3a2+2a)
= (3a2-2a -1)
(3a2-2a-5)
But 3a2-2a -1 = 3a2-3a+a -1 =
3a(a-1)+1(a-1) = (3a+1)(a-1)
3a2-2a-5 = 3a2+3a -5a-5 = 3a(a-1)-5(a+1) = (3a-5)(a+1)
5-(3a2-2a)( 6-3a2+2a) = (3a+1)(a-1) (3a-5)(a+1)
Verification:
1.
Multiply each of the individual terms to expand to check
the correctness.
2.
Check to be sure
that the
answer is correct at least for
value of a=2 : (5 -8*-2) = 21 =
(7*1*1*3)
2.6 Summary of learning
|
No |
Points studied |
|
1 |
Finding
factors of x2+mx +c such
that a+b=m and ab=c where x+a and x+b are its factors |
|
2 |
Split
px2+mx +c such that a+b =m and ab=pc |