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 x^2+mx +c such that a+b=m and ab=c where x+a and x+b are its factors
2	Split px^2+mx +c such that a+b =m and ab=pc