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# Factorising the Quadratic Polynomial (Trinomial)

We have, a = kl , b = (lm +kn) and c = mn, where ac is the product of kl and mn that is, equal to the product of lm and kn which are the coefficient of x. Therefore (kl × mn) = (lm ×kn) .

Factorising the Quadratic Polynomial (Trinomial) of the type ax2 + bx +c, a 0

The linear factors of ax2 + bx +c will be in the form (kx + m) and (lx + n)

Thus, ax2 + bx +c = (kx +m )(lx +n) = klx2 +(lm + kn )x +mn

Comparing coefficients of x2, x and constant term c on both sides.

We have, a = kl , b = (lm +kn) and c = mn, where ac is the product of kl and mn that is, equal to the product of lm and kn which are the coefficient of x. Therefore (kl × mn) = (lm ×kn) .

## Steps to be followed to factorise ax2 + bx +c :

Step 1 : Multiply the coefficient of x2 and constant term, that is ac .

Step 2 : Split ac into two numbers whose sum and product is equal to b and ac respectively.

Step 3 : The terms are grouped into two pairs and factorise.

### Example 3.27

Factorise 2x2 + 15x + 27

Solution

Compare with ax2 + bx +c

we get, a = 2, b = 15, c = 27

product ac = 2 ×27 = 54 and sum b = 15

We find the pair 6, 9 only satisfies “b = 15” and also “ac = 54”. we split the middle term as 6x and 9x

2x 2 + 15x + 27 = 2x 2 + 6x + 9x + 27

= 2x (x + 3) + 9(x + 3)

= (x + 3)(2x + 9)

Therefore, (x + 3) and (2x + 9) are the factors of 2x 2 + 15x + 27 .

### Example 3.28

Factorise 2x2 15x + 27

### Solution

Compare with ax 2 + bx +c

a = 2, b = − 15, c = 27

product ac = 2×27 = 54, sum b=–15 we split the middle term as –6x and –9x

2x2 15x + 27 = 2x2 6x 9x + 27

= 2x(x3)9(x3)

= (x3)(2x9)

Therefore, (x − 3) and (2x − 9) are the factors of 2x2 15x + 27.

### Example 3.29

Factorise 2x2 + 15x 27

### Solution

Compare with ax 2 + bx +c

Here, a = 2, b = 15, c = −27

product ac = 2×–27 = –54, sum b=15 we split the middle term as 18x and –3x

2x2 + 15x 27 = 2x2 +18x 3x 27

= 2x(x + 9)3(x + 9)

= (x + 9)(2x3)

Therefore, (x + 9) and (2x 3) are the factors of 2x2 + 15x 27.

### Example 3.30

Factorise 2x2 − 15x -27

### Solution

Compare with ax 2 + bx +c

Here,a = 2, b = − 15, c = −27

product ac = 2×–27=–54, sum b=–15 we split the middle term as –18x and 3x

2x 2 − 15x -27 = 2x 218x + 3x27

= 2x (x 9) + 3(x 9)

= (x 9)(2x + 3)

Therefore, (x 9) and (2x + 3) are the factors of 2x2 − 15x -27

### Example 3.31

Factorise (x + y )2 + 9(x + y ) + 20

### Solution

Let x + y = p , we get p2 + 9p + 20

Compare with ax2 + bx +c,

We get a = 1, b = 9, c = 20

product ac = 1×20 = 20, sum b=9 we split the middle term as 4p and 5p

p2 + 9 p + 20 = p2 + 4p + 5 p + 20

= p(p + 4) + 5(p + 4)

= ( p + 4)(p + 5)

Put, p = x +y we get, (x + y )2 + 9(x + y ) + 20 = (x +y + 4)(x +y + 5)

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9th Maths : UNIT 3 : Algebra : Factorising the Quadratic Polynomial (Trinomial) | Steps, Example Solved Problems | Algebra | Maths