# Factorise: (i) 12x^{2} - 7x + 1 (ii) 2x^{2} + 7x + 3 (iii) 6x^{2} + 5x - 6 (iv) 3x^{2} - x - 4

**Solution:**

By the splitting of the middle term, we can find factors using the following method.

Find two numbers p, q such that,

p + q = co-efficient of x

p × q = product of the co-efficient of x^{2} and the constant term.

**(i)** 12x^{2} - 7x + 1

p + q = -7 (co-efficient of x)

p × q = 12 × 1 = 12 (co-efficient of x^{2} and the constant term)

By trial and error method, we get p = -4, q = -3

Now splitting the middle term of the given polynomial,

12x^{2} - 7x + 1 = 12x^{2} - 4x - 3x + 1

= 4x(3x - 1) - 1(3x - 1)

= (3x - 1) (4x - 1) [taking (3x - 1) as a common term]

**(ii)** 2x^{2} + 7x + 3

p + q = 7 (co-efficient of x)

p × q = 2 × 3 = 6 (product of co-efficient of x² and the constant term)

By trial and error method, we get p = 6, q = 1.

Now splitting the middle term of the given polynomial,

2x^{2} + 7x + 3 = 2x^{2} + 6x + x + 3

= 2x(x + 3) + 1(x + 3)

= (2x + 1) (x + 3)

**(iii)** 6x^{2} + 5x - 6

p + q = 5 (co-efficient of x)

p × q = 6 × (-6) = -36 (product of co-efficient of x^{2} and the constant term)

By trial and error method, we get p = 9, q = -4.

Now splitting the middle term of the given polynomial,

6x^{2} + 5x - 6 = 6x^{2} + 9x - 4x - 6

= 3x(2x + 3) - 2(2x + 3)

= (3x - 2) (2x + 3)

**(iv)** 3x^{2} - x - 4

p + q = -1 (co-efficient of x)

p × q = 3 × (-4) = -12 (product of co-efficient of x^{2} and the constant term.)

By trial and error method, we get p = -4, q = 3.

Now splitting the middle term of the given polynomial,

3x^{2} - x - 4 = 3x^{2} - 4x + 3x - 4

= 3x^{2} + 3x - 4x - 4

= 3x(x + 1) - 4(x + 1)

= (3x - 4) (x +1)

**☛ Check: **NCERT Solutions Maths Class 9 Chapter 2

**Video Solution:**

## Factorise: (i) 12x² - 7x + 1 (ii) 2x² + 7x + 3 (iii) 6x² + 5x - 6 (iv) 3x² - x - 4

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.4 Question 4:

**Summary:**

The factorized form of 12x^{2 }- 7x + 1, 2x^{2} + 7x + 3, 6x^{2} + 5x - 6, 3x^{2} - x - 4 are (3x - 1)(4x - 1), (2x + 1)(x + 3), (3x - 2)(2x + 3), (3x - 4)(x + 1)

**☛ Related Questions:**

- Determine which of the following polynomials has (x + 1) a factor:i) x3 + x2 + x + 1ii) x4 + x3 + x2 + x + 1iii) x4 + 3x3 + 3x2 + x +1iv) x3 - x2 - (2 + √2)x + √2
- Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:(i) p(x) = 2x3 + x2 - 2x - 1, g(x) = x +1(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2(iii) p(x) = x3 - 4x2 + x + 6, g(x) = x - 3
- Find the value of k, if x - 1 is a factor of p(x) in each of the following cases: (i) p(x) = x2 + x + k (ii) p(x) = 2x2 + kx + √2 (iii) p(x) = kx2 - √2x + 1 (iv) (x) = kx2 - 3x + k
- Factorise:(i) x3 - 2x2 - x + 2(ii) x3 - 3x2 - 9x - 5(iii) x3 + 13x2 + 32x + 20(iv) 2y3 + y2 - 2y - 1

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