Question

factorise: i) 12x²- 7x + 1 ii) 2x²+ 7x + 3 iii) 6x² + 5x-6 iv) 3x²- x- 4​

Answer 1

Answer:

(i) 12x2 - 7x + 1

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

p × q = 12 × 1 = 12 (co-efficient of x2 and the constant term)

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

Now splitting the middle term of the given polynomial,

12x2 - 7x + 1 = 12x2 - 4x - 3x + 1

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

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

(ii) 2x2 + 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,

2x2 + 7x + 3 = 2x2 + 6x + x + 3

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

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

(iii) 6x2 + 5x - 6

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

p × q = 6 × (-6) = -36 (product of co-efficient of x2 and the constant term)

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

Now splitting the middle term of the given polynomial,

6x2 + 5x - 6 = 6x2 + 9x - 4x - 6

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

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

(iv) 3x2 - x - 4

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

p × q = 3 × (-4) = -12 (product of co-efficient of x2 and the constant term.)

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

Now splitting the middle term of the given polynomial,

3x2 - x - 4 = 3x2 - 4x + 3x - 4

= 3x2 + 3x - 4x - 4

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

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