Polynomials
Exercise 2.4 Part 4
Question 4: Factorise:
(i) `12x^2 - 7x + 1`
Answer: By Splitting the middle term method:
Step – 1 – Multiply coefficient of x2 i.e. 12 and constant term 1
By multiplication of 12 and 1 we get 12
Step – 2 – Find out the probable prime factors of 12
Prime factors of 12 are (2 and 6) and (3 and 4)
Step 3 – Now take that prime factor, sum of which becomes equal to middle term of the given polynomial, i.e. coefficient of x which is equal to – 7
Here, 2 + 6 ≠ 7
3 + 4 = 7
So, we will consider the factor (3 and 4) of 12
Now, arrange the given polynomial as given below:
Given, `12x^2 - 7x + 1`
`= 12x^2 - (3x + 4x) + 1`
(Because `3x + 4x = 7x`)
`= 12x^2 - 3x – 4x + 1`
By taking 3x and -1 as common, we get:
`= 3x(4x – 1) – 1(4x – 1)`
By taking (4x – 1) as common, we get:
`= (3x – 1)(4x – 1)`
Factors of given polynomial `= (3x – 1)` and `(4x – 1)`
(ii) `2x^2 + 7x + 3`
Answer: Given, `2x^2 + 7x + 3`
Coefficient of `x^2 = 2`
Constant term = 2
Thus, ab = 6
Prime factors of 6 = 1 and 6
Now, given polynomial can be written as follows:
`2x^2 + x + 6x + 3`
`= x(2x + 1) + 2(2x + 1)`
`= (2x + 1)(x + 3)`
Thus, factor of given polynomial `= (2x + 1)(x + 3)`
(iii) `6x^2 + 5x – 6`
Answer: Given, `6x^2 + 5x – 6`
Here, a = 6 and b = 6
Therefore, ab = 36
Factor of 36 = 9 and 4
Thus, given polynomial can be arranged as follows:
`6x^2 + 9x – 4x – 6`
`= 3x(2x + 3) – 2(2x + 3)`
`= (2x + 3)(3x – 2)`
(iv) `3x^2 - x – 4`
Answer: Given, `3x^2 - x – 4`
Here, ab = 12
Factor of 12 = 3 and 4
Thus, given polynomial can be arranged as follows:
`3x^2 + 3x – 4x – 4`
`= 3x(x + 1) – 4(x + 1)`
`= (x + 1)(3x – 4)`
Thus, factor of given polynomial `= (x + 1)(3x - 4)`