Polynomials

Exercise 2.4 Part 5

Question 5: Factorise:

(i) `x^3 - 2x^2 - x + 2`

Answer: `p(x) = x^3 - 2x^2 - x + 2`

Here constant term = 2
The prime factor of constant term 2 = 1 and 2
By trial method, if p(1) = 0, then, (x – 1) is one of the factor of the given polynomial.

Hence, `p(1) = 1^3 - 2 xx 1^2 - 1 + 2`
`= 1 – 2 - 1 + 2`
Or, `p(1) = 0`

Since, `p(1) = 0`, thus, `(x - 1)` is one of the factor of given polynomial.
Thus, given polynomial can be written as

`x^3 - x^2 - x^2 + x – 2x + 2`
`= x^2(x - 1) – x( x – 1) – 2(x – 1)`
(by taking (x – 1) as common)
`= (x – 1)(x^2 - x – 2)`

[Note: second factor other than (x - 1) can be calculated by using long division. ]

Now, factorise the second factor, ie. `(x^2 - x – 2)`

Using splitting middle term method

`= (x – 1)(x^2 + x – 2x – 2)`
`= (x – 1)[x(x + 1) – 2(x + 1)]`
`= (x – 1)(x + 1)(x – 2)`
Hence, factor of given polynomial `= (x – 1)(x + 1)(x – 2)`


(ii) `x^3 - 3x^2 - 9x – 5`

Answer: Given, `x^3 - 3x^2 - 9x – 5`

Here, prime factors of constant term 5 = 1 and 5
Thus, if p(1) = 0, then (x - 1) will be one of the factor of the given polynomial
Or, if p(5) = 0, then ( x – 5) will be one of the factor of the given polynomial.

Hence, `(1) = 1^3 - 3 xx 1^2 - 9 xx 1 – 5`
`= 1 – 3 – 9 – 5 = - 16`

Since, p(1) ≠ 0, thus, ( x- 1) is not a factor of given polynomial.
Now,

`p(5) = 5^3 - 3 xx 5^2 - 9 x 5 – 5`
`= 125 – 75 – 45 – 5 = 0`

Since, p(5) = 0, thus, (x – 5) is one of the factor of given polynomial.
Now, given polynomial can be written as follows:

`= x^3 - 5x^2 + 2x^2 - 10x + x – 5`
`= x^2(x – 5) + 2x(x – 5) + 1 (x – 5)`
(by taking `(x – 5)` as common)
`= (x – 5)(x^2 + 2x + 1)`

Now, factorise the second factor, i.e. `(x^2 + 2x + 1)` by splitting the middle term.

`= (x – 5)(x^2 + x + x + 1)`
`= (x – 5)[x(x + 1) + 1(x + 1)]`
`= (x – 5)(x+1)(x + 1)`


(iii) `x^3 + 13x^2 + 32x + 20`

Answer: Given, `p(x) = x^3 + 13x^2 + 32x + 20`

Factors of 20 = (1 and 2), (2 and 10), (4 and 5)
Now,

`p(1) = 1^3 + 13 xx 1^2 + 32 xx 1 + 20`
`= 1 + 13 + 32 + 20 = 66`

Since, `p(1) ≠ 0`, thus, `(x – 1)` is not a factor of the given polynomial.
Now,

`p( - 1) = (- 1)^3 + 13 xx( - 1)^2 + 32( - 1) + 20`
`= - 1 + 13 – 32 + 20 = 0`

Since, `p( -1) = 0`, thus, `(x + 1)` is one of the factor of given polynomial.
Now, given polynomial can be written as:

`x^3 + (x^2+ 12x^2 + (12x + 20x) + 20`
`= x^3 + x^2 + 12x^2 + 12x + 20x + 20`
(by taking (x + 1) as common)
`= x^2(x + 1) + 12x(x + 1) + 20(x + 1)`
`= (x + 1)(x^2 + 12x + 20)`
`= (x + 1)[x^2 + 2x + 10x + 20x]`
`= (x + 1)[x(x + 2) + 10(x + 2)]`
`= (x + 1)(x + 2)(x + 10)`

(iv) `2y^3 + y^2 - 2y – 1`

Answer: Given, `2y^3 + y^2 - 2y – 1`

Factors of ab = 2 = (1 and 2)
By trial method we will calculate p(1) and p(2)
Now, Let y = 1

`p(1) = 2 xx 1^3 + 1^2 - 2 xx 1 – 1`
`= 2 + 1 – 2 – 1 = 0`

Since, p(1) = 0 thus, (y - 1) is one of the factor of given polynomial.
Thus, the given polynomial can be written as:

`2y^3 - 2y^2 + 3y^2 - 3y + y – 1`
`= 2y^2(y – 1) + 2y(y – 1) + 1(y – 1)`
`= (y – 1)(2y^2 + 3y + 1)`
`= (y -1)(2y^2 + 2y + y + 1`
`= (y – 1) [2y(y + 1) + 1(y + 1)]`
`= (y – 1)(y + 1)(2y + 1)`



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