Polynomials

NCERT Exercise 2.3

Question 1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) `p(x) = x^3 – 3x^2 + 5x – 3`, `g(x) = x^2 – 2`

Answer:

Here; quotient `= x – 3` and remainder `= 7x – 9`

(ii) `p(x) = x^4 – 3x^2 + 4x + 5`, `g(x) = x^2 + 1 – x`

Answer:

Here, quotient `= x^2 + x – 3` and remainder = 8

(iii) `p(x) = x^4 – 5x + 6`, `g(x) = 2 – x^2`

Answer:

Here; quotient `= - x^2 – 2` and remainder `= - 5x + 10`

Question 2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) `t^2 – 3`, `2t^4 + 3t^3 – 2t^2 – 9t – 12`

Answer:

Here, the first polynomial is a factor of the second polynomial.

(ii) `x^2 + 3x + 1`, `3x^4 + 5x^3 – 7x^2 + 2x + 12`

Answer:

Here, the first polynomial is a factor of the second polynomial.

(iii) `x^3 – 3x + 1`, `x^5 – 4x^3 + x^2 + 3x + 1`

Answer:

Here, the first polynomial is not a factor of the second polynomial.

Question 3: Obtain all other zeroes of `3x^4 + 6x^3 – 2x^2 – 10x – 15`, if two of its zeroes are `(sqrt5)/(3)` and `(-sqrt5)/(3)`

Answer: A quadratic equation can be given as follows:

`x^2 – text((sum of zeroes))x + text(product of zeroes)`

Hence, the equation can be written as follows:

Given polynomial is divided by this equation as follows:

Hence;

`3x^4 + 6x^3 – 2x^2 – 10x – 5`

`= (3x^2 – 5)(x^2 + 2x + 1)`

Roots for the equation `x^2 + 2x + 1` can be calculated as follows:

`x^2 + 2x + 1 = 0`
Or, `x^2 + x + x + 1 = 0`
Or, `x(x + 1) + (x + 1) = 0`
Or, `(x + 1)(x + 1) = 0`

Hence, roots are; - 1 and – 1

Question 4: On dividing `x^3 – 3x^2 + x + 2` by a polynomial g(x), the quotient and remainder were `x – 2` and `– 2x + 4` respectively. Find g(x).

Answer: Subtracting the remainder from the given polynomial we get;

`x^3 – 3x^2 + x + 2 – ( - 2x + 4)`

`= x^3 – 3x^2 + x + 2 + 2x – 4`

`= x^3 – 3x^2 + 3x – 2` …… (1)

Dividing equation (1) by quotient will give the value of g(x)

Hence, `g(x) = x^2 – x + 1`