Polynomials

Exercise 2.2 Part 2

Question 2: Find p(0), p(1) and p(2) for each of the following polynomials:

(i) `p(y) = y^2 - y + 1`

Answer: Given; `p(y) = y^2 - y + 1`

Hence, `p(0) = 0^2 - 0 + 1`

`= 0 – 0 + 1 = 1`

When `y = 1`

Then, `p(1) = 1^2 - 1 + 1`

`= 1 - 1 + 1 = 0 + 1 = 1`

When `y=2'

Then, `p(2) = 2^2 - 2 + 1`

`= 4 – 2 + 1`

`= 2 + 1 = 3`


(ii) `p(t) = 2 + t + 2t^2 - t^3`

Answer: Given, `p(t) = 2 + t + 2t^2 - t^3`

When p(0), i.e. `t = 0`

Then `p(0) = 2 + 0 + 2 x 0^2 - 0^3`

`= 2 + 0 + 0 – 0 = 2`

When, `t = 1`

Then `p(1) = 2 + 1 + 2(1)^2 - 1^3`

`= 2 + 1 + 2 x 1 – 1`

`= 3 + 2 – 1`

`= 5 – 1 = 4`

When `t = 2`

Then `p(2) = 2 + 2 + 2 xx 2^2 - 2^3`

`= 2 + 2 + 2 x 4 – 8`

`= 4 + 8 – 8 = 4`


(iii) `p(x) = x^3`

Answer: Given, `p(x) = x^3`

When `x = 0`

Then `p(0) = 0^3`

Or, `p(0) = 0`

When `x = 1`

Then `p(1) = 1^3`

Or, `p(1) = 1`

When `x = 2`

Then `p(2) = 2^3`

Or, `p(2) = 8`

Hence, 0, 1 and 8 are required answers

(iv) `p(x) = (x – 1)(x + 1)`

Answer: Given, `p(x) = (x – 1)(x + 1)`

When `x = 0`

Then `p(0) = (0 – 1)(0 + 1)`

Or, `p(0) = (- 1) x 1 = - 1`

When `x =1`

Then `p(1) = (1 – 1)(1 + 1)`

Or, `p(1) = 0 xx 2 = 0`

When `x = 2`

Then `p(2) = (2 – 1)( 2 + 1)`

Or, `p(2) = 1 xx 3 = 3`

Hence, - 1, 0 and 3 are required answers



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