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