Question

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = ${{y}^{2}}-y+1$
(ii) p(t) = $2+t+2{{t}^{2}}-{{t}^{3}}$
(iii) p(x) = ${{x}^{3}}$
(iv) p(x) = (x-1)(x+1)

Answer 1

- Hint: To solve this problem, we try to find the respective polynomial values, we input the given values inside the polynomial. Thus, we will put 0, 1 and 2 inside each of these polynomials and find the respective values of these polynomials for each of these cases.

Complete step-by-step solution -

Before solving the problem, we would first try to understand the value of polynomials by taking a small example. Suppose, we are given a polynomial f(x) = 2x+1. Now, we have to find f(1), we simply put 1 inside f(x), thus, we have 2(1)+1 = 3. Now, coming onto the problem in hand, we start by the first part (that is (i) p(y) = ${{y}^{2}}-y+1$).
Thus, to find p(0), that is ${{y}^{2}}-y+1$, we similarly put 0 inside this polynomial to evaluate. We have, p(0) = ${{0}^{2}}-0+1=1$. Now, to find p(1), we have ${{1}^{2}}-1+1=1$. Finally, we find p(2), we have ${{2}^{2}}-2+1=3$. Thus, the values of p(0), p(1) and p(2) are 1, 1 and 3 respectively.

Now for part (ii), that is $2+t+2{{t}^{2}}-{{t}^{3}}$, to find p(0), we have, p(0) = $2+0+2{{(0)}^{2}}-{{(0)}^{3}}=2$. Now, to find p(1), we have $2+1+2{{(1)}^{2}}-{{(1)}^{3}}=4$. Finally, we find p(2), we have $2+2+2{{(2)}^{2}}-{{(2)}^{3}}=4$. Thus, the values of p(0), p(1) and p(2) are 2, 4 and 4 respectively.

Now for part (iii), that is ${{x}^{3}}$, to find p(0), we have, p(0) = ${{0}^{3}}=0$. Now, to find p(1), we have ${{1}^{3}}=1$. Finally, we find p(2), we have ${{2}^{3}}=8$. Thus, the values of p(0), p(1) and p(2) are 0, 1 and 8 respectively.

Now for part (iv), that is (x-1) (x+1), to find p(0), we have, p(0) = (0-1)(0+1) = -1. Now, to find p(1), we have (1-1)(1+1) = 0. Finally, we find p(2), we have (2-1)(2+1) = 3. Thus, the values of p(0), p(1) and p(2) are -1, 0 and 3 respectively.

Note: While solving questions related to finding the value of a polynomial, we should be careful to ensure that the polynomial value doesn’t become indeterminate. For example, if we have f(x) = $\dfrac{1}{x+1}$, we cannot find the value of f(-1). Since, if we put x = -1, the denominator becomes 0, which is not possible.