Find $p (0), p (1)$ and $p (2)$ for $p (t) = (t - 1) (t + 1)$ :

- 2, 1, 4

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- 1, 0, 3

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0, 1, 3

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0, 2, 5

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Solution

The correct option is A $- 1, 0, 3$
Given polynomial is $p (t) = (t - 1) (t + 1)$
$\Rightarrow p (t) = (t^{2} - 1)$
Replace $t$ by $0$ , we get
$p (0) = [(0)^{2} - 1]$
$\Rightarrow p (0) = [0 - 1]$
$\Rightarrow p (0) = - 1$
Replace $t$ by $1$ , we get
$p (1) = [(1)^{2} - 1]$
$\Rightarrow p (1) = [1 - 1]$
$\Rightarrow p (1) = 0$
Replace $t$ by $2$ , we get
$p (2) = [(2)^{2} - 1]$
$\Rightarrow p (2) = [4 - 1]$
$\Rightarrow p (2) = 3$

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Similar questions

p (0), p (1), p (2)

p (t) = 2 + t + 2 t^{2} + t^{3}

Find p(0), p(1) and p(2) for each of the following polynomials:
$1) p (y) = y^{2} - y + 1$
$2) p (t) = 2 + t + 2 t^{2} - t^{3}$ [4 MARKS]

p (y) = y^{2} - y + 1

p (t) = 2 + t + 2 t^{2} - t^{3}

p (0), p (1)

p (2)

p (y) = y^{2} - y + 1

p (t) = 2 + t + 2 t^{2} - t^{3}

p (x) = x^{3}

p (x) = (x - 1) (x + 1)

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