Find $p (0), p (1), p (2)$ for $p (y) = y^{2} - y + 1$

1, 1, 3

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2, 3, 7

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4, 1, 4

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6, 0, 9

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Solution

The correct option is A $1, 1, 3$
Given, $p (y) = y^{2} - y + 1$

$\Rightarrow$ $p (0) = (0)^{2} - (0) + 1 = 1$ ,

$\Rightarrow$ $p (1) = (1)^{2} - (1) + 1 = 1$ ,
$\Rightarrow$ $p (2) = (2)^{2} - (2) + 1 = 4 - 2 + 1 = 3$ .
Answer $(A) 1, 1, 3$

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Similar questions

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 (0), p (1)

p (2)

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

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)

Find p(0),p(1) and p(2) for each of the following polynomials:
(i) p(y)=y²−y+1
(ii) p(t)=2+t+2t²−t³
(iii) p(x)=x³
(iv) p(x)=(x−1)(x+1)

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