For $p (x) = x + 2$ , find $p (1), p (2), p (- 1)$ and $p (- 2)$ . Also hence find the zero of the polynomial $x + 2$ ?

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Solution

The correct option is D $- 2$
Let $p (x) = x + 2$
Replace $x$ by $1$
$p (1) = 1 + 2 = 3$
Replace $x$ by $2$
$p (2) = 2 + 2 = 4$
Replace $x$ by $- 1$
$p (- 1) = - 1 + 2 = 1$
Replace $x$ by $- 2$
$p (- 2) = - 2 + 2 = 0$
Therefore, $1, 2, - 1$ are not the zeroes of the polynomial $x + 2$ . Only $- 2$ is zero of the polynomial.

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For p(x)=x+2, find p(1),p(2),p(−1) and p(−2). Also hence find the zero of the polynomial x+2?

The correct option is D −2Let p(x)=x+2Replace x by 1p(1)=1+2=3Replace x by 2p(2)=2+2=4Replace x by −1p(−1)=−1+2=1Replace x by −2p(−2)=−2+2=0Therefore, 1,2,−1 are not the zeroes of the polynomial x+2. Only −2 is zero of the polynomial.

For $p (x) = x + 2$ , find $p (1), p (2), p (- 1)$ and $p (- 2)$ . Also hence find the zero of the polynomial $x + 2$ ?