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Question

# Verify that (i) 1 and 2 are the zeros of the polynomial p(x) = x2 − 3x + 2. (ii) 2 and −3 are the zeros of the polynomial q(x) = x2 + x − 6. (iii) 0 and 3 are the zeros of the polynomial r(x) = x2 − 3x.

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Solution

## $\left(\mathrm{i}\right)p\left(x\right)={x}^{2}-3x+2=\left(x-1\right)\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒p\left(1\right)=\left(1-1\right)×\left(1-2\right)$ $=0×\left(-1\right)\phantom{\rule{0ex}{0ex}}=0$ Also, $p\left(2\right)=\left(2-1\right)\left(2-2\right)$ $=\left(-1\right)×0\phantom{\rule{0ex}{0ex}}=0$ Hence, 1 and 2 are the zeroes of the given polynomial. $\left(\mathrm{ii}\right)p\left(x\right)={x}^{2}+x-6\phantom{\rule{0ex}{0ex}}⇒p\left(2\right)={2}^{2}+2-6$ $=4-4\phantom{\rule{0ex}{0ex}}=0$ Also, $p\left(-3\right)={\left(-3\right)}^{2}+\left(-3\right)-6\phantom{\rule{0ex}{0ex}}=9-9\phantom{\rule{0ex}{0ex}}=0$ Hence, 2 and $-3$ are the zeroes of the given polynomial. $\left(\mathrm{iii}\right)p\left(x\right)={x}^{2}-3x\phantom{\rule{0ex}{0ex}}⇒p\left(0\right)={0}^{2}-3×0$ Also, $p\left(3\right)={3}^{2}-3×3\phantom{\rule{0ex}{0ex}}=9-9\phantom{\rule{0ex}{0ex}}=0$ Hence, 0 and 3 are the zeroes of the given polynomial.

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