Question

# Verify that: (i) 4 is a zero of the polynomial p(x) = x − 4. (ii) −3 is a zero of the polynomial p(x) = x − 3. (iii) $-\frac{1}{2}$is a zero of the polynomial p(y) = 2y + 1. (iv) $\frac{2}{5}$is a zero of the polynomial p(x) = 2 − 5x. (v) 1 and 2 are the zeros of the polynomial p(x) = (x − 1)(x − 2). (vi) 0 and 3 are the zeros of the polynomial p(x) = x2 − 3x. (vii) 2 and −3 are the zeros of the polynomial p(x) = x2 + x − 6.

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Solution

## $\left(\mathrm{i}\right)p\left(x\right)=x-4\phantom{\rule{0ex}{0ex}}⇒p\left(4\right)=4-4\phantom{\rule{0ex}{0ex}}$ = 0 Hence, 4 is the zero of the given polynomial. $\left(\mathrm{ii}\right)\mathrm{Disclaimer}:\mathrm{In}\mathrm{the}\mathrm{question},\mathrm{instead}\mathrm{of}-3,3\mathrm{should}\mathrm{be}\mathrm{the}\mathrm{zero}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}.\phantom{\rule{0ex}{0ex}}p\left(x\right)=x-3\phantom{\rule{0ex}{0ex}}⇒p\left(3\right)=3-3$ = 0 Hence, 3 is the zero of the given polynomial. $\left(\mathrm{iii}\right)p\left(y\right)=2y+1\phantom{\rule{0ex}{0ex}}⇒p\left(-\frac{1}{2}\right)=2×\left(-\frac{1}{2}\right)+1\phantom{\rule{0ex}{0ex}}=-1+1\phantom{\rule{0ex}{0ex}}=0$ Hence, $-\frac{1}{2}$ is the zero of the given polynomial. $\left(\mathrm{iv}\right)p\left(x\right)=2-5x\phantom{\rule{0ex}{0ex}}⇒p\left(\frac{2}{5}\right)=2-5×\left(\frac{2}{5}\right)$ $=2-2\phantom{\rule{0ex}{0ex}}=0$ Hence, $\frac{2}{5}$ is the zero of the given polynomial. $\left(\mathrm{v}\right)p\left(x\right)=\left(x-1\right)\left(x-2\right)\phantom{\rule{0ex}{0ex}}⇒p\left(1\right)=\left(1-1\right)×\left(1-2\right)$ $=0×\left(-1\right)\phantom{\rule{0ex}{0ex}}=0$ 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. $\left(\mathrm{vi}\right)p\left(x\right)={x}^{2}-3x\phantom{\rule{0ex}{0ex}}⇒p\left(0\right)={0}^{2}-3×0$ 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. $\left(\mathrm{vii}\right)p\left(x\right)={x}^{2}+x-6\phantom{\rule{0ex}{0ex}}⇒p\left(2\right)={2}^{2}+2-6$ $=4-4\phantom{\rule{0ex}{0ex}}=0$ 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.

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