One of the zeroes of the polynomial 2 x^{2}+7 | vedclass

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Question

We have $p(x)=2 x^{2}+7 x+4$

(a) $p(2)=2(2)^{2}+7(2)-4$

$=8+14-4$

$=18 
eq 0$

(b) $p\left(-\frac{1}{2}ight)=2\left(-\frac{1}{2}ight)^

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

We have $p(x)=2 x^{2}+7 x+4$

(a) $p(2)=2(2)^{2}+7(2)-4$

$=8+14-4$

$=18 
eq 0$

(b) $p\left(-\frac{1}{2}ight)=2\left(-\frac{1}{2}ight)^{2}+7\left(-\frac{1}{2}ight)-4$

$=2 	imes \frac{1}{4}-\frac{7}{2}-4=\frac{1}{2}-\frac{7}{2}-4$

$=-3-4$

$=-7 
eq 0$

(C) $p\left(\frac{1}{2}ight)=2\left(\frac{1}{2}ight)^{2}+7\left(\frac{1}{2}ight)-4$

$=2 	imes \frac{1}{4}+\frac{7}{2}-4=\frac{1}{2}+\frac{7}{2}-4=4-4=0$

(d) $p(-2)=2(-2)^{2}+7(-2)-4$

$=8-14-4=-10 
eq 0$

As $p\left(\frac{1}{2}ight)=0,$ we say that $\frac{1}{2}$ is a zero of the polynomial.

Hence, $\frac{1}{2}$ is one of the zero of the polynomial $2 x^{2}+7 x-4.$

Hence, $(c)$ is the correct answer.

One of the zeroes of the polynomial $2 x^{2}+7 x-4$ is

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