# The zeros of the polynomial p(x)

Question:

The zeros of the polynomial $p(x)=2 x^{2}+7 x-4$ are

(a) $4, \frac{-1}{2}$

(b) $4, \frac{1}{2}$

(c) $-4, \frac{1}{2}$

(d) $-4, \frac{-1}{2}$

Solution:

The given polynomial is $p(x)=2 x^{2}+7 x-4$.

Putting $x=\frac{1}{2}$ in $p(x)$, we get

$p\left(\frac{1}{2}\right)=2 \times\left(\frac{1}{2}\right)^{2}+7 \times \frac{1}{2}-4=\frac{1}{2}+\frac{7}{2}-4=4-4=0$

Therefore, $x=\frac{1}{2}$ is a zero of the polynomial $p(x)$.

Putting x = –4 in p(x), we get

$p(-4)=2 \times(-4)^{2}+7 \times(-4)-4=32-28-4=32-32=0$

Therefore, x = –4 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and $-4$ are the zeroes of the given polynomial $p(x)$.

Hence, the correct answer is option (c).

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