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).