Question

The zeros of the polynomial p(x) = 2x² + 7x – 4 are

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

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

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

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

Answer 1

The given polynomial is p(x) = 2x² + 7x – 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

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

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).