Question

The zeros of the polynomial $p\left(x\right)=2x^2+7x-4$ are
(a) $4,\frac{-1}{2}$
(b) $4,\frac{1}{2}$
(c) $-4,\frac{1}{2}$
(d) $-4,\frac{-1}{2}$

Answer 1

$p\left(x\right)=2x^2+7x-4$

Now, factorise $2x^2+7x-4$

= $2x^2+7x-4$

=$2x^2+8x-x-4$

= $2x(x+4)-1(x+4$

=(2x-1)(x+4)

So, zeroes of the polynomial are:

(2x-1)(x+4) = 0

x = $\frac{1}{2}$, -4

Hence, the correct answer is option (c).