MCQ
The zeros of the polynomial $ p(x) = 2x^2 + 7x - 4 $ are:
  • A
    $4,\frac{-1}{2}$
  • B
    $4,\frac{1}{2}$
  • $-4,\frac{1}{2}$
  • D
    $-4,\frac{-1}{2}$

Answer

Correct option: C.
$-4,\frac{1}{2}$
$ p(x) = 2x^2 + 7x - 4$
Now, $p(x) = 0$
$⇒ 2x^2 + 7x - 4 = 0$
$⇒ 2x^2 + 8x - x - 4 = 0$
$⇒ 2x(x + 4) - 1(x + 4) = 0$
$⇒ (x + 4)(2x - 1) = 0$
$⇒ x + 4 = 0$ and $2x - 1 = 0$
$⇒ x = -4 $and $\text{x}=\frac{1}{2}$

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