Question

If x = -1/2 is a zero of the polynomial p (x) = 8x³ - ax² - x + 2, find the value of a.

Answer 1

We have,

p (x) = 8x³ - ax² - x + 2

Put x = \(\frac{-1}{2}\)

p \((-\frac{1}{2})\) = 8 \((-\frac{1}{2})^{3}\) - a \((\frac{-1}{2})^{2}\) - \((-\frac{1}{2})\) + 2

= 8 × \(\frac{-1}{8}\) - a x \(\frac{1}{4}+\frac{1}{2}+2\)

= - 1 - \(\frac{a}{4}+\frac{1}{2}+2\).

\(=\frac{3}{2}-\frac{a}{4}\)

Given that,

x = \(\frac{-1}{2}\) is a root of p (x)

p \((\frac{-1}{2})=0\)

Therefore,

\(\frac{3}{2}.\frac{a}{4}=0\)

\(\frac{3}{2}=\frac{a}{4}\)

2a = 12

a = 6