Question

if x = - 1 upon 2 is a zero of the polynomial P X equals to 8 x cube minus ax square - x + 2 find the value of a

Answer 1

$p(x) = 8x {}^{3 } - ax {}^{2} - x + 2 \\ if \: \frac{ - 1}{2} is \: the \: zero \: of \: p(x) \: so \\ p( \frac{ - 1}{2} ) = 0 \\ 8 \times ( \frac{ - 1}{2} ) {}^{3} - a \times ( \frac{ - 1}{2} ) {}^{2} - ( \frac{ - 1}{2} ) + 2 = 0 \\ 8 \times \frac{ - 1}{8} - \frac{a}{4} + \frac{1}{2} + 2 = 0 \\ - 1 - \frac{a}{4} + \frac{1}{2} + 2 = 0 \\ \frac{ - 4 - a + 2 + 8}{4} = 0 \\ 6- a = 4 \\ - a = 4 - 6 \\ a = 2$

Answer 2

Answer:

The value of a is 6

Step-by-step explanation:

$\text{Given that }\frac{-1}{2}\text{ is a zero of the polynomial }P(x)=8x^3-ax^2-x+2$

we have to find the value of a

$P(x)=8x^3-ax^2-x+2$

$\text{As }\frac{-1}{2}\text{ is the zero therefore by remainder theorem}$

$P(\frac{-1}{2})=0$

$8(\frac{-1}{2})^3-a(\frac{-1}{2})^2-(\frac{-1}{2})+2=0$

$-1-\frac{a}{4}+\frac{1}{2}+2=0$

$\frac{a}{4}=1+\frac{1}{2}$

$\frac{a}{4}=\frac{3}{2}$

$a=\frac{3}{2}\times 4=6$

Hence, the value of a is 6