Question

of 2x-1 is the factor of 2x^3+ax^2-5x+2 b if ab=1 then find a-8b​

Answer 1

Answer:

7

Step-by-step explanation:

We are given that

$2x-1$ is the factor of

$2x^3+ax^2-5x+2b$

$ab=1$

$b=\frac{1}{a}$

We have to find the value of a-8b

$2x-1=0$

$x=\frac{1}{2}$

By factor theorem

$2(\frac{1}{2})^3+a(\frac{1}{2})^2-5(\frac{1}{2})+2b=0$

$\frac{1}{4}+\frac{a}{4}-\frac{5}{2}+\frac{2}{a}=0$

$\frac{a+a^2-10a+8}{4a}=0$

$\frac{a^2-9a+8}{4a}=0$

$a^2-9a+8=0$

$(a-1)(a-8)=0$

a=1 and a=8

$b=1,b=\frac{1}{8}$

When a=1 and b=1

$a-8b=1-8(1)=-7$

When a=8 and b=1/8

$8-8(\frac{1}{8})=8-1=7$

Answer:7