Question

If $\frac{1}{2}$ and 0 are the zeroes of the polynomial $f\left(x\right)=2x^3+a{x}^2+bx+c,$ and b + c = 2, then the value of b(c – a) is

• A. 5
• B. 2
• C. 0
• D. 10

Answer 1

The correct option is D 10

Given polynomial is $f\left(x\right)=2x^3+a{x}^2+bx+c$ ...(i)

Also, b + c = 2 ...(ii)

Here, $\frac{1}{2}$ and 0 are zeroes of polynomial f(x).

Putting 0 in (i), we get

$f\left(0\right)=2(0{)}^3+a(0{)}^2+b\left(0\right)+c=0$

⇒ c = 0 …..(iii)

From (ii), we have

b + c = 2

⇒ b + 0 = 2

⇒ b = 2 ….. (iv)

Again, $\frac{1}{2}$ is a zero of polynomial f(x), we get

$f\left(\frac{1}{2}\right)=2\times {\left(\frac{1}{2}\right)}^3+a\times {\left(\frac{1}{2}\right)}^2+2\times \left(\frac{1}{2}\right)+0=0(\because b=2,c=0)$

$\Rightarrow 2\times \frac{1}{8}+a\times \frac{1}{4}+1=0$

$\Rightarrow \frac{a}{4}+\frac{5}{4}=0$

$\Rightarrow a=-5$ ...(v)

$\therefore b(c-a)=2(0-(-5\left)\right)$ [From (iii), (iv) and (v)]

= 10

Hence, the correct answer is option (d).