Question

If –1, 3 are the two zeroes of the polynomial $p\left(x\right)=2x^3-9x^2+4x+15$ then the third zero of p(x) is

• A. $\frac{5}{2}$
  
• B. 3
• C. 5
• D. $\frac{3}{2}$
  

Answer 1

The correct option is A $\frac{5}{2}$


Given that –1 and 3 are zeroes of $p\left(x\right)=2x^3-9x^2+4x+15.$
For the cubic polynomial ${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+d,$ sum of zeroes is given by $\frac{-b}{a}.$

Let the third zero be $\alpha .$

$\therefore \alpha +(-1)+3=\frac{-b}{a}=\frac{9}{2}$

$\Rightarrow \alpha =\frac{9}{2}-2=\frac{5}{2}$

Hence, the correct answer is option (a).