If the product of zeros of the polynomial

Question:

If the product of zeros of the polynomial $f(x) a x^{3}-6 x^{2}+11 x-6$ is 4 , then $a=$

(a) $\frac{3}{2}$

(b) $-\frac{3}{2}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

Solution:

Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial $f(x)=a x^{3}-6 x^{2}+11 x-6$

$\alpha \beta=\frac{-\text { Constant term }}{\text { Coefficient of } x^{2}}$

So we haveĀ 

$4=-\left(\frac{-6}{a}\right)$

$4=\frac{6}{a}$

$4 a=6$

$a=\frac{6}{4}$

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

The value of $a$ is $\frac{3}{2}$

Hence, the correct alternative is $(a)$.

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