If the zeroes of polynomial $p (x) = 3 x^{3} - x^{2} - a x - 45$ is $3$ , then $a$ =

Open in App

Solution

$p (x) = 3 x^{3} - x^{2} - a x - 45$

zeroes pf $p (x)$ is $3$

$\Rightarrow p (3) = 0 \Rightarrow {3.3}^{3} - 3^{2} - 3 a - 45 = 0$ .

$\Rightarrow 81 - 9 - 3 a - 45 = 0$

$\Rightarrow 3 a = 27$ .

$\Rightarrow a = \frac{27}{3} = 9$ .

$∴ a = 9$ .

Suggest Corrections

Similar questions

3

P (x) = 3 x^{3} - x^{2} - a x - 45

a

p (x) = x^{2} - a x + b

a b

P (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

\frac{- 1}{3}

f (x) = x^{4} - 3 x^{3} - x^{2} + 9 x - 6

- \sqrt{3}, \sqrt{3}

a, - 1, - \frac{1}{a}

p (x) = 3 x^{3} - 5 x^{2} - 11 x - 3

Join BYJU'S Learning Program