$3$ is a zero of $P (x) = 3 x^{3} - x^{2} - a x - 45$ .
Find $a$ .

Open in App

Solution

It is given that $3$ is a zero of the polynomial $p (x) = 3 x^{3} - x^{2} - a x - 45$ which means that $p (3) = 0$ , so let us substitute $x = 3$ in $p (x)$ to find the value of $a$ as shown below:

$p (3) = 3 (3)^{3} - (3)^{2} - a (3) - 45 \Rightarrow 0 = (3 \times 27) - 9 - 3 a - 45 \Rightarrow 0 = 81 - 9 - 3 a - 45 \Rightarrow 0 = 81 - 54 - 3 a \Rightarrow 0 = 27 - 3 a \Rightarrow 3 a = 27 \Rightarrow a = \frac{27}{3} \Rightarrow a = 9$

hence, $a = 9$ .

Suggest Corrections

Similar questions

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

3

a

\sqrt{3}

- \sqrt{3}

(x - 3)

3 x^{3} - x^{2} + p x + q

g (x)

p (x)

p (x) = 2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12 g (x) = x^{2} - 3

Join BYJU'S Learning Program