Let P $(x) = x^{3} - 6 x^{2} + B x + C$ has 1+5i as a zero and B,C real number, then value of (B+C) is

-70

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

138

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A -70
Given $1 + 5 i$ is the root of $P (x) = x^{3} - 6 x^{2} + B x + C$

$⟹ 1 - 5 i$ is also the root of $P (x)$

Let the third root be $h$

$⟹$ sum of roots is $6$

$⟹ 1 + 5 i + 1 - 5 i + h = 6 ⟹ h = 4$

$B = (1 + 5 i + 1 - 5 i) (4) + (1 + 5 i) (1 - 5 i) = 8 + 26 = 34$

$C = - (1 + 5 i) (1 - 5 i) (4) = - 104$

$B + C = 34 - 104 = - 70$

Suggest Corrections

Similar questions

If a,b and c are three nonzero numbers and the polynomial p(x) = x³+ ax² + bx - c factors as (x - a) (x - b ) (x - c ), then the value of p(3) is

x^{3} - 6 x^{2} + 3 x + 10 = (x + a) (x - b) (x - c)

b + c =

a, b, c

\int_{0}^{1} (1 + c o s^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + c o s^{8} x) (a x^{2} + b x + c) d x .

a x^{2} + b x + c = 0

\int_{0}^{1} (1 + {cos}^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + {cos}^{8} x) (a x^{2} + b x + c) d x,

a x^{2} + b x + c = 0

a, b, c

\int_{0}^{1} (1 + {cos}^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + {cos}^{8} x) (a x^{2} + b x + c) d x

a x^{2} + b x + c = 0

Join BYJU'S Learning Program