If $- 1$ is a root of $x^{4} + a x^{3} - 19 x^{2} - 46 x + 20$ , then the value of $a$ is

46

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

47

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

48

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

None of the above

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

Open in App

Solution

The correct option is C $48$
If $- 1$ is the root of the given polynomial then, $x = - 1$ should satisfy it. We have
$x^{4} + a \times x^{3} - 19 \times x^{2} - 46 \times x + 20 = 0$
$∴ (- 1)^{4} + a \times (- 1)^{3} - 19 \times (- 1)^{2} - 46 \times (- 1) + 20 = 0$
$1 - a - 19 + 46 + 20 = 0$
or $a = 48$

Suggest Corrections

Similar questions

x^{4} + \frac{1}{x^{4}} = 47

x^{3} + \frac{1}{x^{3}}

x^{4} - a x^{3} + 11 x^{2} - a x + 1 = 0

a

(m, M)

m + 2 M

lim x \to \infty {(\sqrt{x^{4} + a x^{3} + 3 x^{2} + b x + 2} - \sqrt{x^{4} + 2 x^{3} - c x^{2} + 2 x - d})}

a

x^{4} + a x^{3} + b x^{2} + c x - 1 = 0

x^{4} + a x^{3} + b x^{2} + c x + 1 = 0

a, b, c

a

Join BYJU'S Learning Program