If $2 x^{3} + 4 x^{2} + 2 a x + b$ is exactly divisible by $x^{2} - 1$ Then the value of $a$ and $b$ respectively will be

1, 2

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

- 1, 4

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

1, - 2

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

- 1, - 4

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

Open in App

Solution

The correct option is B $- 1, - 4$
Since $f (x) = 2 x^{3} + 4 x^{2} + 2 a x + b$ is exactly divisible
by $x^{2} - 1 = (x - 1) (x + 1)$
$∴ f (1) = 0$ and $f (- 1) = 0$
These give
$2 + 4 + 2 a + b = 0$
or $2 a + b + 6 = 0$ .....(i)
and $- 2 + 4 - 2 a + b = 0$
or $2 a - b - 2 = 0$ ....(ii)
Solving equations (i) and (ii) we get
$a = - 1, b = - 4$

Suggest Corrections

Similar questions

If 37y4 is exactly divisible by 9, then the least value of y is

(a) 2

(b) 3

(d) 4

x^{4} + 2 x^{3} - 2 x^{2} + x - 1

x^{2} + x - 2

a

b

p (x) = x^{2} + x^{3} + 8 x^{2} - a x + b

x^{2} - 1

If 1A2B5 is exactly divisible by 9, then the least value of (A + B) is

Join BYJU'S Learning Program