$2 x^{2} + 7 x - 1$ completely divides $10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3$ .

True

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

False

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

Open in App

Solution

The correct option is A
True

On dividing, we get

$\begin{matrix} 5 x^{2} - 9 x + 3 2 x^{2} + 7 x - 1 & 10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3 10 x^{4} + 35 x^{3} - 5 x^{2} - - + - 18 x^{3} - 57 x^{2} + 30 x - 3 - 18 x^{3} - 63 x^{2} + 9 x + + - 6 x^{2} + 21 x - 3 6 x^{2} + 21 x - 3 - - + 0 \end{matrix}$

The remainder hence obtained is zero. Therefore $2 x^{2} + 7 x - 1$ divides $10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3$ completely.

Suggest Corrections

Similar questions

10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3

2 x^{2} + 7 x - 1

(10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3)

(2 x^{2} + 7 x - 1)

f (x) = 10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3

g (x) = 2 x^{2} - x + 1

(5 x^{2} - 9 x + 3), (10 x^{4} + 17 x^{3} - 62 x^{2} + 30 x - 3)

Join BYJU'S Learning Program