If $\sqrt{9 x^{4} + 6 x^{3} - 5 x^{2} - m x + 1}$ is a perfect square, then find the value of $m$ .

- 1

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

- 2

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

- 3

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

- 4

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

Open in App

Solution

The correct option is B $- 2$
$\div$ $3 x^{2} + x - 1$ <= Quotient
$3 x^{2}$
$9 x^{4} + 6 x^{3} - 5 x^{2} - m x + 1$
$9 x^{4}$
$6 x^{2} + x$ $6 x^{3} - 5 x^{2}$
$6 x^{3} + x^{2}$
$6 x^{2} + 2 x - 1$ $- 6 x^{2} - m x + 1$
$- 6 x^{2} - 2 x + 1$
Remainder $\Rightarrow 0$
Since the given polynomial is a perfect square, we must have $m = - 2$ .

Suggest Corrections

Similar questions

m - n x + 28 x^{2} + 12 x^{3} + 9 x^{4}

16 x^{4} - 24 x^{3} + 41 x^{2} - m x + 16

4 x^{2} + x (m + 1) + 1

m

9 x^{4} - 6 x^{3} + 7 x^{2} - 2 x + 1

Join BYJU'S Learning Program