Complete factor of $16 x^{3} - 36 x$ is

2 x (4 x + 3) (4 x - 3)

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

4 x (4 x - 3) (4 x - 3)

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

4 x (4 x + 3) (4 x - 3)

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

x (16 x + 3) (16 x - 3)

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

Open in App

Solution

The correct option is C $4 x (4 x + 3) (4 x - 3)$
$16 x^{3} - 36 x$
Take $4 x$ as common, we get
$4 x (16 x^{2} - 9) = 4 x [(4 x)^{2} - 3^{2}]$
We know that $x^{2} - y^{2} = (x + y) (x - y)$
$4 x [(4 x + 3) (4 x - 3)]$
$4 x (4 x + 3) (4 x - 3)$
So, $4 x (4 x + 3) (4 x - 3)$ is a complete factor.

Suggest Corrections

Similar questions

3 x^{3} + 11 x^{2} + 34 x + 106

x - 3

(5 + 3 i) (x + i y) = 3 - 4 i

34 x =

34 x^{3} y^{3} z^{3}

51 x y^{2} z^{3}

34 x^{3} y^{3} z^{3} \div 51 x y^{2} z^{3}

(4 x + 3) (4 x + 3) - (3 x + 6) (3 x - 6) - (2 x + 5) (2 x + 5)

Join BYJU'S Learning Program