Expand:
$(x + 3) (x + 5) (x + 2)$

2 x^{3} + 10 x^{2} + 31 x + 30

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

x^{3} + 20 x^{2} + 31 x + 30

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

x^{3} + 10 x^{2} + 31 x + 20

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

x^{3} + 10 x^{2} + 31 x + 30

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

Open in App

Solution

The correct option is D $x^{3} + 10 x^{2} + 31 x + 30$
Using the Identity
$(x + a) (x + b) (x + c) = x^{3} + (a + b + c) x^{2} + (a b + b c + c a) x + a b c$

We get,
$(x + 3) (x + 5) (x + 2)$
$= (x)^{3} + (3 + 5 + 2) (x)^{2} + (3 \times 5 + 5 \times 2 + 2 \times 3) x + (3) (5) (2)$

$= x^{3} + 10 x^{2} + (15 + 10 + 6) x + 30$

$= x^{3} + 10 x^{2} + 31 x + 30$

Suggest Corrections

Similar questions

$Expand: (x + \frac{1}{x})^{2}$

\frac{5 x + 2}{3 x - 1} - \frac{5 x - 2}{3 x + 7} < 0

\sqrt{3 x^{2} + 5 x + 7} - \sqrt{3 x^{2} + 5 x + 2} > 1.

\frac{3}{5} x, \frac{2}{3} x, \frac{- 4}{5} x

3 x^{2} + 5 x - x^{7}; - 3 x^{2} + 5 x + 8

Join BYJU'S Learning Program