The first step in finding the product $(3 x + 2) (x - 5)$ by use of the distributive property in the form $a (b + c) = a b + a c$ is:

3 x^{2} - 13 x - 10

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

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

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

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

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

3 x^{2} - 17 x - 10

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

3 x^{2} + 2 x - 15 x - 10

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

Open in App

Solution

The correct option is E $(3 x + 2) x + (3 x + 2) (- 5)$
$a (b + c) = a b + b c$
This is distributive property.
$(3 x + 2) (x - 5) = (3 x + 2) x + (3 x + 2) (- 5)$

Suggest Corrections

Similar questions

(1 + \frac{1}{2} x + \frac{1}{\sqrt{3}} x^{2})

(5 - \frac{2}{3} + \sqrt{3} x^{2})

(4 - \frac{1}{3} x - 2 \sqrt{3} x^{2})

1 + \frac{1}{6} x + \frac{10}{\sqrt{3}} x^{2}

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

5 + 2 x

\frac{2 x - 3}{x^{2} - 2 x + 3} = \frac{3 x - 5}{x^{2} - 3 x + 5}

\frac{3 x^{2} + 21 x + 4}{3 x^{2} + 27 x + 5} = \frac{x + 7}{x + 9}

\frac{1 - 2 x - 3 x^{2}}{3 x - x^{2} - 5} > 0

(\frac{1}{3} x^{2} - \frac{1}{2} x + 5)

(\frac{1}{2} x^{2} - \frac{1}{3} x + 1)

Join BYJU'S Learning Program