If 8 27a7b2 - 4 9a5b2 is factorised- then its simplified form is

The correct option is C 4 9a^5b^2( 2 3a^2 + 1) 827 a^7 b^2+49 a^5 b^2 Taking 49a^5b^2 as common factor 49a^5b^2(23a^2 +1) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Comparison of Quantities Using Exponents

If 8 27a7b2...

Question

If $\frac{8}{27} a^{7} b^{2} + \frac{4}{9} a^{5} b^{2}$ is factorised, then its simplified form is

\frac{4}{9} a^{4} b^{2} (\frac{2}{3} a^{2} + 1)

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

\frac{4}{9} a^{4} b (\frac{2}{3} a^{2} + 1)

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

\frac{4}{9} a^{5} b^{2} (\frac{2}{3} a^{2} + 1)

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

\frac{4}{9} a^{4} b^{2} (\frac{2}{9} a^{2} + 1)

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

Open in App

Solution

The correct option is C $\frac{4}{9} a^{5} b^{2} (\frac{2}{3} a^{2} + 1)$
$\frac{8}{27} a^{7} b^{2} + \frac{4}{9} a^{5} b^{2}$
Taking $\frac{4}{9} a^{5} b^{2}$ as common factor
$\frac{4}{9} a^{5} b^{2} (\frac{2}{3} a^{2} + 1)$

Suggest Corrections

Similar questions

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

Q. The equilibrium constant of the reaction

2 A B ⇌ A_{2} + B_{2}

49

. The equilibrium constant of

A B ⇌ \frac{1}{2} A_{2} + \frac{1}{2} B_{2}

is:

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are integers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a < j

for

j = 2, 4, 5, 6, 7.

The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a_{j} < j

for

j = 2, 3, 4, 5, 6, 7

. The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

is-

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

Join BYJU'S Learning Program

If 827a7b2+49a5b2 is factorised, then its simplified form is

The correct option is C 49a5b2(23a2+1)827a7b2+49a5b2Taking 49a5b2 as common factor49a5b2(23a2+1)

If $\frac{8}{27} a^{7} b^{2} + \frac{4}{9} a^{5} b^{2}$ is factorised, then its simplified form is

The correct option is C $\frac{4}{9} a^{5} b^{2} (\frac{2}{3} a^{2} + 1)$
$\frac{8}{27} a^{7} b^{2} + \frac{4}{9} a^{5} b^{2}$
Taking $\frac{4}{9} a^{5} b^{2}$ as common factor
$\frac{4}{9} a^{5} b^{2} (\frac{2}{3} a^{2} + 1)$