Question
Factorise : $9a^2+ 3a - 8b - 64b^2$
✓
Answer
$9a^2+ 3a - 8b - 64b^2$
$= 9a^2- 64b^2+ 3a - 8b$
$= ( 3a )^2 - ( 8b )^2 + 3a - 8b$
$= ( 3a - 8b )( 3a + 8b ) + ( 3a - 8b )$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( 3a - 8b )( 3a + 8b + 1 )$
$= 9a^2- 64b^2+ 3a - 8b$
$= ( 3a )^2 - ( 8b )^2 + 3a - 8b$
$= ( 3a - 8b )( 3a + 8b ) + ( 3a - 8b )$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= ( 3a - 8b )( 3a + 8b + 1 )$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Simplify $ :\frac{8^3 a \times 2^5 \times 2^{2 a}}{4 \times 2^{11 a} \times 2^{-2 a}}$
→Evaluate the following: $\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}$ if $\theta=30^{\circ}$
→Rationalise the denominator of each:-
$\frac{1}{(4+2 \sqrt{3})}$
→$\frac{1}{(4+2 \sqrt{3})}$
In an isosceles triangle, prove that the altitude from the vertex bisects the base.
If one angle of a parallelogram is $90^{\circ}$, show that each of its angles measures $90^{\circ}$.
→Solve the following equations for the unknown: $\frac{5}{3 x-2}-\frac{1}{8}=0, x \neq 0, x \neq \frac{2}{3}$
→Rationalise the denominator of the following:
$\frac{\sqrt{2}}{2 \sqrt{3}}$
→$\frac{\sqrt{2}}{2 \sqrt{3}}$
In the given figure, $AB > AC$. If BO and CO are the bisectors of $\angle B$ and $\angle C$ respectively, prove that $BO > CO$.
→Find three rational numbers between: $-\frac{1}{3}$ and $\frac{1}{4}$
→Factorise:$4 a^2+\frac{1}{4 a^2}-2-6 a+\frac{3}{2 a}$
→