Question
Factorise : $25(2a - b)^2- 81b^2$
✓
Answer
$25(2a - b)^2- 81b^2$
$= [ 5( 2a - b )]^2 - (9b)^2$
$= [ 5( 2a - b ) - 9b ][ 5( 2a - b ) + 9b ]$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ 10a - 5b - 9b ][ 10a - 5b + 9b ]$
$= [ 10a - 14b ][ 10a + 4b ]$
$= 2 \times ( 5a - 7b ) \times 2 \times ( 5a + 2b )$
$= 4( 5a - 7b )( 5a + 2b )$
$= [ 5( 2a - b )]^2 - (9b)^2$
$= [ 5( 2a - b ) - 9b ][ 5( 2a - b ) + 9b ]$
$[ \because a^2 - b^2 = ( a + b )( a - b )]$
$= [ 10a - 5b - 9b ][ 10a - 5b + 9b ]$
$= [ 10a - 14b ][ 10a + 4b ]$
$= 2 \times ( 5a - 7b ) \times 2 \times ( 5a + 2b )$
$= 4( 5a - 7b )( 5a + 2b )$
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 the following:$\frac{2^{ m } \times 3-2^{ m }}{2^{ m +4}-2^{ m +1}}$
→Factorise $: x^2+\frac{1}{(4 x)^2}+1-7 x-\frac{7}{2 x}$
→If $x=\frac{1}{5-x}$ and $x \neq 5$ find $x^3+\frac{1}{x^3}$
→A man invests ₹ 10000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to 11200. Calculate:
(i) the rate of interest per annum;
(ii) the interest accrued in the second year;
(iii) the amount at the end of the third year
(i) the rate of interest per annum;
(ii) the interest accrued in the second year;
(iii) the amount at the end of the third year
The two parallel sides of a trapezium are 58 m and 42 m long. The other two sides are equal, each being 17 m. Find its area.
In $\triangle ABC$, if $AB = AC$ and D is a point on BC .
Prove that $AB ^2- AD ^2= BD \quad CD$.
→Prove that $AB ^2- AD ^2= BD \quad CD$.
In the given figure, ABCP is a quadrant of a circle of radius 14 cm. With AC as diameter, a semi-circle is drawn. Find the area of the shaded region.
Simplify:$(7a +5b)^2 - (7a - 5b)^2$
→Write three irrational numbers between $\sqrt{2}$ and $\sqrt{7}$
→In the figure, $\text{PQR}$ is a straight line and $PS || RT.$ If $QS = 12\ cm, QR = 15\ cm, QT = 10\ cm$ and $RT = 6\ cm,$ find $PQ$ and $PS.$
→