Factorise-i 16a2-254a2ii 16a2b-b16a2iii 100x-y2-81a-b2iv x-12-x-22

We have, nbsp; (i) 16a^2 254a^2 =(4a)^2 ( 52a )^2= ( 4a+52a ) ( 4a 52a ) (ii) nbsp; 16a^2b b16a^2=b ( 16a^2 116a^2 ) =b{ (4a)^2 ( 14a )^2 } =b ( 4a+14a ) ...

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Byju's Answer

Standard VIII

Mathematics

Factors of Algebraic Expressions and Factorisation

Factorise:i 1...

Question

Factorise:
(i) $16 a^{2} - \frac{25}{4 a^{2}}$
(ii) $16 a^{2} b - \frac{b}{16 a^{2}}$
(iii) $100 (x + y)^{2} - 81 (a + b)^{2}$
(iv) $(x - 1)^{2} - (x - 2)^{2}$

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Solution

We have,
(i) $16 a^{2} - \frac{25}{4 a^{2}}$
$= (4 a)^{2} - {(\frac{5}{2 a})}^{2} = (4 a + \frac{5}{2 a}) (4 a - \frac{5}{2 a})$
(ii) $16 a^{2} b - \frac{b}{16 a^{2}} = b (16 a^{2} - \frac{1}{16 a^{2}})$
$= b {(4 a)^{2} - {(\frac{1}{4 a})}^{2}}$
$= b (4 a + \frac{1}{4 a}) (4 a - \frac{1}{4 a})$
(iii) $100 (x + y)^{2} - 81 (a + b)^{2} = {10 (x + y)}^{2} - {9 (a + b)}^{2}$
$= {10 (x + y) + 9 (a + b)} {10 (x + y) - 9 (a + b)}$
$= (10 x + 10 y + 9 a + 9 b) (10 x + 10 y - 9 a - 9 b)$
(iv) $(x - 1)^{2} - (x - 2)^{2} = {(x - 1) + {(x - 2)} {(x - 1) - (x - 2)}$
$= (2 x - 3) (x - 1 - x + 2)$
$= (2 x - 3) \times 1$
$= 2 x - 3$

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Similar questions

Q. Factorise :

(i) 16 a^{2} - \frac{25}{4 a^{2}}

(i i) 16 a^{2} b - \frac{b}{16 a^{2}}

(i i i) 100 (x + y)^{2} - 81 (a + b)^{2}

(i v) (x - 1)^{2} - (x - 2)^{2}

Q. The area between the parabolas

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Q. Find the area common to the circle x² + y² = 16 a² and the parabola y² = 6 ax.

Find the area of the region {(x, y) : y² ≤ 6ax} and {(x, y) : x² + y² ≤ 16a²}.

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(ii) 1,3,5,7,... to 12 terms
(iii) 3, $\frac{9}{2}, 6 \frac{15}{2}$ , ... to 25 terms
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Factorise: (i) 16a2−254a2 (ii) 16a2b−b16a2 (iii) 100(x+y)2−81(a+b)2 (iv) (x−1)2−(x−2)2

Factorise:
(i) $16 a^{2} - \frac{25}{4 a^{2}}$
(ii) $16 a^{2} b - \frac{b}{16 a^{2}}$
(iii) $100 (x + y)^{2} - 81 (a + b)^{2}$
(iv) $(x - 1)^{2} - (x - 2)^{2}$