Factorise- 73102-21102

Using the formula: #160; a^2 b^2=(a+b)(a b) Substituting, a = 7 310 and 2110 in the above formula we get: (7310)^2 (2110)^2=(7310)^2 (2110 ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation Using Algebraic Identities

Factorise: ...

Question

Factorise: $(7 \frac{3}{10})^{2} - (2 \frac{1}{10})^{2}$

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Solution

Using the formula: $a^{2} - b^{2} = (a + b) (a - b)$
Substituting, $a = 7 \frac{3}{10}$ and $2 \frac{1}{10}$ in the above formula we get:

${(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2} = {(\frac{73}{10})}^{2} - {(\frac{21}{10})}^{2}$ $= (\frac{73}{10} + \frac{21}{10}) (\frac{73}{10} - \frac{21}{10}) = \frac{94}{10} \cdot \frac{52}{10} = \frac{4888}{100} = 48.88$

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{(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}

${(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}$

Q. Find

{(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}

Q. Factorise by using formula:

{(7 \frac{3}{10})}^{2} - {(2 \frac{1}{10})}^{2}

48 \frac{22}{25}

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1

and if false then enter

0

Factorise:

(i) $4 a^{2} - 25$ (ii) $x^{2} - \frac{9}{16}$ (iii) $x^{4} - y^{4}$ (iv) $[7 \frac{3}{10}]^{2} - [2 \frac{1}{10}]^{2}$ (v) $(0.7)^{2} - (0.3)^{2}$ (vi) $(5 a - 2 b)^{2} - (2 a - b)^{2}$

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Factorise: (7310)2−(2110)2

Using the formula: a2−b2=(a+b)(a−b)Substituting, a=7310 and 2110 in the above formula we get:(7310)2−(2110)2=(7310)2−(2110)2 =(7310+2110)(7310−2110)=9410⋅5210=4888100=48.88

Factorise: $(7 \frac{3}{10})^{2} - (2 \frac{1}{10})^{2}$