Factorise a2-1-22-2-7 a -7-2A- a -1-2 a -1-2-7B- a -1-2 a -1-a-7C- a -1-22-7D- a -7-22

The correct option is A (a + 1/a) (a + 1/a 7) Given that a^2 + 1/a^2 +2 7a + 7/a = a^2 + 1/a^2 +2 7(a + 1/a) = (a + 1/a)^2 7(a + 1/a) = (a + 1/a) (a + ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation by Common Factors

Factorise a2+...

Question

Factorise $a^{2} + \frac{1}{a^{2}} + 2 - 7 a - \frac{7}{a}$

(a + $\frac{1}{a}$ ) (a + $\frac{1}{a} - 7$ )

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(a - $\frac{1}{a}$ ) (a + $\frac{1}{a} + 7$ )

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(a + $\frac{1}{a})^{2}$ - 7

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(a + $\frac{7}{a})^{2}$

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Solution

The correct option is A
(a + $\frac{1}{a}$ ) (a + $\frac{1}{a} - 7$ )

Given that
$a^{2} + \frac{1}{a^{2}} + 2 - 7 a + \frac{7}{a}$
= $a^{2} + \frac{1}{a^{2}} + 2$ - 7(a + $\frac{1}{a}$ )
= $(a + \frac{1}{a})^{2}$ - 7(a + $\frac{1}{a}$ )
= (a + $\frac{1}{a}$ ) (a + $\frac{1}{a} - 7$ )

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Q. Show that

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,
and

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Factorise a2+1a2+2−7a−7a

The correct option is A (a + 1a) (a + 1a−7) Given that a2+1a2+2−7a+7a = a2+1a2+2 - 7(a + 1a) = (a+1a)2 - 7(a + 1a) = (a + 1a) (a + 1a−7)

Factorise $a^{2} + \frac{1}{a^{2}} + 2 - 7 a - \frac{7}{a}$

The correct option is A
(a + $\frac{1}{a}$ ) (a + $\frac{1}{a} - 7$ )

Given that
$a^{2} + \frac{1}{a^{2}} + 2 - 7 a + \frac{7}{a}$
= $a^{2} + \frac{1}{a^{2}} + 2$ - 7(a + $\frac{1}{a}$ )
= $(a + \frac{1}{a})^{2}$ - 7(a + $\frac{1}{a}$ )
= (a + $\frac{1}{a}$ ) (a + $\frac{1}{a} - 7$ )