Factorize- a-2 - b-2 - 2b -1-a- b -1-a - b - 1-a- b -1-a - b - 1-a - b -1-a - b - 1-a - b -1-a - b - 1-

The correct option is B [a b 1][a + b + 1] ⇒ a^2 b^2 2b 1 ⇒ a^2 (b^2 + 2b +1) ⇒ a^2 (b +1)^2 ∵ (a+b)^2 = a^2 + b^2 + 2ab, and a^2 b^2 = (a b)(a ...

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Method of Common Factors

Factorize: a^...

Question

Factorize:

$a^{2} - b^{2} - 2 b - 1$

$[a - b - 1] [a - b - 1]$

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$[a - b - 1] [a + b + 1]$

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$[a + b + 1] [a - b + 1]$

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$[a - b + 1] [a + b - 1]$

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Solution

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Similar questions

A = [\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}], B = [\begin{matrix} a & 1 \\ b & - 1 \end{matrix}]

If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

Factorize:

$a^{2} - b^{2} - 2 b - 1$

\frac{a^{3} - a - 2 b - \frac{b^{2}}{a}}{(1 - \sqrt{\frac{1}{a} + \frac{b}{a^{2}}}) (a + \sqrt{a + b})} : (\frac{a^{3} + a^{2} + a b + a^{2} b}{a^{2} - b^{2}} + \frac{b}{a - b})

a = 23, b = 22

a = b = 1

a = b

a^{2} = a b

a^{2} - b^{2} = a b - b^{2}

(a + b) (a - b) = b (a - b)

\frac{b (a - b)}{a - b}

a + b = b

1 + 1 = 1

2 = 1

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