Factors of $a^{2} - b^{2} - 4 a c + 4 c^{2}$ are

$(a - 2 c + b)$ and $(a + 2 c - b)$

True

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False

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Solution

The correct option is B
False

$a^{2} - b^{2} - 4 a c + 4 c^{2}$

$a^{2} - 4 a c + 4 c^{2} - b^{2}$

$a^{2} - 2 a c - 2 a c + 4 c^{2} - b^{2}$

$a (a - 2 c) - 2 c (a - 2 c) - b^{2}$

$(a - 2 c)^{2} - b^{2}$

$(a - 2 c + b) (a - 2 c - b)$

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Similar questions

a^{2} + b^{2} + 4 c^{2} + 2 (a b + 2 b c + 2 c a)

(a + b + 2 c)

(\frac{1}{a} + \frac{1}{b} - \frac{2 c}{a b}) (a + b + 2 c) : (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{2}{a b} - \frac{4 c^{2}}{a^{2} b^{2}})

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

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

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

\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}

a^{2} + b^{2} - 2 (a b - a c + b c)

= (a - b) {(a - b) + 2 c} = (a - b) (a - b + 2 c)

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