Find the remiander when we divide $a^{2} + b^{2} + 4 c^{2} + 2 (a b + 2 b c + 2 c a)$ by $(a + b + 2 c)$

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Solution

$a^{2} + b^{2} + 4 c^{2} + 2 (a b + 2 b c + 2 c a) = (a + b c + 2 c)^{2}$
when we divide $(a + b + 2 c)^{2}$ by $(a + b + 2 c)$
we get $(a + b + c)$
$\Rightarrow$ remainder is $z e r o$ .

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Similar questions

(a - b + c)^{2}

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

△ A B C

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

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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 + b - c)^{2}

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

△ A B C, a : b : c = cos A : cos B : cos C

△ A B C

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

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

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

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