Divide $(b^{2} + a c + a b + b c)$ by $(b + c)$

c + b

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c + a

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b + c

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a + b

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Solution

The correct option is D $a + b$
$(b^{2} + a c + a b + b c)$
$= b^{2} + a b + a c + b c$
$= b (b + a) + c (a + b)$
$= b (a + b) + c (a + b)$
$= (a + b) (b + c)$ ...(1)

$∴ (b^{2} + a c + a b + b c) \div (b + c)$

$= \frac{(b^{2} + a c + a b + b c)}{(b + c)} = \frac{(a + b) (a + b)}{(b + c)}$ ...(From 1)

$= a + b$

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Similar questions

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ

Question 10

In $Δ A B C$ ,

a) AB + BC > AC
b) AB + BC < AC
c) AB + AC < BC
d) AC + BC < AB

a \times (b + c) =

a - b + a b, b + c - b c

c - a - a c

2 c + a b - a c - b c

2 c - a b - a c - b c

2 c + a b + a c + b c

2 c - a b + a c + b c

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