If the factors of $a^{2} + b^{2} + 2 (a b + b c + c a)$ are $(a + b + m)$ and $(a + b + n c)$ , then the value of $(m + n)$ is

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $2$
$a^{2} + b^{2} + 2 a b + 2 b c + 2 c a + c^{2} - c^{2}$
$= (a + b + c)^{2} - c^{2}$
$= (a + b) (a + b + 2 c)$
On comparision we get,
$m = 0, n = 2$
$m + n = 0 + 2 = 2$

Suggest Corrections

Similar questions

a + b + c = m

a^{2} + b^{2} + c^{2} = n

2 (a b + b c + c a)

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

1 + 2 a b - (a^{2} + b^{2})

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

a + b + c = 0

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

a + b + c = 5

a b + b c + c a = 4

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

Join BYJU'S Learning Program