One factor of $a^{2} - 2 a b - c^{2} + b^{2}$ is ________.

(a - b + c)

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

(a + b + c)

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

(a + b - c)

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

None of these

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

Open in App

Solution

The correct option is A $(a - b + c)$
Given equation is $a^{2} - 2 a b - c^{2} + b^{2}$
We can write the given equation as,
$\Rightarrow$ $a^{2} - 2 a b + b^{2} - c^{2}$
$\Rightarrow$ $(a - b)^{2} - c^{2}$ ....Since $(a^{2} - 2 a b + b^{2} = (a - b)^{2})$
Also we know,
$m^{2} - n^{2} = (m + n) (m - n)$
Here, $m = (a - b)$ and $n = c$
Thus $a^{2} - 2 a b - c^{2} + b^{2} = (a - b + c) (a - b - c)$
Factors are $(a - b + c)$ and $(a - b - c)$ .

Suggest Corrections

Similar questions

Find:
$(a + b + c) \times (a + b - c) =$

a, b, c

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

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

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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

a, b,

c

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

(a + b + c) (a + b - c)

Join BYJU'S Learning Program