Factorise $2 (a b + c d) - a^{2} - b^{2} + c^{2} + d^{2}$

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

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

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

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

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

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

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

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

Open in App

Solution

The correct option is D
$(c + d + a + b) (c + d - a - b$ )

Given
$2 (a b + c d) - a^{2} - b^{2} + c^{2} + d^{2}$
Opening the bracket , we get
$2 a b + 2 c d - a^{2} - b^{2} + c^{2} + d^{2}$
Re-grouping
$(c^{2} + d^{2} + 2 c d) - (a^{2} - 2 a b + b^{2})$
Using $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ and $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
$(c + d)^{2} - (a + b)^{2}$
Using difference of squares , we have
$(c + d + a + b) (c + d - a - b$ )

Suggest Corrections

Similar questions

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

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)

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

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

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

Join BYJU'S Learning Program