You visited us 1 times! Enjoying our articles? Unlock Full Access!

Zeroes of a Polynomial

Factorize: ...

Question

Factorize: $x^{2} + y^{2} + x + y + 2 x y$

(x + 2 y) (2 x + 2 y + 1)

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

(x + y) (x + y + 1)

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

(2 x + y) (x - y + 1)

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

(2 x + y) (x + y + 1)

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

Open in App

Solution

The correct option is A $(x + y) (x + y + 1)$
$x^{2} + y^{2} + x + y + 2 x y$
$= x^{2} + y^{2} + 2 x y + x + y$
$= (x + y)^{2} + (x + y)$
$= (x + y) (x + y + 1)$

Suggest Corrections

Similar questions

y = y (x)

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

y (2) = 2 / \sqrt{3}

y (x) = sec ({sec}^{- 1} x - \frac{π}{6})

y (x)

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

(x + 2 y) (- 3 x - y) - (x + y) (x - y) + (x - 2 y) (- 2 x + y)

y = y (x)

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

y (2) = \frac{2}{\sqrt{3}}

y (x) = sec (s e c^{- 1} x - \frac{π}{6})

y (x)

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

2 (x + y)^{2} - 6 (x + y)

x^{2} - y^{2} - 2 x + 2 y

(x - y) (x + y - 2)

Join BYJU'S Learning Program