The correct option is B (x 1)(x y) Given polynomial x^2 xy+y x=x^2 xy x+y nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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

Factorisation of Quadratic Polynomials

Factorise: ...

Question

Factorise: $x^{2} - x y + y - x$

(x - 1) (x + y)

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

(x + 1) (x + y)

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

(x - 1) (x - y)

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

none of these

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

Open in App

Solution

The correct option is B $(x - 1) (x - y)$
Given polynomial
$x^{2} - x y + y - x = x^{2} - x y - x + y$
$= x (x - y) - 1 (x - y)$
$= (x - 1) (x - y)$
Option C is correct.

Suggest Corrections

Similar questions

{sin}^{- 1} x, {cos}^{- 1} x, {tan}^{- 1} x

{tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {tan}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{tan}^{- 1} \frac{1 - x}{1 + x} {tan}^{- 1} \frac{1 - y}{1 + y} = {sin}^{- 1} \frac{y - x}{\sqrt{1 + x^{2}} \sqrt{1 + y^{2}}}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

0 < x < 1

\sqrt{1 + x^{2}} {[{x c o s ({c o t}^{- 1} x) + s i n ({c o t}^{- 1} x)}^{2} - 1]}^{1 / 2}

\frac{x}{\sqrt{1 + x^{2}}}

x

x \sqrt{1 + x^{2}}

\sqrt{1 + x^{2}}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} (x + \frac{2}{x}) - {t a n}^{- 1} \frac{4}{x} - {t a n}^{- 1} (x - \frac{2}{x}) = 0

x = . . . . . . . . . .

x = . . . . . . . . (x ϵ R)

{s i n}^{- 1} x + {s i n}^{- 1}

y = 2 π / 3

{c o s}^{- 1} x - {c o s}^{- 1}

y = π / 3

x = . . . . . . . .

y = . . . . . . .

{t a n}^{- 1} y = 4 {t a n}^{- 1} x, (| x | < t a n \frac{π}{8})

t a n (π / 8)

x^{4} - 6 x^{2} + 1 = 0

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{s i n}^{- 1} (1 - x) - 2 {s i n}^{- 1} x = π / 2

{s i n}^{- 1} x + {s i n}^{- 1} (1 - x) = {c o s}^{- 1} x

0, 1 / 2

Join BYJU'S Learning Program