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

GCD of Polynomials

H.C.F. of x...

Question

H.C.F. of $(x^{2} - 3 x + 2)$ and $(x^{2} - 4 x + 3)$ is

(x-1)

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

(x - 2)^{2}

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

(x-1)(x+2)

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

(x-1)(x-3)

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

Open in App

Solution

The correct option is B (x-1)
$x^{2} - 3 x + 2 = x^{2} - 2 x - x + 2$
$x (x - 2) - 1 (x - 2) = (x - 2) (x - 1)$
Now
$x^{2} - 4 x + 3 = x^{2} - 3 x - x + 3$
$x (x - 3) - 1 (x - 3) = (x - 3) (x - 1)$
Thus, the only common factor is (x-1)
Option A

Suggest Corrections

Similar questions

\frac{x^{2} + 3 x + 5}{(x + 1) (x + 2) (x + 3)} =

\frac{A}{(x + 1)} + \frac{B}{(x + 1) (x + 2)} + \frac{c}{(x + 1) (x + 2) (x + 3)}

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

x + \frac{1}{x} = 1

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

x + \frac{1}{x} = x^{2}, x \neq 0

\frac{x^{2} + 5 x + 1}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{(x + 1) (x + 2)} + \frac{C}{(x + 1) (x + 2) (x + 3)}

x^{2} + 4 x + 4

\int \frac{x + 3}{\sqrt{5 - 4 x - x^{2}}} = - \sqrt{5 - 4 x - x^{2}} + {sin}^{- 1} (\frac{x + 2}{3}) + C

\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \frac{x^{2} \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2}}{2} {sin}^{- 1} (\frac{x}{a})

Join BYJU'S Learning Program