Find-gcd-of-x-4-x-3-4x-2-x-5-and-x-3-x-2-9x-9-

Question Number 119600 by bemath last updated on 25/Oct/20

F i n d g c d o f x^{4} + x^{3} - 4 x^{2} + x + 5

a n d x^{3} + x^{2} - 9 x - 9

Answered by TANMAY PANACEA last updated on 25/Oct/20

x^{3} + x^{2} - 9 x - 9

= x^{2} (x + 1) - 9 (x + 1)

= (x + 1) (x + 3) (x - 3)

x^{4} + x^{3} - 4 x^{2} + x + 5

= x^{4} + x^{3} - 4 x^{2} - 4 x + 5 x + 5

= x^{3} (x + 1) - 4 x (x + 1) + 5 (x + 1)

= (x + 1) (x^{3} - 4 x + 5)

g . c . d = (x + 1)

Commented by TANMAY PANACEA last updated on 25/Oct/20

t o T i n k u t a r a w h y i c a n n o t s h a r e q u e s t i o n

Commented by TANMAY PANACEA last updated on 25/Oct/20

Commented by TANMAY PANACEA last updated on 25/Oct/20

n o s h a r e b u t t o n

Commented by bemath last updated on 25/Oct/20

c l i c k t h r e e d o t s s i r

Commented by TANMAY PANACEA last updated on 25/Oct/20

o k s i r

Commented by TANMAY PANACEA last updated on 25/Oct/20

\int_{0}^{2 π} \frac{d x}{2 {c o s}^{2} x + \sqrt{3} {s i n}^{2} x} i c a n n o t p o s t b y + s o i n c o m m e n t

Commented by mindispower last updated on 25/Oct/20

= 2 \int_{0}^{π} \frac{d x}{2 {c o s}^{2} (x) + \sqrt{3} {s i n}^{2} (x)}

= 2 \int_{0}^{\frac{π}{2}} \frac{d x}{2 {c o s}^{2} (x) + \sqrt{3} {s i n}^{2} (x)} + 2 \int_{0}^{\frac{π}{2}} \frac{d x}{2 {s i n}^{2} (x) + \sqrt{3} {c o s}^{2} (x)}

\int_{0}^{\frac{π}{2}} \frac{d x}{{a s i n}^{2} (x) + {b c o s}^{2} (x)}, a, b > 0

= \int_{0}^{\frac{π}{2}} \frac{1}{{b c o s}^{2} (x)} \frac{d x}{(1 + {(\sqrt{\frac{a}{b}} t g (x))}^{2}}

= \frac{1}{\sqrt{a b}} \int_{0}^{\frac{π}{2}} \frac{d (\frac{\sqrt{a}}{\sqrt{b}} t g (x))}{1 + {(\frac{\sqrt{a}}{\sqrt{b}} t g (x))}^{2}}

= \frac{1}{\sqrt{a b}} {[\tan^{- 1} (\sqrt{\frac{a}{b}} t g (x))]}_{0}^{\frac{π}{2}} = \frac{π}{2 \sqrt{a b}}

w e f i n d

2 . \frac{π}{2 \sqrt{2 \sqrt{3}}} + \frac{2 . π}{2 \sqrt{\sqrt{3} . 2}} = \frac{2 π}{\sqrt{2 \sqrt{3}}}

Commented by TANMAY PANACEA last updated on 25/Oct/20

t h a n k y o u s i r

Commented by mindispower last updated on 25/Oct/20

w i t h e p l e a s u r s i r i h o p y o u d o i n g w e l l i d o n t k n o w w h y b u t

i m t h i n k i n g t h a t i s l e c t u r w i t h e r e s i d u s T h e o r e m ?

Commented by TANMAY PANACEA last updated on 25/Oct/20

i h a v e s h a r e d t h i s q u e s t i o n \dots o t h e r q u e s t i o n

s e e m s i m i l a r i n t y p e

Commented by Tinku Tara last updated on 26/Oct/20

If you want to always start in forum

by default . Change setting and

enable start i Q & A forum and save

forum preferences

Commented by Bird last updated on 25/Oct/20

A = \int_{0}^{2 π} \frac{d x}{2 {c o s}^{2} x + \sqrt{3} {s i n}^{2} x} \Rightarrow

A = \int_{0}^{2 π} \frac{d x}{2 \frac{1 + c o s (2 x)}{2} + \sqrt{3} \frac{1 - c o s (2 x)}{2}}

= \int_{0}^{2 π} \frac{2 d x}{2 + 2 c o s (2 x) + \sqrt{3} - \sqrt{3} c o s (2 x)}

= \int_{0}^{2 π} \frac{2 d x}{(2 - \sqrt{3}) c o s (2 x) + 2 + \sqrt{3}}

=_{2 x = t} \int_{0}^{4 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}}

= \int_{0}^{2 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}}

+ \int_{2 π}^{4 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}} (\to t = 2 π + u)

= 2 \int_{0}^{2 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}}

=_{e^{i t} = z} 2 \int_{∣ z ∣= 1} \frac{d z}{i z {(2 - \sqrt{3}) \frac{z + z^{- 1}}{2} + 2 + \sqrt{3}}}

= 2 \int_{∣ z ∣= 1} \frac{2 d z}{i z {(2 - \sqrt{3}) (z + z^{- 1}) + 4 + 2 \sqrt{3})}

= 4 \int_{∣ z ∣= 1} \frac{- i d z}{(2 - \sqrt{3}) (z^{2} + 1) + (4 + 2 \sqrt{3}) z}

= 4 \int_{∣ z ∣= 1} \frac{- i d z}{(2 - \sqrt{3}) z^{2} + (4 + 2 \sqrt{3}) z + 2 - \sqrt{3}}

φ (z) = \frac{- i}{(2 - \sqrt{3}) z^{2} + (4 + 2 \sqrt{3}) z + 2 - \sqrt{3}}

Δ^{^{'}} = {(2 + \sqrt{3})}^{2} - {(2 - \sqrt{3})}^{2}

= 4 + 4 \sqrt{3} + 3 - 4 + 4 \sqrt{3} - 3 = 8 \sqrt{3}

z_{1} = \frac{- 2 - \sqrt{3} + 2 \sqrt{2 \sqrt{3}}}{(2 - \sqrt{3})}

z_{2} = \frac{- 2 - \sqrt{3} + 2 \sqrt{2 \sqrt{3}}}{(2 - \sqrt{3})}

∣ z_{1} ∣< 1 a n d ∣ z_{2} ∣> 1

φ (z) = \frac{- i}{(2 - \sqrt{3}) (z - z_{1}) (z - z_{2})}

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1})

= 2 i π \times \frac{- i}{(2 - \sqrt{3}) 4 \sqrt{2 \sqrt{3}}} \times (2 - \sqrt{3})

= \frac{2 π}{4 \sqrt{2 \sqrt{3}}} = \frac{π}{2 \sqrt{2 \sqrt{3}}} \Rightarrow

A = \frac{4 π}{2 \sqrt{2 \sqrt{3}}} \Rightarrow A = \frac{2 π}{\sqrt{2 \sqrt{3}}}

Commented by MJS_new last updated on 26/Oct/20

seems you are in editor mode

go into menue (“ \equiv^{″} - symbol on left top)

directly after starting the app and then

choose f o r u m

Commented by TANMAY PANACEA last updated on 26/Oct/20

o k s i r

Commented by Tinku Tara last updated on 26/Oct/20

Answered by benjo_mathlover last updated on 25/Oct/20

u s i n g p o l y n o m i a l d i v i s i o n w e f i n d t h a t

x^{4} + x^{3} - 4 x^{2} + x + 5 = x (x^{3} + x^{2} - 9 x - 9) + (5 x^{2} + 10 x + 5)

n e x t w e h a v e t o d i v i d e x^{3} + x^{2} - 9 x - 9 b y 5 x^{2} + 10 x + 5

w e f i n d t h a t x^{3} + x^{2} - 9 x - 9 = (5 x^{2} + 10 x + 5) (\frac{x - 1}{5}) + (- 8 x - 8)

F i n a l l y w e d i v i d e 5 x^{2} + 10 x + 5 b y - 8 x - 8 a n d w e f i n d t h a t

5 x^{2} + 10 x + 5 = (- 8 x - 8) [- \frac{5}{8} (x + 1)]

T h u s g c d (x^{4} + x^{3} - 4 x^{2} + x + 5, x^{3} + x^{2} - 9 x - 9) = x + 1