Find gcd of x^4 +x^3 −4x^2 +x+5 and x^3 +x^2 −9x−9


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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

	$i h a v e s h a r e d t h i s q u e s t i o n . . . o t h e r q u e s t i o n$ $s e e m s i m i l a r i n t y p e$
	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
	$A=∫_0 ^(2π) (dx/(2cos^2 x+(√3)sin^2 x)) ⇒ A=∫_0 ^(2π) (dx/(2((1+cos(2x))/2)+(√3)((1−cos(2x))/2))) =∫_0 ^(2π) ((2dx)/(2+2cos(2x)+(√3)−(√3)cos(2x))) =∫_0 ^(2π) ((2dx)/((2−(√3))cos(2x)+2+(√3))) =_(2x=t) ∫_0 ^(4π) (dt/((2−(√3))cost +2+(√3))) =∫_0 ^(2π) (dt/((2−(√3))cost+2+(√3))) +∫_(2π) ^(4π) (dt/((2−(√3))cost +2+(√3)))(→t=2π +u) =2∫_0 ^(2π) (dt/((2−(√3))cost+2+(√3))) =_(e^(it) =z) 2∫_(∣z∣=1) (dz/(iz{(2−(√3))((z+z^(−1) )/2)+2+(√3)})) =2∫_(∣z∣=1) ((2dz)/(iz{(2−(√3))(z+z^(−1) )+4+2(√3)))) =4∫_(∣z∣=1) ((−idz)/((2−(√3))(z^2 +1)+(4+2(√3))z)) =4∫_(∣z∣=1) ((−idz)/((2−(√3))z^2 +(4+2(√3))z +2−(√3))) ϕ(z) =((−i)/((2−(√3))z^2 +(4+2(√3))z +2−(√3))) Δ^′ =(2+(√3))^2 −(2−(√3))^2 =4+4(√3)+3−4+4(√3)−3 =8(√3) z_1 =((−2−(√3)+2(√(2(√3))))/((2−(√3)))) z_2 =((−2−(√3)+2(√(2(√3))))/((2−(√3)))) ∣z_1 ∣<1 and ∣z_2 ∣>1 ϕ(z)=((−i)/((2−(√3))(z−z_1 )(z−z_2 ))) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) =2iπ×((−i)/((2−(√3))4(√(2(√3)))))×(2−(√3)) =((2π)/(4(√(2(√3))))) =(π/(2(√(2(√3))))) ⇒ A=((4π)/(2(√(2(√3))))) ⇒A =((2π)/( (√(2(√3)))))$
	$A = \int_{0}^{2 π} \frac{d x}{2 {c o s}^{2} x + \sqrt{3} {s i n}^{2} x} \Rightarrow$ $A = \int_{0}^{2 π} \frac{d x}{2 \frac{1 + c o s (2 x)}{2} + \sqrt{3} \frac{1 - c o s (2 x)}{2}}$ $= \int_{0}^{2 π} \frac{2 d x}{2 + 2 c o s (2 x) + \sqrt{3} - \sqrt{3} c o s (2 x)}$ $= \int_{0}^{2 π} \frac{2 d x}{(2 - \sqrt{3}) c o s (2 x) + 2 + \sqrt{3}}$ $=_{2 x = t} \int_{0}^{4 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}}$ $= \int_{0}^{2 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}}$ $+ \int_{2 π}^{4 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}} (\to t = 2 π + u)$ $= 2 \int_{0}^{2 π} \frac{d t}{(2 - \sqrt{3}) c o s t + 2 + \sqrt{3}}$ $=_{e^{i t} = z} 2 \int_{∣ z ∣= 1} \frac{d z}{i z {(2 - \sqrt{3}) \frac{z + z^{- 1}}{2} + 2 + \sqrt{3}}}$ $= 2 \int_{∣ z ∣= 1} \frac{2 d z}{i z {(2 - \sqrt{3}) (z + z^{- 1}) + 4 + 2 \sqrt{3})}$ $= 4 \int_{∣ z ∣= 1} \frac{- i d z}{(2 - \sqrt{3}) (z^{2} + 1) + (4 + 2 \sqrt{3}) z}$ $= 4 \int_{∣ z ∣= 1} \frac{- i d z}{(2 - \sqrt{3}) z^{2} + (4 + 2 \sqrt{3}) z + 2 - \sqrt{3}}$ $φ (z) = \frac{- i}{(2 - \sqrt{3}) z^{2} + (4 + 2 \sqrt{3}) z + 2 - \sqrt{3}}$ $Δ^{^{'}} = {(2 + \sqrt{3})}^{2} - {(2 - \sqrt{3})}^{2}$ $= 4 + 4 \sqrt{3} + 3 - 4 + 4 \sqrt{3} - 3 = 8 \sqrt{3}$ $z_{1} = \frac{- 2 - \sqrt{3} + 2 \sqrt{2 \sqrt{3}}}{(2 - \sqrt{3})}$ $z_{2} = \frac{- 2 - \sqrt{3} + 2 \sqrt{2 \sqrt{3}}}{(2 - \sqrt{3})}$ $∣ z_{1} ∣< 1 a n d ∣ z_{2} ∣> 1$ $φ (z) = \frac{- i}{(2 - \sqrt{3}) (z - z_{1}) (z - z_{2})}$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1})$ $= 2 i π \times \frac{- i}{(2 - \sqrt{3}) 4 \sqrt{2 \sqrt{3}}} \times (2 - \sqrt{3})$ $= \frac{2 π}{4 \sqrt{2 \sqrt{3}}} = \frac{π}{2 \sqrt{2 \sqrt{3}}} \Rightarrow$ $A = \frac{4 π}{2 \sqrt{2 \sqrt{3}}} \Rightarrow A = \frac{2 π}{\sqrt{2 \sqrt{3}}}$
	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

	$u s i n g p o l y n o m i a l d i v i s i o n w e f i n d t h a t$ $x^{4} + x^{3} - 4 x^{2} + x + 5 = x (x^{3} + x^{2} - 9 x - 9) + (5 x^{2} + 10 x + 5)$ $n e x t w e h a v e t o d i v i d e x^{3} + x^{2} - 9 x - 9 b y 5 x^{2} + 10 x + 5$ $w e f i n d t h a t x^{3} + x^{2} - 9 x - 9 = (5 x^{2} + 10 x + 5) (\frac{x - 1}{5}) + (- 8 x - 8)$ $F i n a l l y w e d i v i d e 5 x^{2} + 10 x + 5 b y - 8 x - 8 a n d w e f i n d t h a t$ $5 x^{2} + 10 x + 5 = (- 8 x - 8) [- \frac{5}{8} (x + 1)]$ $T h u s g c d (x^{4} + x^{3} - 4 x^{2} + x + 5, x^{3} + x^{2} - 9 x - 9) = x + 1$
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