Question-121585

Question Number 121585 by Algoritm last updated on 09/Nov/20

Commented by Algoritm last updated on 09/Nov/20

answer : x = 2 \sqrt{2} - 3

Commented by MJS_new last updated on 09/Nov/20

x = r e^{i θ}

if we stay strictly real x \in R \Rightarrow θ = 0 \lor θ = π

usually we define x \in R : x^{1 / 3} =∣ x ∣^{1 / 3} sign (x)

but if {we}^{'} re in C we use the root

x^{1 / 3} = r^{1 / 3} e^{i \frac{θ}{3}} which is \notin R for x \in R^{-}

so we have to decide . if we use the 1^{st} definition

(strictly real) we get

{(x^{3} + 7 x^{2} + 16 x + 5)}^{3} = (- 3 x^{2} - 7 x + 5) {(1 - 2 x)}^{3}

x^{9} + 21 x^{8} + 195 x^{7} + 1030 x^{6} + 3306 x^{5} + 6571 x^{4} + 7637 x^{3} + 4266 x^{2} + 1237 x + 120 = 0

(x + 3) (x^{2} + 6 x + 1) (x^{6} + 12 x^{5} + 68 x^{4} + 187 x^{3} + 295 x^{2} + 159 x + 40) = 0

x_{1} = - 3 - 2 \sqrt{2}

x_{2} = - 3

x_{3} = - 3 + 2 \sqrt{2}

if we use the 2^{nd} definition we get

x_{1} = - 3 + 2 \sqrt{2}

x_{2, 3} \approx - 3.99799 \pm 3 . 26066 i

Commented by Dwaipayan Shikari last updated on 09/Nov/20

S i r, h o w d i d y o u f a c t o r i z e i t ?

Commented by MJS_new last updated on 09/Nov/20

- 3 was easy to find (try factors of the

constant) . the rest usually depends on good

luck . you can approximate the real solutions

and try to see if they “ are {something}^{″} or

you must use software

in this case - 3 + 2 \sqrt{2} was given, so I tried

it and then very often the conjugate is also

a solution

for solutions that cannot be written as

a \pm \sqrt{b} you have to use software . or spend lots

of time \dots

Commented by Dwaipayan Shikari last updated on 09/Nov/20

S u p p o s e a g i v e n x^{5} - x^{3} + x - 2 = 0

A n d w e a r e a s k e d t o f a c t o r i z e i t, t h e n w h i c h m e t h o d w e s h o u l d

a p p l y s i r ?

Commented by MJS_new last updated on 09/Nov/20

first, try \pm 2, \pm 1 \Rightarrow no solution

then find the number of real zeros . plot it or

go on like this :

y^{'} = 5 x^{4} - 3 x^{2} + 1 = 0

this has no real solutions \Rightarrow y has no extremes

\Rightarrow as {it}^{'} s a {(2 n + 1)}^{th} degree polynomial it

must have exactly 1 real solution

if we are lucky it can be written as

y = (x^{2} + a x + b) (x^{3} + α x^{2} + β x + γ)

which gives

(1) a + α = 0

(2) a α + b + β + 1 = 0

(3) a β + b α + γ = 0

(4) a γ + b β - 1 = 0

(5) b γ + 2 = 0

solving this is not easy but possible

we get a “ {nice}^{″} solution

a = β = - 1

b = α = 1

γ = - 2

but in most cases this {won}^{'} t lead to exact

values for these factors .

try

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

Commented by Dwaipayan Shikari last updated on 10/Nov/20

T h a n k i n g y o u