factorise -: x^6 + 5x^3 + 8?

Answer 1

i have used CP300 factor command and this is the result:
(x^2-x+2)(x^4+x^3-x^2+2x+4)

PS:Puggy's equation :(x^3 + 2) (x^3 + 4) equals to x^6+6x^3+8 which is not true factorization

Answer 2

Substitute x^3 = y so that we can have a quadratic equation...

x^6 = (x^3)^2 = y^2

now our equation is as follows

y^2 + 5y + 8

now factors of 8 such that if we add them, we get +5...

no possible real values are possible...

use quadratic equation for ay^2 + by + c

y = (-b + or -sqrt(b^2 - 4ac))/2a

ur answer will be...

y = (-5 + sqrt(-7))/2 or y = (-5 - sqrt(-7))/2

sqrt(-7) is not real... use imaginary unit i...ur final answer will be...

y = (-5 + i*sqrt(7))/2 or y = (-5 - i*sqrt(7))/2

in case, ur given equation was of this form,

x^6 + 6x^3 + 8

u do the same substitution... x^3 = y which will result in following equation...

y^2 + 6y + 8

factors of 8 such that on adding them u get +6... +4 and +2

(y+2)*(y+4)

(x^3 + 2) * (x^3 + 4) will be ur final answer if eq was x^6 + 6x^3 + 8

Answer 3

What you have to do is treat this like a quadratic; what you do instead though is split x^6 into x^3 and x^3

(x^3 + ?) (x^3 + ?)

This doesn't factor properly. I'm open to the possibility that you made a typographical error.

Are you sure you didn't mean to write 6x^3? I'm going to assume you did

x^6 + 6x^3 + 8

(x^3 + 2) (x^3 + 4)

That's what I think you might have meant.

Answer 4

element through "polishing off the sq." ... x^2 + x/4 - a million/8 ... x^2 + x/4 = a million/8 or x^2 + 2(x)(a million/8) = a million/8 or x^2 + 2(x)(a million/8) + (a million/8)^2 = a million/8 + (a million/8)^2 or (x + a million/8)^2 = 9/sixty 4 or x + a million/8 = ± ?(9/sixty 4) or x + a million/8 ± 3/8 = 0 ... x^2 + x/4 - a million/8 = (x + a million/8 - 3/8)(x + a million/8 + 3/8) = (x - a million/4)(x + a million/2)

Answer 5

suppose y=x^3
then
y^2 +5y+8
but b^2 -4ac<0 i.e the discriminant<0
which indicates zero real roots
therefore no factorisation possible

Answer 6

let x^3 = t

then x^6=t^2

t^2+5t+8

delta <0

the value of t is complex number
So no real x values

Answer 7

x^6 +5x^3+8

ANS= (x^2-x+2) (x^4+x^3-x^2+2x+4)

Answer 8

Gals and Guys! You have to factorize, but not just find real roots!
That is do it like this x^6+5x^3+8 = (x^2+ax+b)(x^2+cx+d)(x^2+ex+f);
Yes z=x^3, then be it z^2+5z+8=0, hence z=-5+s*sqrt(25-32) = (-5+s*sqrt(7)*j)/2, where j=sqrt(-1), s=-1, or s=+1;
Now let r^3=0.5*sqrt((-5)*(-5) + sqrt(7)*sqrt(7)) =2sqrt(2), and t=atan(sqrt(7)/5), so that z1 =r^3*exp((2k-1)pi + t)*j) and z2 = r^3*exp((2k+1)pi-t)*j), where k=0, 1, 2, etc;
Hence x1 = r*exp((2k-1)pi +t)j/3), x2 = r*exp((2k+1)pi –t)j/3);

\1\ Let us take pair: x1 and x2 with k=0, then
(x-r*exp((-pi +t)j/3)) * (x-r* exp((pi –t)j/3) = x^2 –r(cos(pi/3-t/3) -jsin(pi/3 -t/3) +cos(pi/3-t/3) +jsin(pi/3-t/3))x +r^2*exp(0) = x^2 +2r*cos(pi/3-t/3) +r^2

/2/ Let us take pair: x1 with k=+1 and x2 with k=-1, then
(x-r*exp((pi/3 +t/3)j)) * (x-r* exp((-pi/3 -t/3)j)) = x^2 –r(cos(pi/3+t/3) +jsin(pi/3 +t/3) +cos(pi/3+t/3) -jsin(pi/3+t/3))x +r^2*exp(0) = x^2 +2r*cos(pi/3+t/3) +r^2

\3\ Let us take pair: x1 with k=-1 and x2 with k=+1, then
(x-r*exp((-pi +t/3)j)) * (x-r* exp((pi –t/3)j) = x^2 –r(cos(pi-t/3) -jsin(pi -t/3) +cos(pi -t/3) +jsin(pi -t/3))x +r^2*exp(0) = x^2 -2r*cos(t/3) +r^2

Let somebody/anybody check and brush it for 10 pts!

Answer 9

put x^3=z
z^2+5z+8=0
z={-5+-sqrt[25-32]}/2
z=-2.5+-i 1.87
There are no real roots.
The factors are
[z-2.5+1.87i][z+2.5-1.87i]
where z=x^3

Answer 10

i cant remember but i can give you the formula, it is somthing like this: a^2+ab+b^2...i hope it helps you
sorry if it is wrong