Factoring out thesae two roots leaves: x^4 - x^3 + 2x + 4?

Question

"Factoring out thesae two roots leaves: x^4 - x^3 + 2x + 4

Which can be sovled closed form as a quartic or factored into two quadratics: (x^2 + x + 1)(x^2 - 2x + 4)"

How do you get this factorization? Please, confirm your address, so that I can contact you


Ana

This is a quote from my other questions answer. Someone did this step and I dont understand it, but, as one of the persons answers here, this is not even right.

To all of you,

Please, if you dont have a way so that I can contact you, contact me in case that I have remarks about your answers

Ana

Answer 1

Let us asume x^4-x^3+2x+4 = (x^2+ax+b)(x^2+cx+d)
= x^4+(a+c)x^3 +x^2(b+d+ac)+x(ad+bc)+db

comparing with your expression

a+c = -1..1

b+d +ac = 0..2

ad+bc = 2... 3

bd = 4... 4

by trial and error
if i take b = -2 and d= -2 3rd equation is consistant with 1(a+c = -1) and ac = -4
these may be solved for a and c also.
Thus it can be factored.
This may be factored in more than one way

becuse say(x+s)(x+t)(x+p)(x+q) which I can take grouping of 2
in more than one way.
These four equations should be used to solve for a to d.
However product of your 2 quadratic terms gives x^2 term as 3 this may kindly be checked

I put the exact product x^4-3x^3+x^2+2x+4

Answer 2

I see that "sebourban" is on the right track, but he throws in too many unknowns.

Suppose x^4 - x^3 +2x + 4 = (x^2 + ax + ...) (x^2 + bx + ...), then we immediately have a + b = -1 to get the x^3 term right. Okay then, the factorisation is (x^2 + ax + c) (x^2 - (a+1)x + d) for some a, c and d. Looking at the x^2 term tells us that 0 = c + d - a(a+1). If they are integers, the only possibilities for (c, d) are (-1, -4), (1, 4), (-2, -2) or (2, 2), allowing a(a+1) to be either -5, 5, -4 or 4, none of which work . . .

. . . BECAUSE your two quadratic factors multiply out to x^4 - x^3 + 3x^2 + 2x + 1, which is not what you typed!

Including the 3x^2 term tells us that 3 = c + d - a(a+1), so the four (c, d) possibilities allow a(a+1) to be either -8, 2, -7 or 1. The only one of those which works is 2, so a(a+1) is either (-2)(-1) or (1)(2), and (c, d) is either (1, 4) or (4, 1). I suppose only one arrangement of these will give the correct coefficient of x, but I think I'll stop there.

Answer 3

Is your question HOW you know to factor the quartic into the two quadratics? Got me.

BUT, I did observer something, which may not be of any help at all.

The first quadratic factor is (x^3 - 1)/(x - 1).
The second quadratic factor is (x^3 + 8)/(x - 2).

again, I have no idea if this is relevant to the question. But is just jumped out at me.

Good luck!

Answer 4

I am not sure I have got it right, and sorry - I did not finish ... here is an attempt ...

you write (ax^2+bx+c)(dx^2+ex+f), you develop and identify with the first equation ... here it goes:
adx^4+(ae+bd)x^3+(be+af+cd)x^2+(bf+ce)x+cf , which leads to
A) ad = 1
B) ae+bd = -1
C) be+af+cd = 0
D) bf+ce = 2
E) cf = 4
since you have 6 unkown for 5 equations, it is safe to pose a = 1, for instance ... which leads to d=1 with A and ...
B') b+e = -1
C') be+f+c = 0
D') bf+ce = 2
E') cf = 4
... remove the f (which is 4/c)
replace E' in c.D' => D") 4b =2c-ec^2
replace E' in c.C' => C") bce+4+c^2 = 0
rest B', C" and D"
... remove b (which is -1-e)
replace D" in 4.B' => B") 2c-ec^2 + 4e = -4
replace ce.B' in C" => C"') -4-c^2+ce^2 = -1
rest C"' and B"

Answer 5

x^4 - X^3 + 2x + 4
if we suppose it's =
(ax² + bx + c ) ( dx² + ex + f )
= adx^4 + aex³ + afx² + bdx³ + bex² + bfx + cdx² + cex + cf
= ad x^4 + (ae + bd )x³ + (af + be + cd ) x² + (bf +ce ) x + cf
ad = 1
ae + bd = -1
af + be +cd = 0
bf + ce = 2
cf = 4
______________when a = d = 1
e + b = -1
f + be + c = 0
bf + ce = 2
cf = 4
_________________multiply the 2nd and the 3rd equations *f
f² + be f + cf = 0 ..................cf = 4
bf² - 2 f+ cfe = 0...................cf =4

_____and solving these equation with comparing
a = 1
b = -2
c = 4
d = 1
e = 1
f = 1
the the equation should be
(x² - 2x + 4 ) (x² + x + 1 )
_______________________________________

if u wanna solve
X^4 - x^3 + 2x + 4 =0
(x² - 2x + 4 ) (x² + x + 1 ) =0
ROOTS :
± √(3) i ...(-1 ± √(3) i ) / 2

Answer 6

(x^2 + x + 1)(x^2 - 2x + 4)
hmmmm
first find all the 4 roots
then, you have to perform the product:
(x - root1)(x - root2)(x- root3)(x - root 4)
my e-mail [e-mail]
I´m brazilian

Answer 7

I think what you're trying to get is:

(x+ (1/2))^2 + (X-1)^2 = 5