x^5 - 7x^2 + 28?

Question

Find all possible rational zeros

Answer 1

I'm assuming you know how to do synthetic substitution, and you understand Descartes Rule of Signs.

p/q = { ±28, ± 14, ±7, ±4, ±2, ±1 }

Descartes:
Positive test: + + + - - +: 2 changes, so 2 real positive roots or none.
Negative test: - - - - - +: 1 change, so must be 1 real negative root.

So try the negative roots first.

-02 | +01 +00 +00 −07 +00 +28
-----| +00 −02 +04 −08 +30 −60
-----|----------------------------------
-----| +01 −02 +04 −15 +30 +20 ....

OK I tried every possible rational root, and none of them worked.

So I used the equation solver at:

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic

And it comes up with 1 negative rational root, and 4 complex roots:

-1.5926
-0.890569 - 2.07391i
-0.890569 + 2.07391i
1.6867 - 0.778276i
1.6867 + 0.778276i

So... I don't think you can solve this one exactly.

Where did you get the problem?

Answer 2

As y=x^5 –7xx +28, then y’= 5x^4 –14x =x(5x^3-14); let p=5^(1/3) and q=14^(1/3); then y’= x(px –q)(ppxx +pqx +qq); thus y(x) may have only 2 extreme points x1=0, x2 =q/p; Now y’’(x) = 20xxx –14; y’’(x1)= y’’(0)= -14 < 0, hence x1=0 s for local max, hence x2=q/p is for local min; thus there could be 1 or 3 real roots; y(x1) = 28 >0 , y(x2) = (q/p)^5 –7(q/p)^2 +28 = (14/5)^(5/3) –7(14/5)^(2/3) +28 =~20 >0; thus there is only 1 root x3 < x1; let us find it:
Approximation by means of tangents yk=y’(xk)*xk +b, hence b= yk -y’(xk)*xk and next x:
x[k+1] = -b/y’(xk) = -yk/y’(xk) +xk, starting with x=-2;

-2.000000000 | –32.00000000
-1.703703704 | -6.672158095
-1.602575876 | -0.548184588
-1.592683631 | -0.004687804
-1.592597569 | -3.51062E-07
-1.592597563 | –0.000000000
result x3= -1.592597563