SOLUTION: find all real and complex zeros of a polynomial P(x) = x^4+x^3+2x-4

Algebra ->  Functions -> SOLUTION: find all real and complex zeros of a polynomial P(x) = x^4+x^3+2x-4      Log On


   

Question 1158973: find all real and complex zeros of a polynomial
P(x) = x^4+x^3+2x-4
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
The expectation may be to use root-checking and synthetic division but the polynomial can also be treated:
x%5E4-4%2Bx%5E3-2x
%28x%5E2-2%29%28x%5E2%2B2%29%2Bx%28x%5E2-2%29
%28x%5E2-2%29%28x%5E2%2Bx%2B2%29
Two of the roots are irrational.
Two other roots are complex, not real. General solution to quadratic formula will help.

Answer by ikleyn(52748) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The  "solution"  by @josgarithmetic is  TOTALLY  INCORRECT,

            so, for your safety,  you simply ignore it . . .

            I came to bring the correct solution.

            I will use grouping,  re-grouping and factoring method.  Watch my steps. 


   x4 + x^3 + 2x - 4 =  (grouping) = (x^4 - 4) + (x^3 + 2x) =  (factoring)  = (x^2-2)*(x^2+2) + x*(x^2+2) =  

= (factor out the common factor (x^2+2) = (x^2+2)*(x^2 - 2 + x) = (regrouping) = (x^2+2)*(x^2 + x - 2) =

= (x^2+2)*(x+2)*(x-1).


Factoring completed.


The real roots are 1 and -2.


The complex roots are  i%2Asqrt%282%29  and  -i%2Asqrt%282%29.

Solved.

---------------- 

    Another way is to guess that 1 and -2 are the roots;

    then divide the original polynomial by (x-1)*(x+2), and then to finish with the quotient.



    Third way would be to plot the graph of the polynomial, which will point the roots 1 and -2.