SOLUTION: Find all zeros of the polynomial. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) P(x) = x^4 − x^2 + 2x + 2

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find all zeros of the polynomial. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) P(x) = x^4 − x^2 + 2x + 2      Log On


   



Question 1204258: Find all zeros of the polynomial. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)
P(x) = x^4 − x^2 + 2x + 2

Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
P%28x%29+=+x%5E4+-+x%5E2+%2B+2x+%2B+2...add 2x%5E3+ and -2x%5E3+, write -+x%5E2 as -2x%5E2%2B+x%5E2, and 2x+ as 4x+-2x
P%28x%29+=+x%5E4+-+2x%5E3+-2x%5E2+%2B2x%5E3-4x%5E2%2B4x+%2B+x%5E2+-+2x+%2B+2....group

P%28x%29+=+x%5E2%28x%5E2+%2B2x%2B+1%29+-2x+%28x%5E2%2B2x%2B1%29+%2B2%28+x%5E2%2B2x++%2B+1%29
P%28x%29+=+%28x%5E2%2B2x+%2B+1%29+%28x%5E2+-+2x+%2B+2%29
P%28x%29+=+%28x+%2B+1%29%5E2+%28x%5E2+-+2+x+%2B+2%29

zeros:

0=+%28x+%2B+1%29%5E2+%28x%5E2+-+2+x+%2B+2%29
+%28x+%2B+1%29%5E2+=0=> x=-1 multiplicity 2
x%5E2+-+2x+%2B+2=0 using quadratic formula

x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A2+%29%29%2F%282%2A1%29+
x+=+%282+%2B-+sqrt%28+4-8+%29%29%2F2+
x+=+%282+%2B-+sqrt%28+-4+%29%29%2F2+
x+=+%282+%2B-+2i%29%2F2+
x+=+%281+%2B-+i%29+
x+=+1+%2B+i+ or x+=+1+-+i+
zeros are:-1,-1,+1+%2B+i+ ,1+-+i+

Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find all zeros of the polynomial. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)
P(x) = x^4 − x^2 + 2x + 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I will factor the polynomial, using the grouping method


    P(x)  =  x^4 − x^2 + 2x + 2 = (x^4 − x^2) + (2x + 2) = x^2(x^2-1) + 2(x+1) = x^2*(x+1)*(x-1) + 2(x+1) =

               // now I see that there is common factor (x+1), and I factor it out

        = (x+1) * (x^2*(x-1) + 2) = (x+1) * (x^3 - x^2 + 2).


Thus, I just extracted one root x= -1 and separated one relating factor (x+1).


Now I will work with the polynomial of the degree 3 in parentheses.


    x^3 - x^2 + 2 = (x^3+x^2) - 2x^2 + 2 = x^2(x+1) - 2(x^2-1) = x^2(x+1) - 2(x-1)*(x+1) = 

               // now I see that there ios common factor (x+1), and I factor it out

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


So, I separated one more root x= -1 and the relating factor (x+1).


Now I will work with the quadratic polynomial in parentheses.

It has the roots

    x%5B1%2C2%5D = %282+%2B-+sqrt%282%5E2+-+4%2A1%2A2%29%29%2F2 = %282+%2B-+sqrt%28-4%29%29%2F2 = 1+%2B-+i.


ANSWER.  The roots are -1, -1,  1+i  and  1-i.

Solved.