SOLUTION: Degree 4; zeros: 5+5i; -4 multiplicity 2

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Degree 4; zeros: 5+5i; -4 multiplicity 2      Log On


   



Question 1129941: Degree 4; zeros: 5+5i; -4 multiplicity 2
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The statement of the problem is not complete. Without the stated requirement that the polynomial have rational coefficients, the fourth root could be any complex number.

Assuming rational coefficients, the four roots are -4, -4, 5+5i, and 5-5i.

The quadratic polynomial with roots -4 and -4 is

%28x%2B4%29%28x%2B4%29+=+x%5E2%2B8x%2B16

The quadratic polynomial with roots 5+5i and 5-5i is



An alternative way to find the quadratic polynomial with roots 5+5i and 5-5i is to use the fact that the quadratic polynomial ax^2+bx+c has roots whose sum is -b/a and whose product is c/a.

The sum of the roots 5+5i and 5-5i is 10; so the linear coefficient in the quadratic polynomial is -10.

The product of the roots 5+5i and 5-5i is 25-25i^2 = 25+25 = 50; so the constant in the quadratic polynomial is 50.

Then the quadratic polynomial with roots 5+5i and 5-5i is x^2-10x+50.

Finally the polynomial with roots -4, -4, 5+5i and 5-5i is

%28x%5E2%2B8x%2B16%29%28x%5E2-10x%2B50%29+=+x%5E4-2x%5E3-14x%5E2%2B240x%2B800