Given that one root of x^4+10x^3+35x^2+50x+34=0 is -4+i, find the other three roots

Question

I have not done an equation like this in a long time and I need help.

Answer 1

Since one root is -4+i then the conjugate -4-i is also a root. Factors would be (x+4-i) and (x+4+i). Multiply them together to get x^2+8x+17. Using long division, divide your given polynomial by x^2+8x+17 to get a quotient of x^2+2x+2. Use quadratic formula on this to get other two roots of -1-i and -1+i. Good luck!

Answer 2

Complex roots come in conjugate pairs. Therefore, a second root is -4-i.

 

Now, the sum of these two roots is (-4+i)+(-4-i)=-8 and the product of these two roots is 4^2+1^2=16+1=17. Therefore, x^2+8x+17 is a polynomial whose roots give that conjugate pair (the number multiplying the x is the negative of the sum of the roots, and the constant is the product of the roots).

 

We can now use long division to divide the given polynomial by x^2 +8x+17 to get the quadratic polynomial x^2 +2x+2. We can then find the roots of that quadratic polynomial by completing the square or using the quadratic formula (factoring will also sometimes work, but it won't work here). We get -1+i and -1-i as the other two roots.