Question

slove the equation x^3+6x+20,one root being 1+3i​

Answer 1

Answer:

Since all coefficients of x3+6x+20=0 are real numbers, if one root is 1+3i, then second root will be 1−3i.

Cubic polynomial will have at least one real root and let that be ‘a’.

We can find the 3rd root in one of the following ways:

Sum of the roots =a+1+3i+1−3i=− Coefficient of x2=0

=>a+2=0=>a=−2

Product of the roots =(a)(1+3i)(1–3i)=− Coefficient of x^0 or (−1)∗Constant

=>a(1−9i2)=−20=>a(10)=−20

=>a=−2

Roots are: (1+3i),(1–3i),(−2)