Find the Roots (Zeros) f(x)=x^4-2x^3+38x^2-2x+37

Problem

ƒ(x)=x^4−2*x^3+38*x^2−2*x+37

Solution

1. Identify the potential rational roots using the Rational Root Theorem, which suggests testing factors of the constant term 37 divided by factors of the leading coefficient 1 The possible rational roots are ±1 and ±37

2. Test x=1 by substituting it into the function: ƒ(1)=1^4−2*(1)^3+38*(1)^2−2*(1)+37=1−2+38−2+37=72 Since ƒ(1)≠0 1 is not a root.

3. Test x=−1 by substituting it into the function: ƒ*(−1)=(−1)^4−2*(−1)^3+38*(−1)^2−2*(−1)+37=1+2+38+2+37=80 Since ƒ*(−1)≠0 −1 is not a root.

4. Rearrange the polynomial to look for patterns or factor by grouping. Notice the coefficients are symmetric: $1, -2, 38, -2, 1$. Group the terms with similar coefficients:

(x^4+37*x^2+37)+(−2*x^3−2*x)+x^2=0

This does not immediately factor, so we test for imaginary roots.

5. Test x=i by substituting it into the function: ƒ(i)=i^4−2*i^3+38*i^2−2*i+37

6. Simplify the powers of i i^4=1 i^3=−i and i^2=−1

ƒ(i)=1−2*(−i)+38*(−1)−2*i+37

ƒ(i)=1+2*i−38−2*i+37

ƒ(i)=0

Since ƒ(i)=0 x=i is a root, and its conjugate x=−i must also be a root.

7. Divide the polynomial by the factor (x−i)*(x+i)=x^2+1 using long division or synthetic division.

(x^4−2*x^3+38*x^2−2*x+37)÷(x^2+1)=x^2−2*x+37

8. Solve the remaining quadratic equation x^2−2*x+37=0 using the quadratic formula x=(−b±√(,b^2−4*a*c))/(2*a)

x=(−(−2)±√(,(−2)^2−4*(1)*(37)))/(2*(1))

x=(2±√(,4−148))/2

x=(2±√(,−144))/2

x=(2±12*i)/2

x=1±6*i

Final Answer

x=i,−i,1+6*i,1−6*i

Want more problems? Check here!