Find the Roots (Zeros) f(x)=x^4-3x^3-2x^2+3x-5

Problem

ƒ(x)=x^4−3*x^3−2*x^2+3*x−5

Solution

1. Identify the possible rational roots using the Rational Root Theorem, which suggests testing factors of the constant term −5 divided by factors of the leading coefficient 1

2. Test the possible rational roots ±1 and ±5 using synthetic division or direct substitution.

3. Evaluate ƒ(1)=1−3−2+3−5=−6

4. Evaluate ƒ*(−1)=1+3−2−3−5=−6

5. Evaluate ƒ(5)=625−375−50+15−5=210

6. Evaluate ƒ*(−5)=625+375−50−15−5=930

7. Observe that there are no rational roots, so we must use the quadratic formula on a depressed polynomial or numerical methods; however, we can check for factors by grouping or using the substitution y=x−3/4 to depress the quartic.

8. Factor the expression by grouping terms: (x^4−5*x^2)+(−3*x^3+3*x)+(3*x^2−5) which does not yield a common factor.

9. Rearrange and group as (x^4−2*x^2−5)−(3*x^3−3*x)=(x^4−2*x^2−5)−3*x*(x^2−1)

10. Apply the quadratic formula to the factors of the form (x^2+a*x+b)*(x^2+c*x+d)=x^4−3*x^3−2*x^2+3*x−5

11. Solve the system of equations for coefficients: a+c=−3 a*c+b+d=−2 a*d+b*c=3 and b*d=−5

12. Test integer values for b and d such as b=1,d=−5 This gives a−5*c=3 and a+c=−3

13. Calculate c=−1 and a=−2 from the system.

14. Verify the middle term: a*c+b+d=(−2)*(−1)+1−5=2+1−5=−2 which matches the original polynomial.

15. Factor the polynomial into two quadratics: (x^2−2*x+1)*(x^2−x−5)=0 Wait, checking a*d+b*c (−2)*(−5)+(1)*(−1)=10−1=9≠3

16. Test b=−1,d=5 This gives 5*a−c=3 and a+c=−3

17. Calculate 6*a=0⇒a=0 and c=−3

18. Verify the middle term: a*c+b+d=(0)*(−3)−1+5=4≠−2

19. Test b=5,d=−1 This gives −a+5*c=3 and a+c=−3

20. Calculate 6*c=0⇒c=0 and a=−3

21. Verify the middle term: a*c+b+d=(−3)*(0)+5−1=4≠−2

22. Test b=−5,d=1 This gives a−5*c=3 and a+c=−3

23. Calculate c=−1 and a=−2

24. Verify the middle term: a*c+b+d=(−2)*(−1)−5+1=2−5+1=−2

25. Verify the linear term: a*d+b*c=(−2)*(1)+(−5)*(−1)=−2+5=3

26. Factor the polynomial as (x^2−2*x−5)*(x^2−x+1)=0

27. Solve x^2−2*x−5=0 using the quadratic formula: x=(2±√(,4−4*(1)*(−5)))/2=(2±√(,24))/2=1±√(,6)

28. Solve x^2−x+1=0 using the quadratic formula: x=(1±√(,1−4*(1)*(1)))/2=(1±√(,−3))/2=(1±i√(,3))/2

Final Answer

x=1+√(,6),1−√(,6),(1+i√(,3))/2,(1−i√(,3))/2

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