Find the Roots (Zeros) f(x)=x^3-10x^2+4x-24

Problem

ƒ(x)=x3−10*x2+4*x−24

Solution

1. Identify the goal, which is to find the values of x such that ƒ(x)=0

x3−10*x2+4*x−24=0

2. Apply the Rational Root Theorem to list potential rational roots by dividing the factors of the constant term −24 by the factors of the leading coefficient 1

Possible roots=±1,±2,±3,±4,±6,±8,±12,±24

3. Test the value x=10 using synthetic division or substitution to see if it is a root.

ƒ(10)=(10)3−10*(10)2+4*(10)−24

ƒ(10)=1000−1000+40−24

ƒ(10)=16

Since ƒ(10)≠0 10 is not a root.

4. Test the value x=6 using synthetic division.

ƒ(6)=(6)3−10*(6)2+4*(6)−24

ƒ(6)=216−360+24−24

ƒ(6)=−144

Since ƒ(6)≠0 6 is not a root.

5. Factor by grouping if possible, or continue testing. Since standard grouping fails, we check for other integer roots. Testing x=−2

ƒ*(−2)=(−2)3−10*(−2)2+4*(−2)−24

ƒ*(−2)=−8−40−8−24=−80

6. Observe that the function does not have simple integer roots. We use the cubic formula or numerical methods to find the real root. Let x=y+10/3 to depress the cubic.

y3−88/3*y−2248/27=0

7. Solve for the real root using the cubic root extraction. The approximate real root is x≈9.68 The other two roots are complex conjugates.

8. Re-evaluate the expression for potential typos. If the expression was x3−10*x2+24*x the roots would be 0, 4, 6.G*i*v*e*n*t*h*e*o*r*i*g*i*n*a*l*e*x*p*r*e*s(s(i))*o*n^3 - 10x^2 + 4x - 24$, the roots are:

(x_1)=10/3+√(3,1124/27+√(,(1263360−681472)/729))+…

(x_1)≈9.682

(x_2,3)≈0.159±1.565*i

Final Answer

x≈9.682,x≈0.159+1.565*i,x≈0.159−1.565*i

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