Find the Roots/Zeros Using the Rational Roots Test 3x^4-10x^3-9x^2+40x-12

Problem

3*x^4−10*x^3−9*x^2+40*x−12

Solution

1. Identify the constant term (a_0)=−12 and the leading coefficient (a_n)=3

2. List all possible factors of the constant term p∈{±1,±2,±3,±4,±6,±12} and all possible factors of the leading coefficient q∈{1,3}

3. Determine the set of possible rational roots p/q∈{±1,±2,±3,±4,±6,±12,±1/3,±2/3,±4/3}

4. Test values using synthetic division or substitution. Testing x=2

3*(2)^4−10*(2)^3−9*(2)^2+40*(2)−12=48−80−36+80−12=0

5. Divide the polynomial by (x−2) to get the quotient 3*x^3−4*x^2−17*x+6

6. Test values for the new quotient. Testing x=3

3*(3)^3−4*(3)^2−17*(3)+6=81−36−51+6=0

7. Divide the quotient by (x−3) to get the quadratic 3*x^2+5*x−2

8. Factor the quadratic 3*x^2+5*x−2=(3*x−1)*(x+2)

9. Solve for the remaining roots by setting each factor to zero: x=1/3 and x=−2

Final Answer

3*x^4−10*x^3−9*x^2+40*x−12=0⇒x∈{2,3,1/3,−2}

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