Find the Roots (Zeros) f(x)=3x^4-11x^3-x^2+19x+6

Problem

ƒ(x)=3*x4−11*x3−x2+19*x+6

Solution

1. Identify possible rational roots using the Rational Root Theorem, which suggests testing factors of the constant term 6 divided by factors of the leading coefficient 3 Possible roots are ±1,±2,±3,±6,±1/3,±2/3

2. Test x=−1 using synthetic division or direct substitution. Since ƒ*(−1)=3*(−1)4−11*(−1)3−(−1)2+19*(−1)+6=3+11−1−19+6=0 x=−1 is a root.

3. Divide the polynomial by (x+1) to get the depressed polynomial 3*x3−14*x2+13*x+6

4. Test x=2 in the new polynomial. Since 3*(2)3−14*(2)2+13*(2)+6=24−56+26+6=0 x=2 is a root.

5. Divide the depressed polynomial by (x−2) to obtain the quadratic 3*x2−8*x−3

6. Factor the quadratic 3*x2−8*x−3 into (3*x+1)*(x−3)

7. Solve for the remaining roots by setting each factor to zero, which gives x=−1/3 and x=3

Final Answer

x=−1,2,3,−1/3

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