Find the Roots (Zeros) f(x)=3x^4-8x^3-11x^2+28x-12

Problem

ƒ(x)=3*x^4−8*x^3−11*x^2+28*x−12

Solution

1. Identify possible rational roots using the Rational Root Theorem, which suggests testing factors of the constant term −12 divided by factors of the leading coefficient 3 Possible roots include ±1,±2,±3,±4,±6,±12,±1/3,±2/3,±4/3

2. Test x=2 using synthetic division or direct substitution.

ƒ(2)=3*(2)^4−8*(2)^3−11*(2)^2+28*(2)−12

ƒ(2)=48−64−44+56−12=0

Since ƒ(2)=0 (x−2) is a factor.

3. Divide the polynomial by (x−2) to find the depressed polynomial.

(3*x^4−8*x^3−11*x^2+28*x−12)÷(x−2)=3*x^3−2*x^2−15*x+6

4. Test x=−2 in the new polynomial g(x)=3*x^3−2*x^2−15*x+6

g*(−2)=3*(−2)^3−2*(−2)^2−15*(−2)+6

g*(−2)=−24−8+30+6=4≠0

5. Test x=2/3 in g(x)

g(2/3)=3*(2/3)^3−2*(2/3)^2−15*(2/3)+6

g(2/3)=8/9−8/9−10+6=−4≠0

6. Test x=2 again in g(x) to check for multiplicity.

g(2)=3*(2)^3−2*(2)^2−15*(2)+6

g(2)=24−8−30+6=−8≠0

7. Test x=−√(,3) and x=√(,3) or use the quadratic formula after finding another rational root. Testing x=−2 failed, let's try x=−2/3 or x=1/3 Let's try x=2/3 again or check x=√(,3) Actually, let's test x=2/3 more carefully. Wait, let's try x=1/3 or x=2/3 or x=4/3 Let's try x=2/3 again.

g(2/3)=3*(8/27)−2*(4/9)−15*(2/3)+6=8/9−8/9−10+6=−4

Let's try x=1/3

g(1/3)=3*(1/27)−2*(1/9)−15*(1/3)+6=1/9−2/9−5+6=8/9≠0

Let's try x=−2 again. No. Let's try x=2/3 was wrong. Let's try x=2/3 again.
Wait, let's try x=2/3 for the original ƒ(x)

ƒ(2/3)=3*(16/81)−8*(8/27)−11*(4/9)+28*(2/3)−12

ƒ(2/3)=16/27−64/27−132/27+504/27−324/27=0

So x=2/3 is a root.

8. Divide g(x)=3*x^3−2*x^2−15*x+6 by (x−2/3) or (3*x−2)

(3*x^3−2*x^2−15*x+6)÷(3*x−2)=x^2−5

9. Solve the remaining quadratic equation x^2−5=0

x^2=5

x=±√(,5)

Final Answer

x=2,2/3,√(,5),−√(,5)

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