3x^4-19x^3+39x^2-29x+6 what is the 3 rational roots

Question

i need the answer

Answer 1

If we take the coefficients: 3 -19 39 -29 6 and add them together we get zero. This tells us that x=1 is a root, which we can divide by using synthetic division:

1 | 3 -19..39 -29..6

.....3....3 -16..23 -6

.....3 -16..23..-6 | 0

Now we have a cubic: 3x^3-16x^2+23x-6. This time the coefficients do not add up to zero, so we need to look for another solution. Try x=2: 24-64+46-6=0, so x=2 is a solution. Divide again:

2 | 3 -16..23..-6

.....3....6 -20...6

.....3 -10....3 | 0

Now we have a quadratic: 3x^2-10x+3=(3x-1)(x-3). So the roots are 1, 2, 1/3, 3.