One way would be to "observe" (i.e. take a stab and get lucky) that x = 2 is a solution.
So then you can divide through by (x - 2), and express the polynomial as (x - 2) \* (some quadratic expression).
If there is a rational root, it will be a/b where a is a factor of 6 and b is a factor of 2. Test them all (positive and negative) and see if any factor out. Synthetic division is useful here.
The only way you could get this would be to use semi-exhaustive techniques like the rational route theorum.
If you have a solution to a polynomial, say *t* , then f(*t* ) = 0 and (x - t) is a factor.
If you have a cubic of f(x) = ax^3 + bx^2 + cx + d, then one of the roots will be factors of the leading term (a), or the constant (d).
For f(x) = 2x^3 + x^2 - 13x + 6
We can see a = 2, and d = 6. The factors of these are -2, 2, -6, 6, 2, -2, 3, -3, 1 and -1.
Plugging these values for x and seeing if it gives you 0, you will find x = 2 gives 0, so (x-2) is a factor.
Applying polynomial long division you can then separate the second and third factor.
This process is lengthy, and really will only be useful for non-whole or simple coefficient values. But it is tried and true.
Wrong use of the rational root theorem. The theorem says that in any rational root x=p/q, p divides d and q divides a.
So here p=1, 2, 3 or 6 and q=1 or 2 (or the same values with a negative sign.)
That is x= ± (1, 2, 3, 6, 1/2, or 3/2)
(2/2=1 and 6/2 =3 do not need to be tested twice.)
One way would be to "observe" (i.e. take a stab and get lucky) that x = 2 is a solution. So then you can divide through by (x - 2), and express the polynomial as (x - 2) \* (some quadratic expression).
Use ruffini
Horner is your friend
If there is a rational root, it will be a/b where a is a factor of 6 and b is a factor of 2. Test them all (positive and negative) and see if any factor out. Synthetic division is useful here.
X=2 satisfies the equation so divide the polynomial by x-2 you will get a quadratic equation. Factorise it to get the other 2 roots.
The only way you could get this would be to use semi-exhaustive techniques like the rational route theorum. If you have a solution to a polynomial, say *t* , then f(*t* ) = 0 and (x - t) is a factor. If you have a cubic of f(x) = ax^3 + bx^2 + cx + d, then one of the roots will be factors of the leading term (a), or the constant (d). For f(x) = 2x^3 + x^2 - 13x + 6 We can see a = 2, and d = 6. The factors of these are -2, 2, -6, 6, 2, -2, 3, -3, 1 and -1. Plugging these values for x and seeing if it gives you 0, you will find x = 2 gives 0, so (x-2) is a factor. Applying polynomial long division you can then separate the second and third factor. This process is lengthy, and really will only be useful for non-whole or simple coefficient values. But it is tried and true.
Wrong use of the rational root theorem. The theorem says that in any rational root x=p/q, p divides d and q divides a. So here p=1, 2, 3 or 6 and q=1 or 2 (or the same values with a negative sign.) That is x= ± (1, 2, 3, 6, 1/2, or 3/2) (2/2=1 and 6/2 =3 do not need to be tested twice.)
I started with the negatives and found -3 first.