Step-by-step explanation:
This is a polynomial so we can use rational roots Theorem to solve the equation.
The rational roots simply states that the roots of a polynomial, in the form of
[tex]p{x}^{n} + ax {}^{n - 1} + bx {}^{n - 2} .....r {}^{0} [/tex]
The possible roots of the polynomial are the factors of
p/r.
We say r^0 to represent the constant and p to represent the leading coeffceint.
So the rational roots states the possible roots of a polynomial is
the factors of leading coeffceint/ the factors of the constant.
In this case, the polynomial leading coeffecient is 3 and its constant is 6 so we do the factors of 3 divided by the factors of 6.
The factors of 3, are plus or minus 1 and 3. divided by factors of 6 which are plus or minus 1,2,3,6. So our possible roots are
positive or negative (1,1/2, 1/3,1/6, 3, 3/2).
Now, we see which of the following roots will that the polynomial, P will equal zero.
It seems that -1 can work so by definition, (x+1) is the a factor of the polynomial. So now we use synetheic or long division to cancel out that factor.
So our factored version of the polynomial is
[tex](x + 1)(3x {}^{3} - 14 {x}^{2} + 13x + 6)[/tex]
Now can we continue and factor the right side of the factors.
3 also works so x-3 is a factor as well so
[tex](3 {x}^{2} - 5x - 2)(x + 1)(x - 3)[/tex]
Now factor the quadratic using factoring by grouping
[tex]3 {x}^{2} - 5x - 2 = 3 {x}^{2} - 6x + x - 2 = 3x(x - 2) + 1(x - 2) [/tex]
So our factor are
[tex](3x + 1)(x - 2)[/tex]
So in conclusion our factors are
[tex](3x + 1)(x - 2)(x + 1)(x - 3)[/tex]
And our x values are -1/3, 2, -1, and 3.
