We can use the Rational Root Theorem to answer this question.
It says that, for a polynomial with integer coefficients, each rational solution:
[tex]x=\frac{p}{q}[/tex]In lowest terms, can be found as:
- p is an integer factor of the constant term of the polynomial.
- q is an integer factor of the leading coefficient of the highest degree term of the polynomial.
That is, given the polynomial:
[tex]6x^4+2x^3-3x^2+2[/tex]p must be a factor of 2 and q must be a factor of 6.
We also need to consider both positive and negative roots.
The factors of 2 are 1 and 2, so these are the options for p.
The factors of 6 are 1, 2, 3 and 6, so these are the options for q.
Now, we need to write every combination of p and q, considering both + and - signs:
[tex]\pm\frac{1}{6},\pm\frac{1}{3},\pm\frac{1}{2},\pm\frac{1}{1},\pm\frac{2}{6},\pm\frac{2}{3},\pm\frac{2}{2},\pm\frac{2}{1}[/tex]Now, we simplify:
[tex]\pm\frac{1}{6},\pm\frac{1}{3},\pm\frac{1}{2},\pm1,\pm\frac{1}{3},\pm\frac{2}{3},\pm1,\pm2[/tex]And we remove the repeating ones:
[tex]\pm\frac{1}{6},\pm\frac{1}{3},\pm\frac{1}{2},\pm\frac{2}{3},\pm1,\pm2[/tex]by comparison, we can see that it corresponds to the third alternative.
