### Instruction: Find all $t$ such that $x-t$ is a factor of $6x^2+13x-5.$

Enter your answer as a list separated by commas.### Response: I need to find the values of $t$ that make the polynomial $6x^2+13x-5$ divisible by $x-t$.
One way to do this is to use polynomial long division and see what the remainder is.
If the remainder is zero, then $x-t$ is a factor.
Alternatively, I could use the factor theorem, which says that if $a$ is a root of a polynomial $p(x)$, then $x-a$ is a factor of $p(x)$.
So I can try to find the roots of $6x^2+13x-5$ and then use them to find the values of $t$.
I think I will use the second method, since it seems faster and easier.
To find the roots of $6x^2+13x-5$, I can try to factor it using the AC method.
I look for two numbers that multiply to $6 \times (-5) = -30$ and add to $13$.
I see that $15$ and $-2$ work.
So I can write $6x^2+13x-5 = 6x^2+15x-2x-5$.
Then I can group the first two terms and the last two terms and factor out the common factors.
I get $6x^2+13x-5 = 3x(2x+5)-1(2x+5)$.
Now I can factor out $(2x+5)$ from both terms and get $6x^2+13x-5 = (3x-1)(2x+5)$.
So the roots of $6x^2+13x-5$ are $x = \frac{1}{3}$ and $x = -\frac{5}{2}$.
That means $x-\frac{1}{3}$ and $x+\frac{5}{2}$ are factors of $6x^2+13x-5$.
Therefore, the values of $t$ that make $x-t$ a factor of $6x^2+13x-5$ are $t = \frac{1}{3}$ and $t = -\frac{5}{2}$.