### 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: To find all $t$ such that $x-t$ is a factor of $6x^2+13x-5,$ I can use the factor theorem, which says that $x-t$ is a factor if and only if $f(t)=0,$ where $f(x)=6x^2+13x-5.$
So I need to find the zeros of $f(x),$ or the solutions to the equation $6x^2+13x-5=0.$
I can try to factor this quadratic equation, or use the quadratic formula, or complete the square.
I'll try factoring first, since it might be easier.
To factor a quadratic equation of the form $ax^2+bx+c,$ I need to find two numbers that multiply to $ac$ and add to $b.$
In this case, $a=6, b=13, c=-5,$ so $ac=-30.$
I can list the factors of $-30$ and see if any pair adds to $13:$

$-30 = (-1)(30) = (-2)(15) = (-3)(10) = (-5)(6) = (-6)(5) = (-10)(3) = (-15)(2) = (-30)(1)$
The only pair that adds to $13$ is $-2$ and $15,$ so I can write:

$6x^2+13x-5 = 6x^2-2x+15x-5 = 2x(3x-1)+5(3x-1) = (2x+5)(3x-1)$
Now I have factored the quadratic equation, so I can set each factor equal to zero and solve for $x:$

$2x+5=0 \implies x=-\frac{5}{2}$

$3x-1=0 \implies x=\frac{1}{3}$
These are the two values of $x$ that make $f(x)=0,$ so by the factor theorem, these are the values of $t$ that make $x-t$ a factor of $f(x).$
Therefore, the answer is $t=-\frac{5}{2}, \frac{1}{3}.$