Consider $t^{2}+4t-20-5t$. Multiply and combine like terms.
$$t^{2}-t-20$$
Consider $t^{2}-t-20$. Factor the expression by grouping. First, the expression needs to be rewritten as $t^{2}+at+bt-20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=1\left(-20\right)=-20$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-20$.
$$1,-20$$ $$2,-10$$ $$4,-5$$
Calculate the sum for each pair.
$$1-20=-19$$ $$2-10=-8$$ $$4-5=-1$$
The solution is the pair that gives sum $-1$.
$$a=-5$$ $$b=4$$
Rewrite $t^{2}-t-20$ as $\left(t^{2}-5t\right)+\left(4t-20\right)$.
$$\left(t^{2}-5t\right)+\left(4t-20\right)$$
Factor out $t$ in the first and $4$ in the second group.
$$t\left(t-5\right)+4\left(t-5\right)$$
Factor out common term $t-5$ by using distributive property.
