For what it's worth:
I'll just address how to factor quadratics.
First, you need to master performing multiplications of the form $(ax+b)(cx+d)$. Factoring is essentially doing this in reverse.
As for factoring:
Here's one way, by example, to factor certain quadratics; namely, where the quadratic factors as $(ax+b)(cx+d)$, where $a,b,c,d$ are all integers.
To factor $2t^2+7t+3,$ I would obtain the factorization by just "guessing and checking":
You first write $\tag{1} 2t^2+7t+3=(at+b)(ct+d). $
You first guess values for $a$ and $c$. This is an educated guess: The product $ac$ has to be 2 because if you multiplied the right hand side of (1) out, the $t^2$ term will be $ac\cdot t^2$ and the $t^2$ term on the left is just $2t ^2$.
Here's the guess: $a=2, c=1$. Let's try that. This gives us
$\tag{2} 2t^2+7t+3=(2t+b)(t+d). $
Next, we guess the values of $b$ and $d$. But, it's an educated guess. Because the product $bd$ has to be 3 (the constant term on the left hand side of $(2)$ is 3), we guess
$\tag{3} b= 3, d= 1, \quad \text{or} \quad b=-3, d=-1 $ (note we made two guesses at the same time. We know since the product is positive 3, that $b$ and $d$ have the same signs. So we guess magnitudes and try all possible sign combinations).
Then we have
$\tag{4} 2t^2+7t+3=(\color{darkgreen}{2t}+\color{maroon}{\pm 3})(\color{maroon}{t}+\color{darkgreen}{\pm1}). $
Here comes the "check" part: The $t$ term on both sides of $(4)$, after expansion, has to be $7t$. This is obvious looking at the left hand side. How is the $t$ term on the right hand side obtained? Well from two parts: it's the sum of the product of the maroon quantities and the product of the green quantities. You need to decide if any choices of sign will give you $7t$. Here we have, taking the products and summing: $ \color{darkgreen}{2t}\cdot (\color{darkgreen}{+1})+(\color{maroon}{+3})\cdot \color{maroon}t=5t ,\qquad \color{darkgreen}{2t}\cdot (\color{darkgreen}{-1})+(\color{maroon}{-3})\cdot \color{maroon}t=-5t . $ None work...
But that's not altogether bad, at least we know the guesses in (3) do not work.
Let's try a different guess:
$ 2t^2+7t+3=(\color{darkgreen}{2t}+\color{maroon}{\pm 1})(\color{maroon}{t}+\color{darkgreen}{\pm3}). $
Here, if we chose $ \color{maroon}{+1}$ and $\color{darkgreen}{+3}$:
$ 2t^2+7t+3=(\color{darkgreen}{2t}+\color{maroon}{ 1})(\color{maroon}{t}+\color{darkgreen}{ 3}) $ works (check)! And we are done.