<s>[INST] How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
 [/INST] For quadratic polynomials of the form $f(x)=ax^2+bx+c$ with $a,b,c$ real numbers, the roots are known to be given by the following:

The Quadratic Formula: The roots of $f(x)=ax^2+bx+c$ are
  $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
  and
  $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

Note that when $b^2-4ac=0$, then the root is "repeated" and when $b^2-4ac<0$ then the roots are non-real complex numbers.
The polynomial then can be factored as $ax^2+bx+c = a(x-x_1)(x-x_2)$

For your specific example, $2x^2-x-3$, this is of the form $ax^2+bx+c$ with $a=2,~b=-1,~c=-3$.
Applying the formula above gives the two roots as:
$x_1=\frac{-(-1)+\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1+\sqrt{1+24}}{4}=\frac{1+5}{4}=\frac{3}{2}$
$x_2=\frac{-(-1)-\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1-\sqrt{1+24}}{4}=\frac{1-5}{4}=-1$
You have then that $2x^2-x-3=2(x-\frac{3}{2})(x-(-1))$, which can be rewritten as $=(2x-3)(x+1)$

note: in some settings we are interested only in factorizations which involve only integers.  It is possible in this setting then that no such factorizations exist and we say the polynomial is irreducible
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