How do you factor a quadratic expression, without using the formula?

I am asked to factor $2x^2 -3x+1=0 $ using factorization, but I run into fractions, and it becomes very messy and complicated to deal with, especially since specifically asked not to use the formula. How exactly would I go on about factorizing this?

The usual method that is taught first should work here: stare at

$$ (ax+b)(cx+d) = 2x^2 - 3x + 1$$

until you see what $a,b,c,d$ have to be. If you need to see the left hand side expanded, it is

$$ (ac) x^2 + (ad+bc) x + (bd) $$

Hint

Write $$2x^2 -3x+1=2\Big(x^2-\frac{3}{2}x+\frac{1}{2}\Big)$$ Now $$x^2-\frac{3}{2}x+\frac{1}{2}=\Big(x-\frac{3}{4}\Big)^2-\frac{9}{16}+\frac{1}{2}=\Big(x-\frac{3}{4}\Big)^2-\frac{1}{16}$$

I am sure that you can take from here.

$$2x^2-3x+1~=~2x^2-2x-x+1~=~2x(x-1)-(x-1)~=~(x-1)(2x-1).$$

Hint $\ $ Reduce to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $=1)$ as follows:

$$\quad\ \ \begin{eqnarray} f \, &\,=\,& \ \, 2\ x^2-\ 3\ x\,\ +\ \ 1\\ \Rightarrow\ 2f\, &\,=\,&\!\ (2x)^2\! -3(2x)+2\\ &\,=\,& \ \ \ \color{#c00}{X^2-\, 3\ X\,\ +\,\ 2},\,\ \ X\, =\, 2x\\ &\,=\,& \ \ (X-2)\ (X-\,1)\\ &\,=\,& \ (2x-2)\,(2x-1)\\ \Rightarrow\ \ f\:=\: 2^{-1}\,(2f)\, &\,=\,& \ \ \, (x- 1)\ (2x\,-1)\\ \end{eqnarray}$$

Remark $\ $ If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply

$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$

Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. This is sometimes called the AC method (cf. below). It works for higher degree polynomials too: we can reduce the problem of factoring a non-monic polynomial to that of factoring a monic polynomial by scaling by a $ $ power of the lead coefficient $\rm\:a\:$ then changing variables: $\rm\ X = a\:x$

$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: X^2 + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\smash[t]{\overbrace{ac}^{\rm\qquad\ \ \ \ \ {\bf AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! =\, g(X),\ \ \ X = a\:x \\ \\ \rm\: a^{n-1}(a\:x^n\! + b\:x^{n-1}\!+\cdots+d\:x + c) &\,=\,&\!\!\rm\: X^n\! + b\:X^{n-1}\!+\cdots+a^{n-2}d\:X + a^{n-1}c \end{eqnarray}$$ After factoring the monic $\rm\,g(X)\, =\, a^{n-1}f(x),\,$ we are guaranteed that the transformation reverses to yield a factorization of $\rm\:f,\ $ since $\rm\ a^{n-1}$ must divide into the factors of $\rm\ g\ $ by Gauss' Lemma, i.e. primes $\,p\in\rm\mathbb Z\,$ remain prime in $\rm\,\mathbb Z[X],\,$ so $\rm\ p\ |\ g_1(x)\:g_2(x)\,$ $\Rightarrow$ $\,\rm\:p\:|\:g_1(x)\:$ or $\rm\:p\:|\:g_2(x).$

This method also works for multivariate polynomial factorization, e.g. it applies to this question.

For more on the ring-theoretic concepts at the heart of this see this answer.