Method to factor an expression As the title says, i want to factorize an expression, but i don't have any clue how to proceed.
Here is the expression :
$2x² -7x +3$
And here is the factorized form :
$(x-3)(2x - 1)$
My question is, which method or rule to use to go from first to second ?
please note that I am a beginner, and the only question that i found which is closer to mine is this post.
Thank you for your help !

In general ,if $x_1$ and $x_2$ are roots of  $$\underbrace{a}_{\neq 0}x^2+bx+c=0$$ then  $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\text{sum of the roots}=-\frac{b}{a}\;\;\&\;\;\text{product of the roots}=\frac{c}{a}$$

So your case, $x_1+x_2= \frac{7}{2}$ and $x_1x_2=\frac{3}{2}$
Now solve these to get $x_1$ and $x_2$ to finish your conclusion

To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-\frac{7}{2}x+\frac{3}{2}=0$$ which means $$x^2-2\left(\frac{7}{4}\right)x=-\frac{3}{2}$$ which is same as $$x^2-2\left(\frac{7}{4}\right)x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}=\frac{25}{16}$$ so $$\left(x-\frac{7}{4}\right)^2=\frac{25}{16}$$ and so $$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}$$ so $$x=\frac{7}{4} \pm \frac{5}{4}=\frac{7\pm 5}{4}$$
