To determine: Factor a) x2−4x+3 b) 2y2−5y+2 c) 6z2−13z+6

Step 1
a) Factoring Quadratics:
We can get a quadraric by multiplying two first-degree polynomials
For example:
${\displaystyle (x+4)(x-2)}$
By using FOIL method
${\displaystyle (x+4)(x-2)=x(x-2)+4(x-2)}$
${\displaystyle =x^2-2x+4x-8}$
${\displaystyle =x^2+2x-8}$
Factoring quadratics needs FOIL backward because factoring is the reverse of multiplication
Given: ${\displaystyle x^2-4x+3}$
We have to find integers b and d such that
${\displaystyle x^2-4x+3=(x+b)(x+d)}$
${\displaystyle =x^2+dx+bx+bd}$
${\displaystyle x^2-4x+3=x^2+(b+d)x+bd}$
Since the constant coefficients on each side of the equation ought to be equal, we must have ${\displaystyle bd=3}$ (i.e) b and d are factors of 3.
Similarly, the coefficients of x must be the same, so that ${\displaystyle b+d=-4}$
The following table shows the possibilities
$$\begin{array}{|cc|}\hline \text{Factors b, d of 3}& \text{Sum}b+d=-4\\ 1\times 3& 1+3=4\\ 1(-3)& 1-3=-2\\ (-1)(-3)& -1-3=-4\\ \hline\end{array}$$
From the above table, the only factors with product 3 and the sum -4 are -1 and -3
So the correct factorization os ${\displaystyle x^2-4x+3=(x-1)(x-3)}$
To check:
${\displaystyle (x-1)(x-3)=x(x-3)-1(x-3)}$
${\displaystyle =x^2-3x-x+3}$
${\displaystyle =x^2-4x+3}$

Step 1
b) Given: ${\displaystyle 2y^2-5y+2}$
We have to find integers a, b, c and d such that
${\displaystyle 2y^2-5y+2=(ay+b)(cy+d)}$
${\displaystyle =ac{y}^2+ady+bcy+bd}$
${\displaystyle 2y^2-55y+2=ac{y}^2+(ad+bc)y+bd}$
Since the coefficient of ${\displaystyle y^2}$ ought to be same on both sides, we find that ${\displaystyle ac=2}$. Likewise, the constant term ${\displaystyle bd=2}$. The positive factors of 2 are 2 and 1. Since the midterm is negative, we consider only negative factors of 2. The possibilities are -2 and -1. Now we have to try various arrangements of these factors until we find the one which gives correct coefficient of y.
${\displaystyle (2y-2)(y-1)=2y(y-1)-2(y-1)}$
${\displaystyle =2y^2-2y-2y+2}$
${\displaystyle =2y^2-4y+2}$
${\displaystyle (2y-1)(y-2)=2y(y-2)-1(y-2)}$
${\displaystyle =2y^2-4y-y+2}$
${\displaystyle =2y^2-5y+2}$
The last trial gives the correct factorization.

$\begin{array}{}\text{Step 1}\\ c)\text{Given:}6z^2-13z+6\\ 6z^2-13z+5=(az+b)(cz+d)\\ =ac{z}^2+adz+bcz+bd\\ 6z^2-13z+6=ac{z}^2+(ad+bc)z+bd\\ \text{Step 2}\\ (3z-1)(2z-6)=3z(2z-6)-1(2z-6)\\ =6z^2-18z-2z+6\\ =6z^2-20z+6\\ (2z-3)(3z-2)=2z(3z-2)-3(3z-2)\\ =6z^2-4z-9z+6\\ =6z^2-13z+6\end{array}$