Juan Hewlett

2021-12-28

To determine: Factor
a) ${x}^{2}-4x+3$
b) $2{y}^{2}-5y+2$
c) $6{z}^{2}-13z+6$

abonirali59

Step 1
We can get a quadraric by multiplying two first-degree polynomials
For example:
$\left(x+4\right)\left(x-2\right)$
By using FOIL method
$\left(x+4\right)\left(x-2\right)=x\left(x-2\right)+4\left(x-2\right)$
$={x}^{2}-2x+4x-8$
$={x}^{2}+2x-8$
Factoring quadratics needs FOIL backward because factoring is the reverse of multiplication
Given: ${x}^{2}-4x+3$
We have to find integers b and d such that
${x}^{2}-4x+3=\left(x+b\right)\left(x+d\right)$
$={x}^{2}+dx+bx+bd$
${x}^{2}-4x+3={x}^{2}+\left(b+d\right)x+bd$
Since the constant coefficients on each side of the equation ought to be equal, we must have $bd=3$ (i.e) b and d are factors of 3.
Similarly, the coefficients of x must be the same, so that $b+d=-4$
The following table shows the possibilities

From the above table, the only factors with product 3 and the sum -4 are -1 and -3
So the correct factorization os ${x}^{2}-4x+3=\left(x-1\right)\left(x-3\right)$
To check:
$\left(x-1\right)\left(x-3\right)=x\left(x-3\right)-1\left(x-3\right)$
$={x}^{2}-3x-x+3$
$={x}^{2}-4x+3$

Juan Spiller

Step 1
b) Given: $2{y}^{2}-5y+2$
We have to find integers a, b, c and d such that
$2{y}^{2}-5y+2=\left(ay+b\right)\left(cy+d\right)$
$=ac{y}^{2}+ady+bcy+bd$
$2{y}^{2}-55y+2=ac{y}^{2}+\left(ad+bc\right)y+bd$
Since the coefficient of ${y}^{2}$ ought to be same on both sides, we find that $ac=2$. Likewise, the constant term $bd=2$. 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.
$\left(2y-2\right)\left(y-1\right)=2y\left(y-1\right)-2\left(y-1\right)$
$=2{y}^{2}-2y-2y+2$
$=2{y}^{2}-4y+2$
$\left(2y-1\right)\left(y-2\right)=2y\left(y-2\right)-1\left(y-2\right)$
$=2{y}^{2}-4y-y+2$
$=2{y}^{2}-5y+2$
The last trial gives the correct factorization.

karton

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