I am trying to find a way to factorise a non-monic quadratic using a system of linear equations. I know there exist various algorithms which can help with this, however I'm not fond of such methods.
I first started by creating a general equation with 4 unkown variables excluding x :
$(ax+b)(cx+d)= acx^2+bcx+adx+bd=acx^2+(bc+ad)x+bd$
Now lets say I was given the quadratic $6x^2-19x+15$. Using the general formula I made, I end up with the simultaneous equations:
$ac=6$
$bc+ad=-19$
$bd=15$
Obviously this system of equations cannot be solved as there must be one more equation. Is there a fourth relationship between the variables which I can add or are there variables which I can eliminate? Any comments or suggestions are much appreciated 🙂
EDIT:
I have tried to remove one variable by using the standard equation:$$a(x-b)(x-c)=ax^2+x(ab+ac)+abc$$
Now using the previous quadratic $6x^2-19x+15$ I end up with 3 equations:
$$a=6$$$$ab+ac=-19$$$$abc=15$$
However after solving these by hand I end up with another quadratic equation similar to my original expression:
$$6b^2+19b+15$$
Best Answer
You can solve it in 2 ways:
FIRST WAY (longer version):
$6x^2-19x+15=0$
$x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}=\frac{-(-19)\pm\sqrt{(-19)^2-4\cdot 6\cdot 15}}{2\cdot 6}$
$x=\frac{19\pm\sqrt{361-360}}{12}=\frac{19\pm\sqrt{1}}{12}=\frac{19\pm 1}{12}$
$x_1=\frac{19+1}{12}=\frac{20}{12}=\frac{5}{3} \quad \text{or} \quad x_2=\frac{19-1}{12}=\frac{18}{12}=\frac{3}{2}$
Formula of factorization by using zero points: $ax^2+bx+c=a(x-x_1)(x-x_2)$
In the given equation:
$a=6\quad x_1=\frac{5}{3}\quad x_2=\frac{3}{2}$
So we get:
$6(x-\frac{5}{3})(x-\frac{3}{2})$
$2\cdot 3\cdot(x-\frac{5}{3})(x-\frac{3}{2})$
$3(x-\frac{5}{3})\cdot 2(x-\frac{3}{2})$
$(3x-5)(2x-3)$
SECOND WAY (shorter version):
$6x^2-19x+15$
$6x^2-9x-10x+15$
$(6x^2-9x)+(-10x+15)$
$3x(2x-3)-5(2x-3)$
$(2x-3)(3x-5)$
In both ways as answer you will get: $(2x-3)(3x-5)$