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:




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 :)


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: $$6b^2+19b+15$$

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    $\begingroup$ You are over-determining it. The coefficients $a,c$ are not independent, so you can assume some arbitrary value for either of them. Or, start with the factorization $a(x-b)(x-c)\,$ to begin with, where $b,c$ are the well known roots of the quadratic . $\endgroup$ – dxiv Mar 4 '17 at 8:20
  • $\begingroup$ @dxiv i have tried to make simultaneous equations using the 3 variable form you suggested. However when solve the equation i end up with the original equation except with a different variable instead of x. I'll elaborate in and edit. $\endgroup$ – billy606 Mar 4 '17 at 10:48
  • $\begingroup$ Right, $b$ and $c$ are the roots of the quadratic (and $a$ is the leading coefficient). You probably know how to solve a quadratic. $\endgroup$ – dxiv Mar 4 '17 at 16:38
  • $\begingroup$ @dxiv yeah I know how to solve and factorise a quadratic using the formula. I was just searching for a way to do it without the formula, usijg simultaneous equations. However it doesnt seem like it can be done. $\endgroup$ – billy606 Mar 5 '17 at 0:43
  • $\begingroup$ Once you factor a quadratic into linear factors, you know its roots. So you are essentially looking for ways to find the roots without using the quadratic formula. That's often times possible using the usual tricks - rational root theorm, completing the square, by inspection - but any general method (including indeterminate coefficients) is equivalent to the quadratic formula in the end, since they produce the same information. $\endgroup$ – dxiv Mar 5 '17 at 0:46

You can solve it in 2 ways:

FIRST WAY (longer version):


$x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}=\frac{-(-19)\pm\sqrt{(-19)^2-4\cdot 6\cdot 15}}{2\cdot 6}$

$x=\frac{19\pm\sqrt{361-360}}{12}=\frac{19\pm\sqrt{1}}{12}=\frac{19\pm 1}{12}$

$x_1=\frac{19+1}{12}=\frac{20}{12}=\frac{5}{3} \quad \text{or} \quad x_2=\frac{19-1}{12}=\frac{18}{12}=\frac{3}{2}$

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:


$2\cdot 3\cdot(x-\frac{5}{3})(x-\frac{3}{2})$

$3(x-\frac{5}{3})\cdot 2(x-\frac{3}{2})$


SECOND WAY (shorter version):






In both ways as answer you will get: $(2x-3)(3x-5)$

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  • $\begingroup$ Thankyou for your reply. In my post i mentioned that i knew of these methods. I dont think i was clear enough though. Both these methods work quite well however if we have a very complicated non-monic quadratic for example $$2.3x^2+4x-5$$ it gets harder to use the quadratic formula and the second method you suggested is almost impossible to use. $\endgroup$ – billy606 Mar 4 '17 at 10:33
  • $\begingroup$ Np. Yeah it was little bit unclear. But have you tried to change decimals into fractions? $\endgroup$ – RedRose Mar 4 '17 at 10:52
  • $\begingroup$ And give me some really complicated example which you would think that couldn't be solved with the following 2 ways? $\endgroup$ – RedRose Mar 4 '17 at 10:59
  • $\begingroup$ Any quadratic can be factorised using the quadratic formula however I was looking for a quicker way to factorise without using it; hence my attempt at building the simultaneous equations. $\endgroup$ – billy606 Mar 4 '17 at 11:12
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    $\begingroup$ Haha yeah everything boils down to the quadratic formula eventually so your answer was quite fitting. $\endgroup$ – billy606 Mar 5 '17 at 3:11

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