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Firstly, the title may be a little hard to understand so could someone please suggest a better one and make up for my 'ignorance.'

Onto the question. If I have a quadratic equation: $$ax^2+bx+c$$ Is there a way for me to factor this to $(nx+r)(tx+s)$ by just using an equation. For example say I have the equation: $$2x^2+x-6$$ Can I factor this to $(2x-3)(x+2)$ only using an equation not the traditional method of look, guess, test : if right stop, if wrong repeat.

The reason I ask this is because I have made a formula to do what I have described and I am not sure if any mathematicians have done this before. I want to know if I have made a discovery or rediscovered something already found.

The formula is $d$ = quadratic formula, $ax^2+bx+c = (x-d)*(x -(ad+b))$

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  • $\begingroup$ See my answer as a plain example so that you can do more of those, the other answers provided are more formal (and mathematically arguably better). $\endgroup$
    – imranfat
    Commented Sep 30, 2015 at 3:17
  • $\begingroup$ Always check for factors of the intercept coefficent ($c$) if an equation does factor easily you will find that they form the majority of the brackets. $\endgroup$
    – Chinny84
    Commented Sep 30, 2015 at 17:29

5 Answers 5

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Given the quadratic polynomial $ax^2+bx+c$, with $a\neq 0$, by using the quadratic formula you can find its roots, lets say $$r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\quad\text{and}\quad r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

So the polynomial can be factorised as $$ax^2+bx+c=a(x-r_1)(x-r_2).$$

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This is well known.

It is called the quadratic formula.

The roots of $ax^2+bx+c$ are $\frac{-b\pm \sqrt{d}}{2a} $ where $d = b^2-4ac$.

If $d > 0$, there are two real roots.

If $d < 0$ there are two complex roots.

If $d = 0$ there is one repeated root.

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  • $\begingroup$ I am well aware of the quadratic formula. I am not finding roots. I am factoring the equation. On a quadratic when a>1 you are no longer factoring, you are finding roots. $\endgroup$
    – Stephen Fratamico
    Commented Sep 30, 2015 at 3:25
  • $\begingroup$ As Mario G stated, finding the roots is equivalent to factoring. Let's see your formula. I will be very surprised if it is new. $\endgroup$
    – marty cohen
    Commented Sep 30, 2015 at 3:27
  • $\begingroup$ d = quadratic formula, sgn(x) = en.wikipedia.org/wiki/Sign_function, so ax^2+bx+c = (x-sgn(d)d)*(x - sgn(a(d)+b)(ad+b)) $\endgroup$
    – Stephen Fratamico
    Commented Sep 30, 2015 at 3:49
  • $\begingroup$ sorry ignore the sgn() function I wasn't following my work for a second. ax^2+bx+c = (x-d)*(x -(ad+b)) $\endgroup$
    – Stephen Fratamico
    Commented Sep 30, 2015 at 4:07
  • $\begingroup$ Factoring and finding the roots is the exact same thing. A polynomial is reducible, that means it can be factored further, if it has at least one root. Thus, finding roots is how you factor a polynomial. $\endgroup$
    – Auclair
    Commented Sep 30, 2015 at 17:43
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Given a quadratic $f(x)=ax^2+bx+c$, it is a bit ambiguous what you mean by $d = quadratic formula$.

$$d=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The quadratic gives two roots, so rewrite your equation to be more accurate. Perhaps call them $d_1$ and $d_2$ in accordance with the $\pm$ that appears in the quadratic formula. Now let's consider your formula.

$$(x-d)(x -(ad+b))$$ $$ = x^2 -dx -(ad+b)x +d(ad+b)$$ $$ = x^2 -(d + ad+b)x +d(ad+b)$$

Regardless of the ambiguity of $d$, this is not giving us the $a$ coefficient for the quadratic term. If you correct your equation, it'll likely match the following procedure:

Use the quadratic equation to find each root, make the linear factors, and multiply the two factors by $a$.

$$f(x)=a(x-d_1)(x-d_2)$$

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Here is an example: Factor $x^2+8x+2$ It is not nicely factorable. So use the quadratic formula to find solutions to $x^2+8x+2=0$ Verify that the solutions are $x=-4 \pm \sqrt14$ So working backwards, the factors came from $(x+4+\sqrt14)(x+4-\sqrt14)$. It is maybe not the cleanest way of doing, but it works...

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  • $\begingroup$ The area this starts to fall apart at is when a>1. When you use the quadratic formula past this point you are no longer factoring just finding roots. I have made a formula to factor a quadratic, not solve it. $\endgroup$
    – Stephen Fratamico
    Commented Sep 30, 2015 at 3:24
  • $\begingroup$ @Stephen. My example was simple. If it doesn't come out nice, you have fractions as well as a radical in your factorization. The method works nonetheless...I am essentially now referring to MArio's answer $\endgroup$
    – imranfat
    Commented Sep 30, 2015 at 3:26
  • $\begingroup$ I think I may not be getting my point across and that is on me. The method I have allows a computer to factor an equation without 'thinking.' My method is a formula that factors the equation without contingencies. Also, I am not finding roots I am factoring. (3x^2+8x+2) does not equal (x + (8 (+or-) sqrt(40))/4) $\endgroup$
    – Stephen Fratamico
    Commented Sep 30, 2015 at 3:36
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See also "completing the square". For example, $x^2+6x+8=(x+3)^2-1$,

so $x^2+6x+8=y$ translates to $(x+3)^2=y+1$ or simply $x=\pm \sqrt{y+1}-3$

However, I also encourage you to play around on your own, as that's how you develop your own understanding of these concepts.

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