Is there an equation for factoring an quadratic equation.

Question

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

Answer 1

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 $$ax^2+bx+c=a(x-r_1)(x-r_2).$$

Answer 2

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...

Answer 3

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.

Answer 4

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 $a$.

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

Answer 5

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.