I was working on a problem about induction, and I got stuck with a factorization: $$(2n^2+7n+6)$$ If I find It's roots, I get $$x=-2$$ and $$x=-3/2$$. Then I can write it as: $$a(x+2)(x+\frac{3}{2})$$ But how could I find $$a$$? I don't have a solid background in this topic as I mostly ignored middle school mathematics and now I'm interested in it, been trying to search "how to factor using quadratic formula", the only thing I got right is that if I decide to factor the $$2$$, I get $$(x+2)(2x+3)$$ which now is the same parabola. What core topic I'm missing?

• Try setting $x=0$ in both expressions (or rather, $n=0$ in the first and $x=0$ in the second, but I think you mean to have $x=n$.) and solving for $a$. Commented Oct 8, 2022 at 22:24
• The only choice for $a$ is for $a$ to be the coefficient of $x^2$. There nothing more or less to it than that. In $2n^2 + 7n + 6$ we have the coefficient of $n^2$ is $2$ so $a = 2$. That's it. No more. Commented Oct 8, 2022 at 22:55
• " What core topic I'm missing" that $2x + 3 = 2(x + \frac 32)$. That's all. If you can factor by guessing that is fine but if you can't guess you'll have to either complete the square or memorize the quadratic formula. Is complete the square $2x^2 + 7x + 6=0$ the $2$ just gets in your way so you factor it out. Divide both sides by two and get $x^2 +\frac 72x + 3 = 0$. Then continue with completing the square. Commented Oct 8, 2022 at 22:59
• Oh god how blind, thank you all. Commented Oct 8, 2022 at 23:38
• I guess what you are asking about how to recreate the $a$ is: You don't have to. The $a$ is completely irrelevant and nobody gives a tinker's patootie what a is. If $ax^2 + bx + c = 0$ then you can multiply or divide both sides by anything to get something. You can divide by $a$ to get rid of the $a$ and $x^2 + \frac bax + \frac ca = 0$. If you get the two roots and get $(x-r_1)(x-r_2) = x^2 + \frac bax + ca = 0$ and you can multiply both sides by anything to get anything. You can multiply both sides by $a$ but.... who cares? You don't have to. Commented Oct 8, 2022 at 23:41

If you know how to factor $$ax^2 + bx + c$$ to $$(k_1x + w_1)(k_2x + w_2)$$ where $$a = k_1k_2$$ and $$b= k_1w_2 + k_2w_1$$ and $$c =w_1w_2$$ by guessing or using the rational root theorem and guessing, then that's just fine. You can do that.

And if $$ax^2 + bx + c = (k_1x+w_1)(k_2x + w_2) = 0$$ then you know that either $$k_1x + w_1 =0$$ or $$k_2x + w_2 = 0$$ and if you solve those you get the solutions that either $$x = -\frac {w_1}{k_1}$$ or $$x = -\frac {w_2}{k_2}$$.[1]

That is IF factoring is easy and obvious and something easy to see.

It's when they are not easy and obvious to factor that you need to resort to "completing the square" or using the quadratic equation (which is just a formula for completing the square when you realize that completing the square is the exact same thing every single time so you don't need to reinvent the wheel every time and you can just remember the formula).

Then the idea of completing the square is to convert $$ax^2 + bx + c=0 \to (x -v)^2 = M$$ and therefore we can conclude $$x-v =\pm \sqrt M$$ and $$x = v\pm \sqrt M$$.

To do $$ax^2 +bx + c=0\to (x-v)^2 = M$$ we must first get rid of that stupid $$a$$ that does nothing be get in the way. We get rid of it by dividing both sides by $$a$$ (the assumption is $$a \ne 0$$.)

$$ax^2 + bx + c = 0 \iff$$

$$\frac 1a(ax^2 + bx + c) = \frac 1a \cdot 0 \iff$$

$$x^2 + \frac ba x +\frac ca = 0$$.

The next step is to try to get $$x^2 +\frac bax + \frac ca=0 \to (x-v)^2 = M$$. Now we know that $$(x-v)^2 = x^2 - 2vx + v^2$$ so we need to coefficient of $$x$$ to be $$-2v$$. But we have the coefficient of $$x$$ must be $$\frac ba$$ so we must have $$\frac ba=-2v$$.

Okay, but we are jumping ahead of ourselves.

We have $$x^2 +\frac ba x + \frac ca = 0 \iff$$

$$x^2 + \frac bax = -\frac ca$$. Now we want $$\frac ba = -2v$$ and so we rewrite $$\frac ba$$ as $$2\cdot \frac b{2a}$$.

$$x^2 + \frac bax = -\frac ca \iff$$

$$x^2 + 2\cdot \frac b{2a}x = -\frac ca$$.

Now we want $$(x-v)^2 = x^2 -2vx + v^2$$ and we have $$x^2 +2\cdot \frac b{2a} x$$ but we are missing the $$v^2$$ term. But as $$v =-\frac b{2a}$$ we must have $$v^2 = (\frac b{2a}^2$$. So to get that, we add $$(\frac b{2a})^2$$ to each side.

$$x^2 + 2\cdot \frac b{2a} x = -\frac ca \iff$$

$$x^2 + 2\cdot \frac b{2a} x + (\frac b{2a})^2 = (\frac b{2a})^2 -\frac ca \iff$$

$$(x + \frac b{2a})^2 = \frac {b^2}{4a^2} -\frac ca \iff$$

$$(x+\frac b{2a})^2 = \frac {b^2 - 4ac}{4a^2}$$.

And we just solve......

$$(x+\frac b{2a})^2 = \frac {b^2 - 4ac}{4a^2} \implies$$

$$x + \frac b{2a} = \pm \sqrt{\frac {b^2 - 4ac}{4a^2}} = \pm\frac {\sqrt{b^2 -4ac}}{2a}\implies$$

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

Thus we have determined if $$ax^2 + bx + c=0$$ then the two roots $$r_1, r_2$$ are $$r_ 1 = \frac {-b + \sqrt{b^2 -4ac}}{2a}$$ and $$r_2 = \frac {-b - \sqrt{b^2 -4ac}}{2a}$$

Now you seem to be asking about getting back to the equation $$(x-r_1)(x-r_2) = 0$$.

Well, if the two roots are $$r_1$$ and $$r_2$$ then EITHER $$x=r_1$$ or $$x = r_2$$ and if so then EITHER $$x-r_1 = 0$$ or $$x-r_2 =0$$. In either case we have $$(x-r_1)(x-r_2) = 0$$ as one of them is zero, so the product of both must be zero.

Now you seem to be asking how we get from $$(x-r_1)(x-r_2) = 0$$ to $$a(x-r_1)(x-r_2)=0$$. We'll the answer to that is we just multiply both sides by $$a$$. $$(x-r_1)(x-r_2) = 0 \iff a(x-r_1)(x-r_2) = 0$$ (assuming $$a\ne 0$$). You seem to be asking how do we know what $$a$$ is. The answer to that is they TOLD us what $$a$$ was when they gave us $$ax^2 + bx + c = 0$$. $$a$$ is what we were given as the coeficient of $$x^2$$.

Maybe you are asking if we were given only the roots and $$(x-r_1)(x-r_2) = 0$$ how would we know what $$a$$ is. Well, the answer to that is we wouldn't. And we wouldn't care. $$a$$ could be have been anything (except $$0$$) and we just don't care. After all $$(x-r_1)(x-r_2) = 0 \iff a(x-r_1)(x-r_2) = 0;a\ne 0 \iff 58901(x-r_1)(x-r_2) = 0 \iff e^{\sqrt \pi}(x-r_1)(x-r_2) = 0$$.

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[1] Might be interesting to not that if $$a=k_1k_2$$ and $$b=k_1w_2 + k_2w_1$$ and $$c =w_1w_2$$ then

$$x =-\frac {w_1}{k_1}$$ would imply

$$x = -\frac {w_1k_2}{k_1k_2}=$$

$$-\frac {b-w_2k_1}{a}=\frac {-b +w_2k_1}{a}=$$

$$\frac {-b - b+2w_2k_1}{2a}=$$

$$\frac {-b -\sqrt{(-b+2w_2k_1)^2}}{2a} =$$

$$\frac {-b -\sqrt{b^2 -4w_2k_1b + 4w_2^2k_1^2}}{2a}=$$

$$\frac {-b -\sqrt{b^2 - 4w_2k_1(k_1w_2 +k_2w_1)+4w_2^2k_1^2}}{2a}=$$

$$\frac {-b-\sqrt{b^2 - 4k_1k_2w_1w_2}}{2a} =$$

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

and by similar reasoning $$x =-\frac {w_2}{k_2}$$ would imply $$x = \frac {-b+\sqrt{b^2 -4ac}}{2a}$$ so this verifies that the roots are what we expected them to be.

The coefficient $$a$$ is the coefficient of the degree-2 term, which tells you that $$a=2$$. Another way to see is evaluating at some point other than zero. For instance if you evaluate both expressions at 1, you get $$2+7+6=a\times 3\times \frac52=\frac{15a}2.$$ This shows that $$15=15a/2$$ and so $$a=2$$.