Factoring quadratics - why does this method work?

Question

Given a quadratic like this:

$2x^2 + 7x + 3$

I need to find two numbers $a$ and $b$ that have these properties:

$a + b = 7$

$a \cdot b = 2 \cdot 3$

Which in this case is 1 and 6.

So I can rewrite the equation as:

$ 2x^2 + 1x + 6x + 3$

$= x(2x + 1) + 3(2x + 1)$

$= (2x + 1) (x + 3)$

I wanted to know why this method works. I found a proof in a khan academy video that goes like this, based on the multiplication of two binomials:

$(fx + g) (hx + j)$

$= fhx^2 + ghx + jfx + gj$

$= fhx^2 + x(gh + jf) + gj$

Then let's assume that $gh = a$ and $jf = b$ this would be the proof for finding the middle coefficients. I understand this part.

The video then tells me that therefore $a \cdot b = gh \cdot jf$ which can be rewritten as $a \cdot b = fh \cdot gj$.

I don't understand this part - because for me it doesn't really prove that I need to find a number that multiplies the first coefficient with the constant (the last number) in the equation. Is there maybe another way to explain that?

You've got to divide your equation by $2$ and then you need $a+b=\frac{7}{2}$, $ab=\frac{3}{2}$. And you need to check out your other calculations as well. — ancient mathematician, Mar 20 '17 at 08:03
@ancientmathematician Not sure what you are referring to, I'm showing the example from the website I linked. — BMBM, Mar 20 '17 at 08:21
Just before you say " I don't understand" you say correctly that $ab=(fh)(gj)$. So you want $a, b$ whose product is $fh$ (the coefft of $x^2$ times $gj$ (the constant coefficient). So .... ? — ancient mathematician, Mar 20 '17 at 09:42

N. F. Taussig · Accepted Answer · 2017-12-13T09:29:17.797

Let's assume $ax^2 + bx + c = (rx + s)(tx + u)$. Then \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = rtx^2 + rux + stx + su\\ & = rtx^2 + (ru + st)x + su \end{align*} is an algebraic identity that holds for every real number $x$. In particular, it holds if $x = 0$, $x = 1$, and $x = -1$. If $x = 0$, we obtain $$c = su \tag{1}$$ If $x = 1$, we obtain $$a + b + c = rt + ru + st + su \tag{2}$$ Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side to obtain $$a + b = rt + ru + st \tag{3}$$ If $x = -1$, we obtain $$a - b + c = rt - ru - st + su \tag{4}$$ Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side to obtain $$a - b = rt - ru - st \tag{5}$$ Adding equations 3 and 5 yields $$2a = 2rt \tag{6}$$ Dividing both sides of equation 6 by 2 yields $$a = rt \tag{7}$$ Since $a = rt$, we can cancel $a$ from the left hand side of equation 3 and $rt$ from the right hand side of equation 3 to obtain $$b = ru + st \tag{8}$$ Furthermore, observe that $$ac = (rt)(su) = (ru)(st) \tag{9}$$ Hence, we can factor a polynomial with rational coefficients with respect to the rational numbers if there exist two numbers ($ru$ and $st$) with product $ac$ and sum $b$.

I worked through your proof with pen and paper, this was very interesting for me, thanks a lot! — BMBM, Mar 22 '17 at 06:10

score 0 · Answer 2 · answered Mar 20 '17 at 09:52

This method uses four parameters ($f,g,h,j$) when the initial trinomial only has three, making things a little confusing. I am showing a simpler version.

From $$px^2+qx+r$$

you can factor out the coefficient $p$, giving

$$px^2+qx+r=p\left(x^2+\frac qpx+\frac rp\right).$$

Now it should be obvious that the factorization is of the form

$$px^2+qx+r=p(x+a)(x+b)=p(x^2+(a+b)x+ab).$$

With the given example

$$2\left(x^2+\frac72x+\frac32\right),$$ and you need to solve

$$\begin{cases}a+b=\dfrac72,\\a\,b=\dfrac32\end{cases}.$$

Then, notice that

$$(a+b)^2-4ab=(a-b)^2.$$

Using this relation,

$$a-b=\pm\sqrt{\frac{7^2}{2^2}-4\cdot\frac32}=\pm\frac52.$$

Then combining with the known sum,

$$a,b=\frac12,3$$

and

$$2x^2+7x+3=2\left(x+\frac12\right)\left(x+3\right)=(2x+1)(x+3).$$

This does not explain why the algorithm given on the Khan academy video works. — Ted Shifrin, Mar 20 '17 at 15:10

score 0 · Answer 3 · answered Mar 20 '17 at 15:27

Suppose that we have $$ax^2+bx+c$$

One strategy is to restate this as a quadratic a leading coefficient equal to unity.

$$\begin{align} ax^2+bx+c&=\frac{a}{a}(ax^2+bx+c)\\ &=\frac{1}{a}(a^2x^2+bax+ac)\\ &=\frac{1}{a}((ax)^2+b(ax)+ac)\\ &=\frac{1}{a}(t^2+bt+ac)\\ \end{align}$$ Where $t=ax$

Now, we're just looking for two numbers that add together to make $b$ and multiply together to make $ac$. Afterwards, just replace the $t$s with $ax$s.

Factoring quadratics - why does this method work?

3 Answers3