# Intuitive understanding of factoring quadratic equations [duplicate]

When factoring a second degree equation $ax^2 + bx + c$ you find the roots then take $a(x - \text{root})(x - \text{root})$. I am wondering why this works. Sorry if poorly phrased question.

## marked as duplicate by user147263, quid♦, Peter Woolfitt, David K, user99914 May 17 '15 at 2:46

This is actually a nontrivial question, at least compared to the rest of intermediate algebra in which one is first exposed to factoring polynomials. One can verify the fact directly by solving the general quadratic and then plugging into the factorization, but this is really a more general fact that applies to all polynomials for common reasons. There are a few key elements to know about.

One element is long division. We can long divide one polynomial by another, and wind up with a so-called "remainder" in the end. If $f(x)$ is a polynomial and $g(x)$ is another, there exists a unique quotient $q(x)$ and remainder $r(x)$ such that $f(x)=q(x)g(x)+r(x)$ with $\deg r<\deg g$. This can be compared to the case for integers: for all integers $n$ and $m$ there exists a quotient $q$ and a remainder $r$ such that $n=qm+r$ with $r<m$. The division algorithm is the same.

A second element is that roots correspond to linear factors. That is, if $z$ is a root of any polynomial $f(x)$, then we can factor $f(x)=q(x)(x-z)$ for some polynomial $q(x)$. This fact can be deduced from the division algorithm. Write $f(x)=q(x)(x-z)+r$ for some quotient $q(x)$ and remainder value $r$. (The only way for $\deg r$ to be less than $\deg(x-z)=1$ is if $r$ is a constant.) Then plug in the value $x=z$ and get $f(z)=q(z)(z-z)+r$. Since $f(z)=0$, this simplifies to $0=r$, so we have deduced $f(x)=q(x)(x-z)$ for some $q(x)$. Note that the coefficients of $q(x)$ may be more complicated than the ones we started with in $f(x)$.

Thus, given a root $z$ of $f(x)$, we can factor $f(x)=q(x)(x-z)$. Then, finding another a root of $q(x)$, we can factor $q(x)$ that way, and so on. We can keep doing this until we end up with a constant times the last linear factor. Thus we have $f(x)=a(x-z_1)(x-z_2)\cdots(x-z_n)$ for some values $a,z_1,\cdots,z_n$. The values $z_1,\cdots,z_n$ are all roots of $f(x)$ (plug one in and see what happens in the factorization) and $a$ is the leading coefficient (if you expand the factorization back out, the leading term is $ax^n$).

Note that the fundamental theorem of algebra guarantees every nonconstant polynomial has a root, possibly a complex number. So the procedure outlined here won't get "stuck."

• Thanks for the reply. If f(x) is a polynomial and g(x) is another, there exists a unique quotient q(x) and remainder r(x) such that f(x)=q(x)g(x)+r(x) with deg r<deg g. Could you explain why deg r < deg g – plebian May 19 '15 at 14:45
• @plebian There exist many $q(x)$ and $r(x)$ for which $f(x)=q(x)g(x)+r(x)$, but only one pair for which $\deg r < \deg g$. When you divide a number $n$ by another number $m$, do you ever want the remainder to equal $m$? Of course not - you keep going until the remainder is smaller than $m$. For any polynomial $f_1(x)$, if $\deg f_1\ge\deg g$, then one multiply $g(x)$ by a constant times a power of $x$ and subtract from $f_1(x)$ in order to get a new polynomial $f_2(x)$ that has strictly lesser degree than $f_1(x)$. Thus, one can keep decreasing the degree until it is below $g(x)$'s degree. – anon May 19 '15 at 14:57

We can show the solutions to $ax^2+bx+c=0$ is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ by completing the square and solving for $x$. So that means the factored form of $ax^2+bx+c$ can be written as $a(x-\frac{-b + \sqrt{b^2-4ac}}{2a})(x-\frac{-b - \sqrt{b^2-4ac}}{2a})$. Expanding this gives: $a(x+\frac{b - \sqrt{b^2-4ac}}{2a})(x+\frac{b + \sqrt{b^2-4ac}}{2a})=a(x^2+(\frac{b- \sqrt{b^2-4ac}}{2a}+\frac{b+\sqrt{b^2-4ac}}{2a})x+\frac{b^2-(b^2-4ac)}{4a^2})=(x^2+\frac{b}{a}x+\frac{4ac}{4a^2})=ax^2+bx+c$

• While technically correct, verification isn't usually as enlightening as deeper theory and motivation. – anon May 16 '15 at 4:22

One can just as well treat the case of general polynomials. (This requires slightly more work than an explicit factorization using the quadratic formula, but the argument is conceptual, and has the advantage that it applies to higher-degree polynomials, for which formulas are either badly complicated, or, in the case of degree $> 4$, in a profound sense unavailable.)

First, the Fundamental Theorem of Algebra says that any polynomial $p$ of degree $n$ has $n$ roots, but one must interpret this appropriately: In particular, it counts each multiple ("repeated") root separately toward this total, and it counts nonreal complex roots.

Now, for any root $r$ of $p$ we can divide $p(x)$ by $x - r$ using polynomial long division to produce a quotient polynomial $q(x)$ and a remainder $a$ (since $x - r$ has degree $1$, this remainder has degree at most $0$, that is, it's simply a number); this is equivalent to writing $$p(x) = q(x) (x - r) + s.$$ Since $r$ is a root of $p$, substituting $x = r$ gives $$0 = p(r) = q(r) ((r) - r) + s = s,$$ and thus $$p(x) = q(x) (x - r);$$ in particular $x - r$ is a factor of $p$. Now, for each other root $r' \neq r$ of $p$ we have $$0 = p(r') = q(r') (r' - r),$$ and hence $q(r') = 0$, that is, $r'$ is a root of the factor polynomial $q$. So, applying this procedure inductively to the collection of $n$ roots guaranteed by the F.T.A. (and, in light of the fact that the observation in the previous sentence uses that $r' \neq r$, taking some care in handling multiple roots) lets us factor $p$ into a product of $n$ linear factors: $$p(x) = a (x - r_1) \cdots (x - r_n)$$ as desired. (Again, some of these roots may be complex. The analogous real statement is that any polynomial over $\Bbb R$ admits an essentially unique factorization into a product of linear and quadratic polynomials over $\Bbb R$.)

• You are using $a$ to mean two different things in your argument, namely as the coefficient of the leading term and as the remainder term. – N. F. Taussig May 16 '15 at 19:22
• @N.F.Taussig Cheers, I've fixed this. – Travis May 17 '15 at 9:58