# How does factoring a quadratic equation relate to the parabola on the graph and to common sense?

I just learned how to factor quadratic equations (at least, how to factor them when they're simple and easy; I don't know how to factor when $$a$$ isn't equal to $$1$$). I'm fine at doing this, and I get that the outcome gives me the roots of the parabola. However, I would love to know how the process itself (as in the factoring and the creation of two linear equations) relates to graphing parabolas and to the general behavior of quadratic equations. Is there any visual representation of what's happening?

By the way, I am somewhat familiar with other ways to solve quadratic equations such as completing the square and using the quadratic formula, and I know how to find the vertex; I would love to know how factoring relates to these things (conceptually) as well.

Thank you, and I'm sorry for the vagueness of my question.

• Note that if you enclose the math in \$symbols, it will look okay. For example, \$x^2\$is rendered as$x^2$. You can read up on the MathJax help when you are writing a question on this site. – MPW Jun 1, 2020 at 22:32 • The suggestion above should fix the formulas in this question, but if anything doesn't look right (or if you want to add something new and it doesn't come out nicely), you can get more information through this link: math.stackexchange.com/help/notation Jun 1, 2020 at 22:36 • L.B. the best thing you can do is to get some actual graph paper aqnd practice drawing parabolas simply by plotting a half dozen points, it will often be enough to plot just when$x\$ is an integer. If you cannot find any graph paper, you can download pdf's from printablepaper.net/category/graph and print out as needed. It really helps. Use a non-graphing calculator (programmable if possible) to find points. Jun 1, 2020 at 23:24
• Your question is indeed vague. If you are looking for ways to visualize a quadratic equation, then you can use free graphing websites such as desmos.com . Visualizing equations does give a much better understanding than the straightforward formulas. Jun 2, 2020 at 14:51

It's not entirely clear how you use your method to factor a quadratic polynomial, but the usual goal of factorizing a general quadratic,

$$ax^2 + bx + c,$$

is that you end up with something like this:

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

(That is, to deal with the case where $$a \neq 1,$$ you can factor $$a$$ out of the polynomial as the very first step, and then you just have to factor something where the $$x^2$$ term has coefficient $$1$$.)

The numbers $$r_0$$ and $$r_1$$ are the roots of the quadratic, that is, they are the $$x$$ coordinates where the graph of $$y = ax^2 + bx + c$$ crosses the $$x$$ axis.

The $$x$$ coordinate of the vertex is midway between $$r_0$$ and $$r_1$$. The $$y$$ coordinate is not obvious from the factorization; basically I would find the $$x$$ coordinate and then plug it into either the original polynomial or the factorization in order to compute the $$y$$ coordinate.

There is also something called the vertex form of a quadratic equation, which is somewhat related to factorization (and can help you with the factorization), but it is not actually a factorization:

$$ax^2 + bx + c = a(x - h)^2 + k.$$

Now the coordinates of the vertex are $$(h,k),$$ but the $$x$$ coordinates where the graph crosses the $$x$$ axis are not as easy to find as in the factorization.

The vertex form is not a factorization in general because it is not the product of some expressions multiplied together; it is the sum of $$a(x-h)^2$$ and $$k.$$ In the case where $$k = 0$$ you are left with just $$a(x-h)^2,$$ which is a factorization, but in other cases the $$k$$ prevents you from using $$x - h$$ as a factor of the quadratic.

Suppose a function $$F: R -> R$$ has n-roots $${r1, r2, r3 ... rn}$$ then the function F can be written as $$F(X) = (X - r1)*(X - r2)*(X - r3) ... *(X - rn)$$ .

So basically when we factor a function $$F(x)$$ in simpler degree-1 terms like $$(X - ri)$$ we are actually finding root ri that contributes to the function F.

But why is thing true ?

Lets see what remainder theorem is :
Suppose : $$F(x) = G(x) + rem$$
Now dividing by $$(X - r1)$$ on both sides gives:
$$F(x)/(x - r1) = Q(x) + rem/(x - r1)$$
$$=> F(x) = Q(x)(X - r1) + rem$$
If $$X = r1$$, then $$F(r1) = rem$$.

So we can see if $$rem = 0$$ or $$F(r1) = 0$$, then $$r1$$ is a root of F(x). Similarly $$Q(x)$$ can be factored sown to get all the remaining roots.

References : https://brilliant.org/wiki/polynomials/#factoring-polynomials
Try this video. Will surely help. https://www.youtube.com/watch?v=MHXO86wKeDY.

• You say "function" but you apparently mean "monic polynomial function." Oct 11, 2020 at 2:25