Why Do We Factor Polynomials
We factor polynomials because they give us points on their curve. These points allow us to analyze functions and solve equations for the values we need.
Why We Factor Polynomials
In this article, I want to continue to talk about how we factor polynomials and why we need to do so.
Division of Polynomials
We already know how to work with quadratic equations. That subject was covered previously. We saw that once factored, we could analyze their “zeros” and see where the equations were equal to zero. This gives us valuable information.
The curves equal zero where the lines intercepted or crossed the x-axis. This will give us the factors of the equation that we need.
The problem is, what if they give us an equation that is not nice and neat?
For example, some polynomial that is not a quadratic and needs work done to it in order for us to solve it. This is where division of a polynomial can be handy.
Sometimes, we can see on a graph where a curve equals zero at one place but it is not clear at other points. This is why we need to divide that single point out of the existing polynomial so we can analyze the rest of it.
Let’s start with some easy examples:
Example 1
\[ \frac{3x^4-2x^2-1}{3x^3} \]
The denominator is a monomial, and we are dividing each term by the denominator.
\[ \frac{3x^4}{3x^3} \longrightarrow x \]
\[ \frac{-2x^2}{3x^3} \longrightarrow \frac{-2}{3x} \]
\[ \frac{-1}{3x^3} \longrightarrow \frac{-1}{3x^3} \]
We now put all those terms together to get:
\[ x - \frac{2}{3x} - \frac{-1}{3x^3} \]
Example 2
\[ \frac{5x^4-15}{10x} \]
As before, we divide each term into the numerator by the single term in the denominator.
\[ \frac{5x^4}{10x} \longrightarrow \frac{1}{2x^3} \]
We do the same for the last term in the numerator.
\[ \frac{-15}{10x} \longrightarrow \frac{-3}{2X} \]
Example 3
\[ \frac{6x^3-3x^2+2}{2x^2} \]
Again, we have a single monomial in the denominator.
We divide each term into the numerator by the denominator.
\[ \frac{6x^2}{2x^2} \longrightarrow 3x \]
Now, we do the same for the second term.
\[ \frac{3x^2}{2x^2} \longrightarrow \frac{-3}{2} \]
Once more, we go to the last term.
\[ \frac{2}{2x^2} \longrightarrow \frac{1}{x^2} \]
Division Remainders
When doing division upon polynomials, you will often end up with a remainder.
It may not be intuitive, but there is a rule for this. When doing division, the remainder is set over the original term we are dividing by.
Let’s use the previous problem as an example again.
\[ \frac{6x^3-3x^2+2}{2x^2} \]
If we ended up with a remainder in the above problem, say 3, it would look like this:
\[ \frac{3}{2x^2} \]
That would be the last part of our answer. Of course, we did not have a remainder in the last example, but that happens when you encounter them.
Example 4
Our function is:
\[ f(x) = 5x^2 -3x +1 \]
Divide this by :
\[ x-1 \]
and say what the remainder is.
We start out by dividing the first term by x-1.
\[ \frac{5x^2}{x} = 5x \]
Now we multiply by “x” and “-1”. Don’t forget to change the sign.
We do addition on “-3x” and “5x”. Then we divide “2x” by “x”.
\[ \frac{2x}{x} = 2 \]
We multiply through again and get a remainder of 3 which is the correct answer.
Factoring Polynomials
At the beginning of this section, I declared why we factor equations. Remember, factoring gives us information we need about an equation. The factors give us the “zeroes” of that equation’s curve. The “zeroes” are where the curve intercepts the x-axis.
A polynomial has a factor if the factor equals 0. This is always how we can check our work. Substitute that factor back into the equation. If the equation turns out to be 0, then we do indeed have a factor.
With all of this information, we can give a complete factored form. A complete factored form involves the function, any coefficients, and the factors. If you do not know the factors, determine those first.
We can give you a graph or just the function. We can model the algebraic function using software or a graphing calculator. This is usually acceptable or even preferred. Once you have the function graphed, you can see the “zeroes” on the graph. Use the coefficient information that you have to give the complete factored form.
Example 5
Write the complete factored form of:
\[ f(x) = 2x^2 - 25x + 77 \]
The zeros are:
\[ \frac{11}{2} \text{and} 7 \]
Our coefficient is 2, and they gave us the zeros. This gives us:
\[ 2(x-\frac{11}{2})(x-7) \]
Graphs and Factors
If you haven’t noticed, polynomials can have even or odd levels of exponents. Polynomials behave differently in graphs depending on this multiplicity. An even multiplicity is 0,2,4 and so on. Odd multiplicities are 1,3,5.
The odd multiplicities cross the x-axis at its zero point. Even multiplicities just intercept, or touch the x-axis at its zero point. I’m sure you have seen graphs that exhibit this behavior, and this is why.
Rational Zeros
This is an interesting little test. It isn’t always applicable, but it can be useful. If a polynomial has a zero that is also a rational number, then you can use the rational zero test with it.
First, write out the factors of your coefficient and constant terms. It is easier to write them on separate lines or in a table so we can compare easily them.
When you can see them easily, make a list of all combinations of factors. Substitute these factors into the original equation and evaluate the function. Any of them that makes the function equal zero are a real factor. A second degree polynomial will have two factors and a third degree polynomial will have three factors.
Solving Polynomials
Factoring and using the quadratic equation are the two principal methods of solving polynomials. The problem with the quadratic equation is the polynomial needs to be in quadratic form. If it isn’t, you need to see if you can get it there.
This is where we go back to our factoring tool. If you have a cubic or greater function, then you can factor an “x” or more out of it to get it into quadratic form. This concept will continue to be important from here on out.
Example 6
Solve and find the (x) values.
\[ x^3 + x^2 - 6x = 0 \]
Start by factoring an (x) out of the equation.
\[ (x)(x^2 + x - 6 = 0) \]
This is a cubic function so we will have three (x) values.
We now factor the quadratic portion. We can see the factors of this equation.
\[ (x)(x-2)(x+3) = 0 \]
So, our answers are:
\[ x= 0,2,-3 \]
Summary
When doing division with a monomial, divide the denominator into every term in the numerator. Polynomial division is like long division. The remainder theorem is when f(x) is divided by “x-a”, so the remainder is f(a).
The factor theorem, (x-a), is only a factor if f(a) - 0. When you have a zero with odd multiplicity, the graph crosses the x-axis at this zero. Conversely, when you have a zero with even multiplicity, the graph intercepts but does not cross the x-axis.
Hopefully, I have made it clear why we factor polynomials. We factor them to find the points that make the equation equal zero. These are factors and they are points where a graph will cross or intercept the x-axis.
We then discussed the tools that we used to factor polynomials. These include division, regular factoring, and the quadratic equation.
Thanks for Reading
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