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Root

Finding the root of a polynomial is probably easier than many students assume. That being said, the difficulty of this process depends on the complexity of the polynomial. The good news is that by applying the same set of rules to these math problems, we can find both complex and simple roots of polynomials by following the same basic steps.

What is a root?

You might have heard the term "root" being used in various ways. One of the most common "roots" is a square root. But is this related at all to other "roots?"

Not quite. A "square root" and a root are actually two different things (even though we sometimes say "root" as a quick way of saying "square root."

The definition of a root in math is as follows:

Let's consider the following polynomial:

$f\left(x\right)={x}^{2}-4$

What are the roots of this polynomial? We can see by inspection that if we substitute -2 or 2 for x, we get zero. Therefore, we assume that these are our two roots, but are there others?.

According to the fundamental theorem of algebra: a polynomial in one variable will have a number of roots equal to the degree of the polynomial. Some of these roots may be "double" or "multiple roots".

Practicing our knowledge of roots

Let's hone our understanding of these principles and try a few practice questions:

Consider the following polynomial:

$f\left(x\right)={x}^{2}-5x+6$

Our first step is to "equate the polynomial to zero."

$f\left(x\right)={x}^{2}-5x+6=0$

This particular equation can easily be factored:

$\left({x}^{2}-2x\right)+\left(-3x+6\right)=0$

$x\left(x-2\right)-3\left(x-2\right)=0$

$\left(x-2\right)\left(x-3\right)=0$

Note that this step allows us to quickly see our two roots. All we need to do is make the values inside the brackets equal zero. If the value in one set of brackets is zero, multiplying by the other value must also be zero. Why? Because zero times any number equals zero.

Because of this, we can safely say that $x=2$ or $x=3$ . And there we have it: Our two roots.

Note that by looking at the degree of our polynomial, we can quickly find out how many roots are possible. Since this particular polynomial has degree 2, we know that there are 2 possible roots. Remember that the degree of a polynomial is the value of its highest exponent of a variable.

This is a relatively simple example since both of our roots are positive integers. But as we will soon find out, our roots can be a little more complicated. For example, we might get radicals and or complex numbers as our roots!

Let's try another practice question. Consider the following polynomial:

$f\left(x\right)=2{x}^{3}+2{x}^{2}+3x$

We can see right away that the degree of this polynomial is 3. This means that we're looking for a total of 3 roots.

Let's start with exactly the same step as before, which is to equate the polynomial to zero:

Can we factor this number? You bet!

$x\left({2x}^{2}+2x+3\right)=0$

We are left with two values here that can equal zero:

$x$ or ${2x}^{2}+2x+3$

But we're not finished just yet. As you may recall, we're looking for a total of three roots -- and so far we've only established that one root is $x=0$ . So what about the other two? In order to find these roots, we're going to need to use the quadratic formula:

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

As we may recall, the quadratic formula gives us the quickest way to solve for the value of x when we have a quadratic equation in its proper form. The quadratic formula is derived from the process of completing the square.

Let's plug in our values: $a=2,b=2,c=3$

$x=\frac{-2±\sqrt{{2}^{2}-4\left(2\right)\left(3\right)}}{2\left(2\right)}$

Let's simplify:

$x=\frac{-2±\sqrt{-20}}{4}$

$x=\frac{-2±2i\sqrt{5}}{4}$

$x=-\frac{1}{2}±\frac{i\sqrt{5}}{2}$

Now we know that the polynomial root has 1 real root and 2 complex roots.

Let's try one more.

Consider the following polynomial:

$6{x}^{2}+7x-2$

We can start by equating the polynomial to zero:

$6{x}^{2}+7x-2=0$

Let's factor this equation:

$2x\left(3x+2\right)-1\left(3x+2\right)=0$

Now we know that the two possible roots are $-\frac{2}{3}$ and $\frac{1}{2}$ . Substitute these values into each bracket, and you'll see that they caused the polynomial to equal zero.

Last one!

${\left(x-3\right)}^{2}+4$

Let's equate the polynomial to zero:

${\left(x-3\right)}^{2}+4=0$

If ${\left(x-3\right)}^{2}=-4$ then $x-3=±2i$ and $x=3±2i$

Now we know that our two roots or "zeros" are $3+2i$ and $3-2i$ .

Visualizing roots

It might help to understand roots if we visualize them. The best way to do this is by graphing the polynomial. When you see the graph of the equation intersect the x-intercept, you know that you're looking at a root. For example, the polynomial may intercept the x-intercept at -2 and 2. This means that $y=0$ at these points. In other words, the function equals zero at these coordinates.

We often use the term "root" to describe the solution of any expression, whether it equals zero or not. But the proper definition involves making a polynomial equal zero. For example, we can say that pi is a root of the equation x sin(x) = 0.

Why do we need to find the zeros of polynomials?

But what exactly is the point of all this? Why do we need to find the zeros of polynomials? The truth is that this process can help us find roots more easily. Consider the following polynomial:

${x}^{2}+2x+100=99$

In order to solve this equation, we need to find the value of x. But wouldn't this become a much simpler process if we equated the polynomial to zero? What happens if we subtract 99 from both sides of the equation?

${x}^{2}+2x+1=0$

Now we're working with something that seems familiar! All we need to do now is use the quadratic formula to find the value of x. This trick works with any polynomial, and it's always easier the find the root if we first equate it to zero. Of course, the quadratic formula only works if we see a polynomial arranged in a certain way. But with a few steps, we can rearrange polynomials so that they fit:

• Step One: Arrange our equation into the form (quadratic = 0)
• Step Two: Arrange the terms in the equation in decreasing order (squared term, then x-term, then linear term)
• Step Three: Pull out the numerical parts of the above three terms (a, b, and c in our quadratic formula)
• Step Four: Plug these values into the formula
• Step Five: Simplify the results to get our answers

The quadratic formula is a very important part of finding roots in certain common polynomials. It's worth revisiting this concept if you're still not quite sure how it works.

Flashcards covering the Root

Algebra 1 Flashcards

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