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# Factoring Real Number Coefficients

## Key Questions

• Factoring polynomials is usually a very simple and straightforward process, but when you get polynomials of a higher degree (i.e. with the highest power being something large, like 5), then you start to run into problems. The best way to solve those types of problems is to use synthetic division to condense your equation down to a quadratic, and then factor normally.

Here's a video that explains synthetic division:

Now let's get to actually factoring.

Now factoring is quite simply writing a polynomial as the product of a set of factors. It's the same thing as writing 6 as the product of 4 and 8.

How is this useful, you may ask? Let me show you. So for starters, try solving this linear equation:

$3 x - 1 = 0$

Add 1 to both sides and divide by 3, and your answer is $\frac{1}{3}$. Easy peasy right? Now try doing this:

${x}^{2} + 6 x + 9 = 0$

A little more complicating right? The solution doesn't stick out that easily.

Now we get to how factoring would help us. If we could rewrite this equation as the product of 2 linear equations, then we could solve each one individually, and it would make life so much easier. So let's look at that quadratic again:

${x}^{2} + 6 x + 9 = 0$

At first sight it looks impossible to figure out a way write this as a product of two factors. But fear not, there is a way. We're going to solve it using grouping. We're going to rewrite it as 4 terms, and then work with two at a time, like we did with the linear equation. We will do this by splitting the middle term into two terms. But what will we split it into? Well, we're going to split it into two numbers that multiply to equal the 3rd term. So hence, we'd get:

${x}^{2} + 3 x + 3 x + 9$.

Sidenote: Notice that we're not actually changing the value of the equation. We're simply manipulating it's look to suit our purpose.

Then, we deal with each pair separately. Let's look at the first two:

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

Is there anything we can do to this? Any GCF we can pull out? Well there is. We can pull out an $x$, and rewrite it as:

$x \left(x + 3\right)$

The second set:

$3 x + 9$

Here we can pull out a 3, and rewrite it as:

$3 \left(x + 3\right)$.

Now let's write the whole thing out:

$x \left(x + 3\right) + 3 \left(x + 3\right)$

Now here, we see that we have a new GCF: $\left(x + 3\right)$ itself. So let's pull that out:

$\left(x + 3\right) \left(x + 3\right) = 0$

Now it looks much simpler to solve doesn't it? You've just factored a polynomial.

In case you want a video that explains all of what I just wrote, here is one that I made (you may need to go through about 9 minutes to get to the factoring part):

Hope that helped :)

See below:

#### Explanation:

Almost al coefficients that you'll see will be real numbers. Coefficients are simply numbers in front of variables.

In the monomial $3 {x}^{4}$, the coefficient is $3$, because it is what's multiplying the variable.

Simply put, real numbers are numbers that can be plotted along a number line, excluding any imaginary parts.

Numbers we deal with everyday like $3 , 6 , 41 , 279$ and even a million, are all examples of real numbers.

Hope this helps!

An irreducible polynomial is one that cannot be factored into simpler (lower degree) polynomials using the kind of coefficients you are allowed to use, or is not factorisable at all.

#### Explanation:

Polynomials in a single variable

${x}^{2} - 2$ is irreducible over $\mathbb{Q}$. It has no simpler factors with rational coefficients.

${x}^{2} + 1$ is irreducible over $\mathbb{R}$. It has no simpler factors with Real coefficients.

The only polynomials in a single variable that are irreducible over $\mathbb{C}$ are linear ones.

Polynomials in more than one variable

If you are given a polynomial in two variables with all terms of the same degree, e.g. $a {x}^{2} + b x y + c {y}^{2}$, then you can factor it with the same coefficients you would use for $a {x}^{2} + b x + c$.

If it is not homogeneous then it may not be possible to factor it. For example, ${x}^{2} + x y + y + 1$ is irreducible.

## Questions

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