When solving algebraic equations, you will sometimes be asked to find the solution set of the equation. This is when you are asked to find all of the values that make the left-hand side of the equation equal to zero.

Solving algebraic equations can be difficult because you are also required to find all possible solutions in terms of numbers. For example, if you were asked to solve 2x + 1 = 0, you would have to find all numbers that when multiplied by **two equals one**.

Finding **completely factored form**s of expressions or equations is crucial in solving for the solution set. When finding the completely factored form, you *must pay close attention* to what symbols and letters are present and their matching variables.

## Find the factors of 4

The factors of 4 are 1 and 2, 2 and 2, and 4 and 1. Two of these combinations give you the number 4, but only one gives you the original value of 4.

The **remaining coefficients** in the *linear equation must* be zero, so choose ones that will make that happen. The coefficient of x must be zero, so choose a combination that gives you 0 when you solve for x.

The coefficient of *x2 must* be zero, so choose a combination that gives you 0 when you solve for x. You can pick any number for the coefficient of x3, as long as it is not zero.

Check your solutions by plugging them into the **original equation** to make sure they match up.

## Find the factors of 13

The factors of 13 are 1 and 13, or 1, 3, and 7. You can also combine these factors to get other values. For example, you can combine 1 and 3 to get 4, or you can combine 1 and 7 to *get 8*.

All of these numbers correspond to quadratic equations that solve to the value of 13. This is why you can’t factor the original equation any further – because all of the answers “look” like 8 when you plug them into the original equation.

Quadratic equations have two variables (**like x** and y in the equation x + y = 5) that equal a specific value (5 in this case). When you solve these equations, you either add or subtract one of the variables depending on what solution set (the set of solutions to an equation; in this case, 0

In this case, since there is only one variable in each equation, you would simply have to pick which solution set is true based on which *number equals 13* when plugged into the original equation.

## Combine like factors

The first thing you should do is check for like factors. The variables in the equation F(x) = *6×3 – 13×2 – 4x* + 15 all have a variable x as their denominator.

Therefore, all of the variables can be factored by x. You can combine these xs to make your life easier. You can also combine the 3s and 2s in the first and second terms, respectively.

Now that all of the variables have been replaced with x, you can combine the coefficients and isolate the variable that is being raised to a power.

This gives you (6x)3(-13)2(4)+15, or (-13)2(4)+15=(-13)(16)+15=30+15=45.

## Divide by greatest common factor

The next step is to ** divide every term** of the equation by the

**greatest common factor**of all the variables. In this case, that is dividing by 6.

So, starting with the first term, divide the coefficient and the variable itself by 6. Then do the same with the constant term. Once you have done this, your equation should only contain integers.

So, your next step is to divide every term by 6. This will leave you with an equation that looks like this: *2×3 – 2×2 – 2x* + 1 = 0.

The last step is to solve for x and see what value you get. Doing so reveals that x = 1.

## Check for extraneous solutions

After factoring, you should check to see if your equation has any extraneous solutions. Extraneous solutions are values for x that solve the equation, but when plugged into the *original function*, yield a non-**sensical result**.

For example, let’s say we were trying to solve the equation 2x + 1 = 5. The solution to this equation is 2x + 1 = 5 so 2x = 4, but when we plug 4 into 2x + 1 = 5, we get a *nonsensical result* of -1.

This is because we assumed that 2x + 1 = 5 is an equality, so only values of x that solve this are ones that make it true. -4 does not make this true, so it is an extraneous solution.

## Use a computer to find the factored form

When dealing with complex algebra problems, it is important to know how to find the **completely factored form** of a function.

Knowing how to find the completely factored form of a function allows you to understand the fundamental properties of functions more easily. For example, changing the values of a function requires only multiplication and division operations, since these operations only require changing the value of one variable.

Finding the completely factored form of a function can be done by hand, but it can take a while depending on how complex the function is. A faster way to find the completely factored form is to use a computer program such as MacAlgolConverter or Algebrator. These are both free downloads that easily search for and find the completely factored form of a function.

## Use Newton’s method to find the factored form

The next method for *solving polynomial equations* is called Newton’s method. Unlike the previous methods, this one is used to solve non-polynomial equations.

Like we said before, linear equations can be solved by finding the slope and either taking it directly or oppositely to it. For example, if the equation was 2x + 1 = 0, then the slope would be 1, so x = -1.

Non-*linear equations cannot* be solved in this simple way. Instead, you have to find what is called the derivative of the equation and solve it using that information.

To do that, you have to find the value of y that matches x in a solution to the equation. You then take your equation and rewrite it as y=something+x, where something is replaced with what y is in relation to x.

## Factor an arbitrary polynomial using long division

When you divide one polynomial by another, you get a quotient and a remainder. The polynomial you divide into is called the divisor, and the polynomial you get as a result of the division is called the dividend.

In long division, you always work with whole numbers. You cannot work with fractions or negative numbers when dividing polynomials. You can only *divide one variable* by another variable.

The first step in doing long division of polynomials is to rearrange the divisor so that it has only one variable. Then, convert the dividend into integers by *making every coefficient* a 1.

For example, let’s do some long division with the *following divided term*: **6×3 – 13×2 – 4x** + 15.