Solving for Variables by Factoring

When given an equation with a polynomial and asked to solve for a value, one way you may be able to do this is by factoring. Previously when solving for variables, you would have been told to isolate the variable on one side, but this cannot be done as easily with certain polynomials.

Example:

x^2 - 3x - 40 = 0

With this equation, it will be hard to isolate x, as there are x's to the power of both 1 and 2. If you tried to isolate, you would likely end up trying to use a square root and moving things around but eventually give up. Instead we can try factoring it:

(x + 5)(x - 8) = 0

We know that the equation equals zero, so we need to find the values that make it true. Remember that the two brackets are being multiplied together, technically making both of them factors of a larger term. We also know that to make any number 0 using multiplication, we just multiply them to zero. This means that at least one of those brackets will work out to zero, so we have to find the values of x that make what is in the brackets equal zero. We do this by setting whatever is inside the brackets to equal zero and solving for x.

(x + 5)(x - 8) = 0

x + 5 = 0

x = -5

(x + 5)(x - 8) = 0

x - 8 = 0

x = 8

You will notice that there are two values of x. Both values are correct as both make the statement true. To check this step, you can simply plug it back into your original equation and see if it is true. Another way you can think of this last step is just to reverse the sign of the number that is not x in the brackets, and that would be your answer. However, this become slightly more complicated when x is multiplied to a number.

Example:

3x^2 + 4x + 1 = 0

(x + 1)(3x + 1) = 0

You can probably tell that one of the answers is -1 easily, but the equation in the other set of brackets may not be as obvious. Instead if we were to look at the equation of the brackets as something like:

(ax + b)

The value of x would be determined by the expression -b/a. In the previous example, if we used this for (3x + 1), we would be left with x = -1/3. (b = 1, a = 3)

For all the previous examples, our polynomial was equal to zero, but what if it was not? What if the polynomial equaled a different number? What if equaled another polynomial? In that case, you would shift every value over to one side of the equation and factor from there.

Example:

x^2 - 3x - 4 = 3x^2 + 4x + 1

First, shifting all values to one side:

3x^2 + 4x + 1 - (x^2 - 3x - 4) = 0

3x^1 + 4x + 1 - x^2 + 3x + 4 = 0

2x^2 + 7x + 5 = 0

Now, the polynomial equals 0, and you can solve it normally. Keep in mind that when doing this step, you may want to shift it in a way that the term with the greatest degree is positive to make solving easier.

(x + 1)(2x + 5) = 0

x = -1, -5/2