Factoring Polynomials By Cole Funk

Factoring Factoring is a method by which parts of an expression are separated For example, you can factor all non-prime numbers This can be used for making division simpler and simplifying an expression Factoring polynomials is a method by which the different roots or solutions of the polynomial can be found

Let’s see how helpful factoring can be! Divide 5130 by 54 using long division Divide 5130 by 54 by factoring each and cancelling common factors Simplify (5056 x 70784) / 4096 by multiplying within the parentheses, then dividing Simplify the same expression by factoring and cancelling

Factoring polynomials There are many ways to factor polynomials. I will show you three basic ways. There may be more ways that you find online or just by figuring it out on your own. You can use these other ways (unless otherwise told) as long as you show your work! Solving for coefficients Grouping Dividing by likely factors using long or synthetic division

Assumptions you can make ALL coefficients will be whole numbers. This will be true in this class because some of the methods depend on it. In the natural world, this may not be the case. Except for a few special cases that you will be taught, no imaginary numbers will be used Except for a few special cases that you will be taught, there will be no square cube or any other power rooted coefficients

Before we start, let’s cover foiling and distributing FOILing is a method used to multiply out two two-term mathematical statements FOILing is actually just distributing. Whenever you multiply two mathematical statements, all terms must of one must multiply by all terms of the other.

Distributing practice! x(x + 4) (3x +2)(x +4) (3x2 + x +2)(x+4) (3x2 + x +2)(x2 - 4x + 4)

Why is this relevant??? Factoring is just the reverse of distributing! The only problem is that, just like brownies, it’s much harder to separate the ingredients than mix them.

Factoring by solving for coefficients First, let’s distribute (ax + b)(cx + d)  acx2 +(ad +cb)x + bd Next, relate this to what you’re trying to factor (x2 + 5x +6) and make some equations based on the relation ac = 1, ad + cb = 5, bd = 6 Notice, three equations and four variables. How are we supposed to solve for the coefficients numerically??? This is where our assumptions are important! We know that all of them are real, whole numbers. This forces a = 1 and c = 1 (or both -1). Note, these won’t always both equal 1! d + b = 5, bd = 6 (which two numbers add to 5 and multiply to 6?) d = 2, b = 3 or the other way around

Let’s Practice! Factor by solving for coefficients x2 + 3x + 2

Factoring by solving for coefficients can be done with all polynomials! Set up so a degree 1 binomial is multiplying with a degree (n-1) expression with n terms, where n is the highest power of what is to be factored Example: x3 + 3x2 + 3x + 1  (ax + b)(cx2 + dx + e) Next multiply out expression acx3 + (ad +cb)x2 + (ae + bd)x + be Relate to original ac = 1, ad + cb = 3, ae + bd = 3, be = 1 Use your assumptions and math to solve the problem!

Factoring by Grouping This method uses your knowledge of distributing to factor. Take (x2 + 2)(x + 3) for example. If you distribute, you may first get to x2(x + 3) + 2(x + 3), then x3 + 3x2 + 2x + 6. Factoring by grouping is just the opposite steps!

Factoring by long division Factoring by long division uses the well-known method of long division to factor an expression First, you must come up with an expression that you want to factor out. Simple problems just require you to look at the last term and ask what can be multiplied to become it For example, factoring x2 + 5x + 6, the factors of the last term are 1, 2, 3, 6, -1, -2, -3, and -6 so some possible factors may be (x + 1), (x + 2), (x + 3), etc. Next, set up long division with what you’re factoring on the inside and your possible factor on the outside. Divide like you would do normal long division (an example is on the next slide). If you get zero at the very end, the possible factor goes into the expression evenly and is indeed a factor. What you got on top is what is left of the expression (either another factor or something that can be factored further). If you get something other than zero, then the possible factor does not go into the factor evenly and is not a factor. Try a different one!

Factoring by Synthetic division Factoring by synthetic division is easier than long division Set up the problem by writing the coefficients of the expression to be factored in a line. Next, skip a line for some space to put numbers below those. Just below the empty line, draw a sum line (like in addition). In the top left corner above it all, make a box. The box contains a possible value for x that will make the whole expression zero, which is actually called a root. To find some possibilities, put all of the factors of the last term over all the factors of the coefficient of the first term in a positive or negative (+) fraction (example on the next slide). Once you have a possible root, put it in the box. Bring down the first coefficient. Multiply what’s in the box by what you have on the bottom and put that beneath the next coefficient. Sum the coefficient and the number you just produced to get what goes underneath the sum line. Repeat the last step until you have no more coefficients to sum with. If the last number is zero, the possible root is indeed a root. What you have below the sum line is what is left of the factored expression. You just need to put x’s in it (example in a few slides). The root is set equal to x then solve for zero to get your factor.

Finding possible roots