Polynomials

Numerical expressions, such as 8 + 12 + 20, and 3 + 39, can be factored.First, a common divisor must be found. In the two expressions above, a common divisor can be found. A common divisor is a number that divides each number of the expression. Lets look at the two examples above:8 + 12 + 20:common divisor is 4.Thus, 4 (2 + 3 + 5) is the answer.

3 + 39:common divisor is 3.Thus, 3 (1 + 13) is the answer.Factoring Numerical Expressions

Main MenuVariable expressions such as x3 + 2x2 + x and x4 - x2 can be factored.First, a common divisor, must be found. In the expressions above, a common divisor can be found. In this case, the common divisor is not a number, it is a variable that divides the polynomial expression. Lets look at the two examples above:x3 + 2x2 + x:common divisor is xThus, x (x2 + 2x + 1) is the answer.

x4 - x2:common divisor is x2Thus, x2 (x2 1) is the answer.Factoring Variable Expressions

Main MenuWhen we look at a polynomial expression, like x2 3x 4, we see that the variable, A, before the monomial expression, x2 is 1. Thus, A = 1. We will take an in-depth look at how to factor when A = 1.Lets use the example above for practice. The form that we will use when factoring these expressions, as well as those from here on out, is (q r) (s t).Since we know A = 1, we need to find what two variables, q and s, make x2. Those two variables are x and x. Both x and x will be plugged into the form to make (x r) (x t).We now need to find what two numbers, r and t multiplied make -4, and add up to -3. Those happen to be -4 and 1. We plug -4 and 1 into the two remaining spaces in the form to get (x + 1) (x - 4).Thus, the factorization of x2 3x 4 = (x + 1) (x - 4).Factoring When A = 1

Main MenuThe same form will be used to factor expressions, such as 2x2 9x 5. Lets look at that example.Since we see A > 1, we can note that 2x x = 2x2. So, we can appropriately enter both q = 2x and s = x into the form for factoring and get (2x r) (x t).Now, we need to find two numbers such that -5 2 = r t, such that r x t 2x = -9. We can see that when r = 1 and t = -5, x 10x = -9x.Thus, when q = 2x, r = 1, s = x, and t = -5, 2x2 9x 5 factored is (2x + 1) (x 5)Factoring When A > 1

Main MenuThe same form will be used to factor expressions, such as -2x2 + 11x 5. Lets look at that example.Since we see A < 0, we can note that -2x x = -2x2. So, we can appropriately enter both q = -2x and s = x into the form for factoring and get (-2x r) (x t).Now, we need to find two numbers such that -5 2 = r t, such that r x t (-2x) = 11. We can see that when r = 1 and t = -5, x 10x = 11x.Thus, when q = -2x, r = 1, s = x, and t = -5, -2x2 9x 5 factored is (-2x + 1) (x 5).Factoring When A < 0

Main MenuFACTORING is breaking down a large polynomial expression into smaller, manageable parts, which we will later solve.Factoring involves manipulation of numbers and variables. Most polynomial expressions will have forms like:

Ax2 + Bx + C, where a, b, and c are integers.

We will be dealing with large polynomial expressions, such as:

x2 6x + 9andx3 8What is Factoring?: A Basic Principle

ContinueSums and Differences of Cubes

Main MenuWe will go to the extent of solving polynomial equations such as x2 + 5x + 6 = 0.To solve this equation, we must use our factoring knowledge to factor the polynomial expression, x2 + 5x + 6. From previous knowledge, we see that (x + 2) (x + 3) = 0.We now need to set each independent binomial equal to zero such that x + 2 = 0 and x + 3 = 0. We now need to solve for x.By simple algebra, we see that x = -3 and -2.

Note: If after factoring, you get a factorization of x (x + 2) (x 2)=0, the independent x by itself is equal to zero, thus obtaining x = 0.Solving Elementary Polynomial Equations

Main MenuSolve:x3 6x2 + 8x = 0.

Choose the best answer:A. x = -2, 0, 2B. x = 0, 2, 4C. x = 2, 4 D. x = 6, 8Short Practice