Factoring Polynomials

FactoringPolynomials An Introduction to Factoring and Solving Polynomial Expressions Mr. Rowinski 9th Grade Algebra I Continue

What is Factoring?: A Basic Principle • FACTORING 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 + 9 and x3 – 8 Continue

Types of Factoring • Sums and Differences of Squares • Sums and Differences of Cubes • Solving Elementary Polynomial Expressions • Short Practice • Factoring Numerical Expressions • Factoring Variable Expressions • Factoring When A = 1 • Factoring When A > 1 • Factoring when A < 0

Factoring Numerical Expressions • 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. Let’s 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. Main Menu

Factoring Variable Expressions • Variable 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. Let’s look at the two examples above: • x3 + 2x2 + x: common divisor is x Thus, x (x2 + 2x + 1) is the answer. • x4- x2: common divisor is x2 Thus, x2 (x2 – 1) is the answer. Main Menu

Factoring When A = 1 • When 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. • Let’s 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). Main Menu

Factoring When A > 1 • The same form will be used to factor expressions, such as 2x2– 9x – 5. • Let’s 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 – t2x = -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) Main Menu

Factoring When A < 0 • The same form will be used to factor expressions, such as -2x2 + 11x – 5. • Let’s 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). Main Menu

Sums and Differences of Squares • Sums of Squares occur when same factors of C add up to make the coefficient, B in the term Bx of a polynomial expression, Ax2 + Bx + C. Let’s look at a brief example, x2 + 6x + 9. • In this example, we first have to find like factors of 9 that add up to 6. Thus we find that 3 × 3 = 6. Hence we have the C variables picked out. • Since we know that x × x = x2 and 3 × 3 = 9 and 3 + 3 = 6, we can say that (x + 3) × (x + 3), or as we write that factorization as (x + 3)2, and (x + 3)2 = x2 + 6x + 9. • Differences of Squares occur when operational signs differ in the factorizations such that we obtain (q + r) × (s – t) or vice versa. Let’s look at another brief example, x2 – 25. • In this example, we first have to find like factors of 25, one positive and one negative that add up to 0. Remember, there is no Bx term in this situation. We see that 5 and -5 multiplied together give -25, and 5 – 5 = 0. Thus we have found our r and t variables, which are 5 and -5 respectively. • Since we know that x × x = x2, we can say that (x + 5) × (x – 5) is the factorization of x2 – 25. Main Menu

Sums and Differences of Cubes • There are two important formulas in deciphering sums and differences of cubes: • a3+ b3= (a+ b)(a2– ab + b2)  Sums of Cubes • a3– b3= (a– b)(a2+ ab + b2)  Differences of Cubes • Knowing this, let’s look at an example: x3 – 27 • By definition, this is a difference of cubes problem, so we have to find = 3, and = x. Thus we have the first expression made, which is (x – 3). We now need to find the second, bigger expression to be multiplied. • We know that we need to use x2 and 9 as our a2 and b2 terms respectively, which looks like (x2 + ____ + 9), but we still have to find the empty space. To do this, we have to examine what happens when -3 is distributed across the bigger term. We see that we have to have a 3x2 and a 9x that cancels leaving us with x3 – 27.So, the remaining option left is to use 3x as the middle term. • So by simple algebra, we can see that (x – 3) × (x2 + 3x + 9) = x3 – 27. Main Menu

Solving Elementary Polynomial Equations • We 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. Main Menu

Short Practice Solve: x3 – 6x2 + 8x = 0. Choose the best answer: A. x = -2, 0, 2 B. x = 0, 2, 4 C. x = 2, 4 D. x = 6, 8

Short Practice Choice A TRY AGAIN Check your algebra. Look at all expressions when set equal to zero. Click here to try again Solve: x3 – 6x2 + 8x = 0. Choose the best answer: A. x = -2, 0, 2 B. x = 0, 2, 4 C. x = 2, 4 D. x = 6, 8

Short Practice Solve: x3 – 6x2 + 8x = 0. Choose the best answer: A. x = -2, 0, 2 B. x = 0, 2, 4 C. x = 2, 4 D. x = 6, 8 Choice B CORRECT!!!!!!!!!!!!!!!!!! Continue

Short Practice Choice C TRY AGAIN Look closely at each expression set equal to zero after factorization. Click here to try again Solve: x3 – 6x2 + 8x = 0. Choose the best answer: A. x = -2, 0, 2 B. x = 0, 2, 4 C. x = 2, 4 D. x = 6, 8

Short Practice Choice D TRY AGAIN Did you not factor the original expression? Click here to try again Solve: x3 – 6x2 + 8x = 0. Choose the best answer: A. x = -2, 0, 2 B. x = 0, 2, 4 C. x = 2, 4 D. x = 6, 8

Thank You • I hope you enjoyed your tutorial on how to factor and solve polynomial expressions. • If you would please reset the presentation by clicking on the red arrow at the bottom of the screen for the next person, that would be greatly appreciated. • Thank You! - Mr. Rowinski Back to Start-Up Screen

Factoring Polynomials