SOLVING POLYNOMIALS By Nathan, Cole, Austin and Luke

AUSTIN-SOLVING CUBIC EQUATIONS  A cubic equation is an expression of the form ax^3+bx^2+cx+d.  Here are some examples  8x^3+125  2x^3+2x^2-12x  Here is how you would solve one of these problems  2x^3+2x^2-12x 2x(x^2+x-6) *take out a 2x from everything[divide]* 2x(x+3)(x-2) *factor (x^2+x-6)* X=-3, X=2, X=0 *now find out what x equals [x-2, x=2]*

AUSTIN-SOLVING CUBIC EQUATIONS  To solve 8x^ here is what you do  First you take the cubed root of the equation and put it into the first parenthesis as so. (2x+5)  Then you put x^2 into the first part of the next parenthesis, flip the sign from the first equation and put (ax + b) next, then you ADD the square root of (b) from your first parenthesis. You should get something like this. (x^2 – 10x + 25)  Now when you get both of those you just put both parenthesis next to each other and you’re done! (2x+5)(x^2-10x+25)

AUSTIN-SOLVING CUBIC EQUATIONS  Lets do a problem! Who would like to come up?  (100x^3+13,824)  5X^3+25x^2-250x

NATHAN-DIVIDING POLYNOMIALS  Divide 3x 3 – 2x 2 + 3x – 4 by x – 3 3x^2+7x+24+68/x-3 Long ShlongSynthetic

NATHAN-DIVIDING POLYNOMIALS  Synthetic Division 1. Carry down the leading coefficient 2. Multiply the number on the left and carry it to the next column 3. Add down the column 4. Multiply the number on the left and carry it to the next column 5. Repeat until you get a remainder/0

NATHAN-DIVIDING POLYNOMIALS  Long Division 1. Set up the problem 2. Divide the first two numbers by the divisor 3. Move the next number down next to your remainder 4. Divide that by the divisor 5. Repeat until you end up with a remainder or 0

COLE-FACTORING POLYNOMIALS  When factoring polynomials, you are finding numbers or polynomials that divide out evenly from the original polynomials. But in the case of polynomials, you are dividing numbers and variables out of expressions, not just dividing numbers out of numbers.

COLE-FACTORING POLYNOMIALS  So lets try it!!!  3x-12  So the only thing that you can take out of this problem would be a three. 3x-12=3( )  Next you moved the three to the other side, and when you divide 3 out of 3x you get x so you put that in the parentheses.  3x-12=3(x )  And when you divide the -12 by three you get negative 4 and you place that into the parentheses and then that’s your final answer!!!1  3(x - 4)

COLE-FACTORING POLYNOMIALS  So now you do some!!!!  12y^2 – 5y  7x - 7

LUKE-PASCAL’S TRIANGLE  ROWS  The 1 st number in each row represents the number row that it is. (Remember that the 1 at the beginning of each row is the 0 th number)  The 1 at the very top of the triangle is the 0 th row.  1,3,3,1is the 3 rd row. The 1 st number is 3, meaning the 3 rd row.  1,5,10,10,5,1 is the 5 th row. The 1 st number is 5, meaning that it is the 5 th row.

LUKE-PASCAL’S TRIANGLE  Sums of the Rows  If you do 2^4 power, that will equal the sum of the 4 th row.  =16 (4 th row)  2^4=16 (4 th row)  =64 (6 th row)  2^6=64 (6 th row)

LUKE-PASCAL’S TRIANGLE  The Hockey System  If you take however many diaganals of numbers, add them up, the sum of the numbers is the number that is diaganally downward.  Take 1, 6, 21, and 56 for example. Circle those numbers and at 56, turn the opposite direction, going diaganally down and the number 84 is the sum. Here is a picture to represent what I am talking about.

Unit 2 Be ready to go when the bell rings! Example 1: Name according to degree and number of terms. a) 2x 2 + 3x + 1 b) 2x 4

Unit 2 Example 2: Factor. 8x 3 – 27

Unit 2 Example 3: Solve. x 4 – 2x 2 – 48 = 0

Unit 2 Example 4: Divide x 3 – 7x + 12 by x + 3

Unit 2 Example 5: Completely factor f(x) = x 3 – -x 2 - 4x + 4 given x-1 is a factor

Unit 2 Example 6: Expand (x + 2y) 3

Unit 2

Example 8: Write a function in standard form given the roots: 3, -5i

Unit 2 Example 9: Factorsx-interceptsSolutionsZerosStandard Form 9) x 3 + 3x 2 – x - 3

MSL (Common Exam)

MSL (Common Exam)

MSL (Common Exam)

MSL (Common Exam)

Want more practice??? Go to Mrs. Bell’s website.