Warm-up Simplify. 5x + 2 + 7x – 1 2. 3a + 2b – (a – 2b) Multiply. 5(x + 4) 4. 3x(2x + 7)   x2(x3 + y) 6. 3y2(2x3 + 6x +2) 7. Find the degree and leading coefficient of 15x5 + 3x7 + 18.

Warm-up Answers Simplify. 5x + 2 + 7x – 1 2. 3a + 2b – (a – 2b) 12x + 1 2a + 4b Multiply. 5(x + 4) 4. 3x(2x + 7) 5x + 20 6x2 + 21x 5. x2(x3 + y) 6. 3y2(2x3 + 6x +2) x5 + x2y 6x3y2 + 18xy2 + 6y2 Find the degree and leading coefficient of 15x5 + 3x7 + 18. Degree = 7 L.C. = 3

HW Answers 7a + 1 2. 0 3. 4p2 – 9p + 10 4. 2x2 5. 0 6. 5a2 + 4ab – 7ab2 + b2 7. 7x2 – x – 3 8. 7x + 1.5y – 6.75 9. 5x2 – xy – y2 10. 2x2 + 7x – 13 11. 1.7x2 – 2.6x – 1.3 12. 6p2q2 + pq + 19

Math Blooper for Binomial theorem ACTIVATOR Math Blooper for Binomial theorem

4.3B Multiplying Polynomials and the Binomial Theorem Standard: MCC9-12.A.APR.1 Essential Question: How do I add, subtract and multiply with polynomials?

There are numerous ways to set up the multiplication of two binomials There are numerous ways to set up the multiplication of two binomials.  The first three methods shown here work for multiplying ALL polynomials, not just binomials.  All methods, of course, give the same answer.  Multiply (x + 3)(x + 2)

1. "Distributive" Method: The most universal method 1. "Distributive" Method: The most universal method. Applies to all polynomial multiplications, not just to binomials. Start with the first term in the first binomial - the circled blue X. Multiply (distribute) this term times EACH of the terms in the second binomial. (x + 3)(x + 2) = x ∙ x + x ∙ 2

Add the results: x•x + x•2 + 3•x +3•2 Now, take the second term in the first binomial - the circled red +3 (notice we take the sign also).  Multiply this term times EACH of the terms in the second binomial.  (x + 3)(x + 2) = 3 ∙ x + 3 ∙ 2 Add the results:  x•x + x•2 + 3•x +3•2 x² + 2x +3x + 6 x² + 5x + 6  Answer Do you see the "distributive property" at work? (x + 3)(x + 2) = x(x + 2) + 3(x + 2)

Before we move on to the next set up method, let's look at an example of the "distributive" method involving negative values Notice how the negative sign is treated as part of the term following the sign. (x - 4)(x - 5) = x ∙ x + x ∙ -5 (x - 4)(x - 5) = -4 ∙ x + -4 ∙ -5 Add the results:   x•x + x•(-5) +(-4)•x +(-4)•(-5) x² - 5x - 4x + 20 x² - 9x + 20  Answer

3x + 6 multiply "3" from bottom term times "x+2“ 2.  "Vertical" Method: x + 2     x + 3     3x + 6 multiply "3" from bottom term times "x+2“   x² + 2x       multiply "x" from bottom term times "x+2“ x² + 5x + 6 add the like terms Be sure to line up the like terms.

O: (x + 3)(x + 2) x · 2 = 2x I: (x + 3)(x + 2) 3 · x = 3x For Binomial Multiplication ONLY! "FOIL" Method:   multiply First Outer Inner Last The words/letters used to describe the FOIL process pertain to the distributive method for multiplying two binomials.  These words/letters do not apply to other multiplications such as a binomial times a trinomial.   F:  (x + 3)(x + 2)  x · x = x2 O:  (x + 3)(x + 2) x · 2 = 2x I:   (x + 3)(x + 2)  3 · x = 3x L:  (x + 3)(x + 2)  3· 2 = 6 Add the four terms together: x2 + 2x + 3x + 6 = x2 + 5x + 6 The drawback to using the FOIL lettering is that it ONLY WORKS on binomial multiplication.

The following are special multiplications involving binomials that you will want to try to remember. Be sure to notice the patterns in each situation. You will be seeing these patterns in numerous problems. Don't panic! If you cannot remember these patterns, you can arrive at your answer by simply multiplying with the distributive method. These patterns are, however, very popular. If you can remember the patterns, you can save yourself some work.

Let's examine these patterns: Squaring a Binomial - multiplying times itself (a + b)² = a² + 2ab + b² (a - b)² = a² - 2ab + b² Notice the middle terms in both of these problems. In each problem, the middle term is twice the multiplication of the terms used to create the binomial expression.

Example 1: (x + 3)² = (x + 3)(x + 3) = x² + 3x + 3x + 9 Distributive method = x² + 6x + 9 * Notice the middle term. Example 2: (x - 4)² = (x - 4)(x - 4) = x² - 4x - 4x + 16 Distributive method = x² - 8x + 16 * Again, notice the middle term.

Product of Sum and Difference (notice that the binomials differ only by the sign between the terms) (a + b)(a - b) = a² - b² Notice that there appears to be no "middle" term to form a trinomial answer, as was seen in the problems above. When multiplication occurs, the values that would form the middle term of a trinomial actually add to zero.

Example 3: (x + 3)(x - 3) = x² - 3x + 3x - 9 Distributive method = x² - 9 *Notice how the middle term is zero. Example 4: (2x + 3y)(2x - 3y) = 4x² - 6xy + 6xy - 9y² Distributive method = 4x² - 9y² *Again, notice how the middle term is zero.

Multiplying a Binomial and a Trinomial Put the "distributive method" of multiplication to work. Multiply each term from the binomial times each term of the trinomial. You can do this by using either a horizontal method or a vertical method. Note: FOIL will not work in this problem. Horizontal Method: (x + 3)(x² + 2x + 4) = x(x² + 2x + 4) + 3(x² + 2x + 4) = x³ + 2x² + 4x + 3x² + 6x + 12 = x³ + 2x² + 3x² + 4x + 6x + 12 Group like terms. = x³ + 5x² + 10x + 12 Combine like terms.

Vertical Method: x² + 2x + 4 x + 3 3x² + 6x + 12 x³ + 2x² + 4x Line up like terms x³ + 5x² + 10x + 12 Add like terms. Cubing a Binomial Expand: (x + 3)³ If you take your time, this problem is really very simple. All you need to do is multiply twice. Multiply (x + 3) times (x + 3), and then multiply that answer by (x + 3). Simple!

(x + 3)³ = (x + 3)(x + 3)(x + 3) = (x² + 3x + 3x + 9)(x + 3) the first multiplication is done = (x + 3) (x² + 6x + 9) combine like terms and switch order = x³ + 6x² + 9x + 3x² + 18x + 27 multiply again = x³ + 9x² + 27x + 27 combine like terms

The "distributive method" of multiplication was used in this problem The "distributive method" of multiplication was used in this problem. Note:  FOIL will not work for the second multiplication in this problem.

Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 The first five rows of Pascal’s triangle are shown. The triangle can be generated recursively.

Binomial expansion Pascal’s triangle The numbers in Pascal’s triangle can be used to find coefficients in binomial expansions (a + b)n where n is a positive integer. Binomial expansion Pascal’s triangle (a + b)0 = 1 1 (n = 0) (a + b)1 = 1a + 1b 1 1 (n = 1) (a + b)2 = 1a2 + 2ab + 1b2 1 2 1 (n = 2) (a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3 1 3 3 1 (n = 3) (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4 1 4 6 4 1 (n = 4)

Example: Use the Binomial Theorem and Pascal’s triangle to write the binomial expansion of (x + 3)5 Solution: The binomial coefficients from the fifth row of Pascal’s triangle are 1,5,10,10,5,1. So the expansion is as follows. (x + 3)5 = 1(x)5 + 5(x)4(3) + 10(x)3(3)2 + 10(x)2(3)3 + 5(x)(3)4 + 1(3)5 = 1(x5) + 5(x4)(3) + 10(x3)(9) + 10(x2)(27) + 5(x)(81) + 1(243) = x5 + 15x4 + 90x3 + 270x2 + 405x + 243

So the expansion is as follows. Example: Use the Binomial Theorem and Pascal’s triangle to write the binomial expansion of (4x + 3)4. Solution: The binomial coefficients from the fourth row of Pascal’s triangle are 1,4,6,4,1. So the expansion is as follows. (4x + 3)4 = 1(4x)4 + 4(4x)3(3) + 6(4x)2(3)2 + 4(4x)(3)3 + 1(3)4 = 1(256x4) + 4(64x3)(3) + 6(16x2)(9) + 4(4x)(27) + 1(81) = 256x4 + 768x3 + 864x2 + 432x + 81