Presentation on theme: "For Common Assessment Chapter 10 Review"— Presentation transcript:
1 For Common Assessment Chapter 10 Review For Common AssessmentAdding and Subtracting PolynomialsMultiplying PolynomialsFactoring Polynomials
2 Adding & Subtracting Polynomials To add or subtract polynomials,1) Align The Like Terms2) Add/Subtract The Like Terms*Subtracting is the same as adding the opposite!!** When adding or subtracting, EXPONENTS STAY THE SAME!!
3 There are two ways to add and subtract polynomials There are two ways to add and subtract polynomials. You can do it horizontally or vertically.Horizontal example:Simplify (2z + 5y) + (3z – 2y)(2z + 5y) + (3z – 2y) = 2z + 5y + 3z – 2y = 2z + 3z + 5y – 2y = 5z + 3y
4 Add the following polynomials (9y – 7x + 15a) + (-3y + 8x – 8a) Line up your like terms.9y – 7x + 15a+ -3y + 8x – 8a_________________________6y+ x+ 7a
5 Add the following polynomials (3a2 + 3ab – b2) + (4ab + 6b2) _________________________3a2+ 7ab+ 5b2
6 Add the following polynomials (4x2 – 2xy + 3y2) + (-3x2 – xy + 2y2) Line up your like terms.4x2 – 2xy + 3y2+ -3x2 – xy + 2y2_________________________x2 - 3xy + 5y2
7 Subtract the following polynomials (9y – 7x + 15a) – (-3y +8x – 8a) Line up your like terms and add the opposite.9y – 7x + 15a+ (+ 3y – 8x + 8a)12y– 15x+ 23a
8 Subtract the following polynomials (7a – 10b) – (3a + 4b) 4a– 14b
12 Find the sum or difference. (5a – 3b) + (2a + 6b)
13 Find the sum or difference. (5a – 3b) – (2a + 6b)
14 (5x2 - 3x + 7) + (2x2 + 5x - 7)= 7x2 + 2x(3x3 + 6x - 8) + (4x2 + 2x - 5)= 3x3 + 4x2 + 8x - 13Example… this is called the horizontal method…where you keep the problem written horizontally and mentally group the terms together by degree. You can also write the rearrangement down (by commuting the terms).Just be sure to combine the terms with the correct terms. For example, the second example, it is common for student to accidentally combine the cubic term and the quadratic term.
15 (2x3 + 4x2 - 6) – (3x3 + 2x - 2) (2x3 + 4x2 - 6) + (-3x3 + -2x - -2) More examples… this time with subtraction. You have to remember that every term of that second polynomial is being subtracted. So it is useful to change the problem to an addition one before moving on. You basically have to change every sign in the polynomial that you are subtracting. In essence, you are distributing a negative one.(7x3 - 3x + 1) + (-x3 - -4x2 - -2)= 6x3 + 4x2 - 3x + 3
16 7y2 – 3y y2 + 3y – 42x3 – 5x2 + 3x – 1 – (8x3 – 8x2 + 4x + 3)= 15y2Now, some people like to arrange their addition and subtracting in the “old school” way… which is how you were taught to add and subtract in elementary school. You just write the problem so that the like terms line up with each other vertically… then add or subtract as necessary. I put the subtraction in parenthesis to remind myself that every term needs to be subtracted. Sometimes if you don’t , you just think that the coefficient of the first term is negative.–6x3 + 3x2 – x – 4
17 (7y3 +2y2 + 5y – 1) + (5y3 + 7y) 12y3 + 2y2 + 12y – 1 More examples… the zeroes are placed only as place holders. They are unnecessary
19 MULTIPLYING POLYNOMIALS Remember that when you multiply two powers with the same bases, you add the exponents.To multiply two monomials, multiply the coefficients and add the exponents of the variables that are the same.(5m2n3)(6m3n6)5 · 6 · m2+3n3+630m5n9Pre-Algebra
20 Multiplying Monomials A. (2x3y2)(6x5y3)(2x3y2)(6x5y3)Multiply coefficients and addexponents.12x8y5B. (9a5b7)(–2a4b3)(9a5b7)(–2a4b3)Multiply coefficients and addexponents.–18a9b10Pre-Algebra
21 –21x6y7 Try This Multiply. A. (5r4s3)(3r3s2) (5r4s3)(3r3s2) Multiply coefficients and addexponents.15r7s5B. (7x3y5)(–3x3y2)(7x3y5)(–3x3y2)Multiply coefficients and addexponents.–21x6y7
22 Multiplying a Polynomial by a Monomial A. 3m(5m2 + 2m)3m(5m2 + 2m)Multiply each term inparentheses by 3m.15m3 + 6m2B. –6x2y3(5xy4 + 3x4)–6x2y3(5xy4 + 3x4)Multiply each term inparentheses by –6x2y3.–30x3y7 – 18x6y3
23 Multiplying a Polynomial by a Monomial C. –5y3(y2 + 6y – 8)–5y3(y2 + 6y – 8)Multiply each term inparentheses by –5y3.–5y5 – 30y4 + 40y3Pre-Algebra
24 Insert Lesson Title Here Try This: Example 2A & 2BMultiply.A. 4r(8r3 + 16r)4r(8r3 + 16r)Multiply each term inparentheses by 4r.32r4 + 64r2B. –3a3b2(4ab3 + 4a2)–3a3b2(4ab3 + 4a2)Multiply each term inparentheses by –3a3b2.–12a4b5 – 12a5b2
25 Insert Lesson Title Here Example 2Multiply.C. –2x4(x3 + 4x + 3)–2x4(x3 + 4x + 3)Multiply each term inparentheses by –2x4.–2x7 – 8x5 – 6x4Pre-Algebra
26 Multiply. (2x + 3)(5x + 8) Using the Distributive property, multiply 2x(5x + 8) + 3(5x + 8).10x2 + 16x + 15x + 24Combine like terms.10x2 + 31x + 24Another option is called the FOIL method.
27 F.O.I.LF irstO uterI nnerL ast( x + 2 ) ( x + 5 )
28 EXAMPLES ( x + 4 ) ( x + 8 ) = ( x + 5 ) ( x – 6) = x2 + 8x + 4x + 32 + 32x2 + 12x + 32x2− 6x+ 5x− 30x2 − x − 30
29 PRACTICE ( x + 10 ) ( x + 3 ) = ( x − 7 ) ( x − 4 ) =
47 FACTORINGGCF Sum + Product Factor by Grouping 4 Terms Special Products
48 Techniques of Factoring Polynomials 1. Greatest Common Factor (GCF). The GCF for a polynomial is the largest monomial that divides each term of the polynomial.Factor out the GCF:
49 Factoring Polynomials - GCF Write the two terms in theform of prime factors…They have in common 2yyThis process is basically the reverse of the distributive property.
50 Factoring - GCFThe simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term.Example: Factor 18x3 + 60x.GCF = 6xFind the GCF.18x3 + 60x = 6x (3x2) + 6x (10)Apply the distributive law to factor the polynomial.= 6x (3x2 + 10)Check the answer by multiplication.6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x
57 Factoring – Sum and Product To factor a trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example,x2 + 10x + 24 = (x + 4)(x + 6).4 and 6 add up to 104 and 6 multiply to 24
58 Factoring Trinomials factors of 6 that add up to 7: 6 and 1
59 Factoring – Sum and Product Example: Factor x2 – 8x + 15= (x + a)(x + b)= x2 + (a + b)x + abTherefore a + b = – 8and ab = 15.It follows that both a and b are negative.x2 – 8x + 15 = (x – 3)(x – 5).
60 Factoring – Sum and Product Example: Factor x2 + 13x + 36.= (x + a)(x + b)= x2 + (a + b) x + abTherefore a and b are two positive factors of 36 whose sum is 13.x2 + 13x + 36= (x + 4)(x + 9)
61 Factoring 4 Terms by Grouping There is no GCF for allfour terms.In this problem we factor GCFby grouping the first twoterms and the last two terms.
62 Factoring – By Grouping 4 Terms Some polynomials can be factored by grouping terms to produce a common binomial factor.Examples: Factor 2xy + 3y – 4x – 6.= (2xy + 3y) – (4x + 6)Group terms.= (2x + 3)y – (2x + 3)2= (2x + 3)( y – 2)Factor each pair of terms.Factor out the common binomial.
64 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 8b2 + 2b – 3Now factor by grouping
65 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 2x2 + 19x - 10Now factor by grouping
66 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 6y2 – 11y - 10Now factor by grouping
67 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 2x2 – x – 3Now factor by grouping
68 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 3t2 + 16t + 5Now factor by grouping
69 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 5x2 + 2x – 3Now factor by grouping
70 Factoring a trinomial when a ≠ 1 Multiply 8 -3, and break up the middle termFactor 6b2 – 11b – 2Now factor by grouping
71 Factoring the Difference of Two Squares a2– ab + ab – b2= a2 – b2(a + b)(a – b) =a2 – b2 = (a + b)(a – b)FORMULA:The difference of two bases being squared, factors as the product of the sum and difference of the bases that are being squared.
72 Factoring the difference of two squares a2 – b2 = (a + b)(a – b)Factor x2 – 4y2Factor 16r2 – 25(4r)2(5)2(2y)2(x)2DifferenceOf two squaresDifferenceof two squares(x – 2y)(x + 2y)(4r – 5)(4r + 5)
76 Factoring – Special Products A difference of squares can be factoredusing the formulaa2 – b2 = (a + b)(a – b).Example: Factor x2 – 9y2.= (x)2 – (3y)2Write terms as perfect squares.= (x + 3y)(x – 3y)