Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Algebra 1 – Unit 10 – Reading Workbook Section 6.3 Multiplication of Polynomials Section 6.3 Objective A: To multiply a polynomial by a monomial 1) What property is used to multiply a polynomial by a monomial? The distributive property 2) Give your own example of multiplying a polynomial by a monomial. x(x^2 + 2) = x^3 + 2x Section 6.3 Objective A Concept Check: CC#1) Find the area of a rectangle that has a length of 5x m and a width of (2x – 7) m. (Where m stands for meters) Multiply: 5x(2x-7) = 10x^2 – 35x square meters CC#2) The base of a triangle is 4x m and the height is (2x +5) m. Find the area of the triangle in terms of the variable x. Area = ½ * b * h = ½ * 4x * (2x+5) = 2x(2x+5) = 4x^2 + 10x square meters CC#3) An athletic field has dimensions of 25 yd by 50 yd. An end zone that is w yards wide borders each end of the field. Express the total area of the field and the end zones in terms of the variable w. Area = length * width = 25 * (50 + 2w) = 1250 + 50w square yards Section 6.3 Objective B: To multiply two polynomials 1) What property is used to multiply a polynomial by another polynomial? The FOIL property 2) Give your own example of multiplying a polynomial by another polynomial. (x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12 Section 6.3 Objective B Concept Check: CC#1) Find the numerical value of the “?” in the following: (3x2 + 2x +5)(x + 3) = 3x2(?) + 2x(?) + 5(?) ? = (x+3) Section 6.3 Objective C: To multiply polynomials that have special products 1) Tell what each letter in FOIL stands for and how it is used in the multiplication of two binomials. First, outside, inside, last It tells you the order for multiplying terms. You multiply the first two terms, then the outside terms, then the inside terms, and finally the last terms. 2) Tell about the special product given by multiplying two binomials that have a sum and difference of two terms. Give your own example using numbers as the bases. When you multiply binomials with a sum and a difference, the result is a difference of their squares. For example: (a-b)(a+b) = a^2 – b^2 With numbers: (10-3)(10+3) = 10^2 – 3^2 = 100 – 9 = 91 3) Tell about the special product give by squaring a binomial. Give your own example using numbers as the bases. You get the first term squared, plus twice the product of the two terms, plus the last term squared. The formula is: (a+b)^2 = a^2 + 2ab + b^2 With numbers: (2+3)^2 = 2^2 + 2*2*3 + 3^2 = 4 + 12 + 9 = 25 Section 6.3 Objective C Concept Check: Simplify: CC#1) (x – 3)(x – 6) = x^2 – 6x – 3x + 18 = x^2 – 9x + 18 CC#2) (x +4)(x + 5) = x^2 + 5x + 4x + 20 = x^2 + 9x + 20 CC#3) (x – 6)(x + 2) = x^2 + 2x – 6x – 12 = x^2 – 4x - 12 CC#4) (x + 4)(x – 7) = x^2 – 7x + 4x – 28 = x^2 – 3x - 28 CC#5) (x + 3)(x – 3) Difference of squares: = x^2 - 9 CC#6) (x + 4)2 Square: = x^2 + 8x + 16 CC#7) (5b – 7c)(5b + 7c) Difference of squares: = 25b^2 – 49c^2 CC#8) (4y – 5z)2 Square: = 16y^2 – 40yz + 25z^2 Section 6.3 Objective D: To solve application problems Take some time and make sure you can work application problems before continuing. Section 6.3 Objective D Concept Check: CC#1) The length of a rectangle is 6 in. The width is 4 in. Find the area of the rectangle. Area = length*width Area = 6*4 = 24 square inches CC#2) The length of a rectangle is (x + 3) in. The width is (x – 2) in. Find the area of the rectangle in terms of the variable x. Area = length*width = (x+3)(x-2) = x^2 – 2x + 3x – 6 = x^2 + x – 6 square inches Section 6.4 Division of Polynomials Section 6.4 Objective A: To divide a polynomial by a binomial 1) Describe two methods of simplifying the expression 12 + 36 6 You could add the two numbers on the top, and then divide by 6. Or, you could split this into two fractions, 12/6 and 36/6, simplify those, and then add the result. Either way, the answer is 8 2) Explain how to use multiplication to check that 8x5 + 12x3 equals 2x3 + 3x 2 4x You can multiply the denominator by the supposed result, and hope you get the numerator: 4x^2(2x^3+3x) = 8x^5 + 12x^3 You do, so it works. Section 6.4 Objective A Concept Check: CC#1) What is the quotient of 8x2y + 4xy and 2xy? Factor the top: 2xy(4x+2)/2xy Cancel: 4x+2 CC#2) What is 6a2b2 – 9a2b + 18ab divided by 3ab? Factor out 3ab from the top: 3ab(2ab – 3a + 6)/3ab Cancel 2ab – 3a + 6 Section 6.4 Objective B: To divide polynomials 1) What equation is good to use as a check of your answer when dividing polynomials? You should check your answer by multiplying the result by the denominator of the original fraction, to ensure that you get back the numerator. 2) When you are dividing polynomials, how do you want your terms arranged in your divisor and dividend? The terms should be arranged in order of decreasing degree. Section 6.4 Objective B Concept Check: CC#1) Given that x3 + 1 = x2 – x + 1 x+1 name two factors of x3 + 1. The factors are then: (x+1) and (x^2-x+1) CC#2) 3x + 1 is a factor of 3x3 – 8x2 – 33x – 10. Find a quadratic factor of 3x3 – 8x2 – 33x – 10. Divide: x^2 - 3x -10 3x + 1 | 3x^3 – 8x^2 – 33x – 10 3x^3 + x^2 -9x^2 -33x -10 -9x^2 -3x -30x – 10 -30x -10 0 The factor is x^2 – 3x - 10 CC#3) 4x – 1 is a factor of 8x3 – 38x2 + 49x – 10. Find a quadratic factor of 8x3 – 38x2 + 49x – 10. Divide: 2x^2 - 9x + 10 4x – 1 | 8x^3 – 38x^2 + 49x – 10 8x^3 – 2x^2 - 36x^2 +49x -10 - 36x^2 + 9x 40x -10 40x -10 0 The factor is: 2x^2 – 9x + 10 CC#4) Is 2x – 3 a factor of 4x3 + x - 12? Explain your answer. Divide: 2x^2 + 3x + 5 2x – 3 | 4x^3 + 0x^2 + x – 12 4x^3 - 6x^2 6x^2 + x – 12 6x^2 - 9x 10x -12 10x – 15 3 NO, it is not a factor, since there is a remainder in the division. Section 6.4 Objective C: To divide polynomials using synthetic division 1) When using synthetic division which term must be in the form x - a, the divisor or the dividend? The divisor (bottom of the fraction) must be in that form. 2) Why can the variables be omitted when using synthetic division? As long as we keep all of the terms in order, and leave “zeros” for terms with 0 coefficients, we can eliminate the variables. The ordering maintains them throughout the process, and we can place the variables back in at the end. 3) How is the degree of the quotient determined by the degree of the dividend and the divisor? For synthetic division problems, the degree of the quotient is one less than the degree of the dividend (since the divisor is always degree 1). For generic division problems, the degree of the quotient = degree of the dividend – degree of the divisor. 4) Why would you want to replace the constant term in the divisor by its additive inverse? We have to replace the constant term by its additive inverse since we are trying to find whether (x-a) is 0, in which case, x = a, not –a. Section 6.4 Objective C Concept Check: CC#1) a) b) c) d) e) Which of the following divisions can be performed using synthetic division? (x2 + 3x +1)/(x – 2), yes (x6 – 8x4 + 3x2 – 9)/(x + 9), yes (x4 – 5x3 – x2 + 7x +3)/(x2 + 1), no (x8 – 2)/(x2 – 4), no (2x2 + 6x + 7)/(5 – x), yes Section 6.4 Objective D: To evaluate a polynomial using synthetic division 1) What are you doing when you evaluate a polynomial? When you “evaluate” a polynomial, you are determining the numerical result that you get from plugging a number in to each “x” (or other variable) in the equation. 2) If you wanted to evaluate a polynomial for x = 5 using synthetic division, what binomial would you use as your divisor in the synthetic division? (x-5) 3) If you wanted to evaluate a polynomial for x = -5 using synthetic division, what binomial would you use as your divisor in the synthetic division? (x+5) Section 6.4 Objective D Concept Check: CC#1) State the Remainder Theorem. The Remainder Theorem says that f(a) is equal to the remainder when you divide f(x) by (x-a). CC#2) If the polynomial 3x4 – 8x2 + 2x + 1 is divided by x + 2 and the remainder is 13, what do we know about f(-2) for the function f(x) = 3x4 – 8x2 + 2x + 1? We know that f(-2) is equal to 13, by the remainder theorem.