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MA 1128: Lecture 10 – 6/10/13

MA 1128: Lecture 10 – 6/10/13. Addition, Subtraction, and Multiplication of Polynomials. Polynomials. A polynomial in one variable x is an expression which is the sum of only constant terms, x -terms, x 2 -terms, x 3 -terms, etc.

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MA 1128: Lecture 10 – 6/10/13

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  1. MA 1128: Lecture 10 – 6/10/13 Addition, Subtraction, and Multiplication of Polynomials

  2. Polynomials A polynomial in one variable x is an expression which is the sum of only constant terms, x-terms, x2-terms, x3-terms, etc. For example, 7x4 – 2x2 + x + 3 is a polynomial. The variable doesn’t have to be x, of course, and the book talks about polynomials in several variables. In several variables, we might have y3-terms and x2y4-terms. We’ll stick to one variable, for the most part. You could say that a “nomial” is the same as a term, so “polynomial” means many terms. We’ll actually take “poly” to mean any number of terms, so 3x2 by itself would be a polynomial. Occasionally, the words monomial, binomial, and trinomial are used to explicitly describe a polynomial with one, two, or three terms. Next Slide

  3. Adding Polynomials We’ll need to be able to add and subtract polynomials. We’ve done this already, and called it simplifying. Pay special attention to combining like terms. Example. (3x2 – 2x + 1) + (7x3 – 2x2 + x + 1). With addition, the parentheses don’t really do anything, since we can add the terms in any order. Dropping the parentheses, we get = 3x2 – 2x + 1 + 7x3 – 2x2 + x + 1 Combining like terms we get: = 7x3 + x2 – x + 2 Next Slide

  4. Practice Problems • Add the polynomials 4x4 – 2x2 + x + 7 and 3x3 – 2x + 8. Click for answers: 4x4 + 3x3 2x2 – x + 15 Next Slide

  5. Subtraction of Polynomials Example. (7x2 + 2x – 1) – (3x2 – 2x + 8). Here the second set of parentheses are doing something very important. They tell us that every term inside is being subtracted. We must subtract 3x2, subtract 2x, and subtract 8. = 7x2 + 2x – 1 – 3x2 + 2x – 8 Note that the signs changed on each term in the second polynomial. = 4x2 + 4x – 9. Example. (5x2 – 2x + 7) – (10x3 – 7x2 + 8x – 13) = 5x2 – 2x + 7 – 10x3 + 7x2 – 8x + 13 = 10x3 + 12x2 – 10x + 20. Next Slide

  6. Practice Problems • Subtract: (x2 + 2x – 3) – (7x – 4). • Subtract: (x2 – 4) – (3x3 – 2x2 + x – 7). Answers: 1) x2 – 5x + 1 2) 3x3 + 3x2 – x + 3 Next Slide

  7. Multiplying Polynomials We’ve done a little of this, but not too much. Consider the multiplication (x2 – 2x + 2)(x + 1). Recall that multiplication distributes over addition. A careful application of this concept leads to the fact that Every term in the first polynomial needs to be multiplied times every term in the second. Next Slide

  8. Example In (x2 – 2x + 2)(x + 1), each of the terms x2, 2x, and 2 must be multiplied times each of the terms x and 1. Let’s be systematic: x2 times x and x2 times 1 2x times x and 2x times 1 2 times x and 2 times 1. This gives us the new terms: x3 and x2 2x2 and 2x 2x and 2 These get added together: x3 + x2 – 2x2 – 2x + 2x + 2 = x3 – x2 + 0 + 2 = x3 – x2 + 2 Next Slide

  9. Example (3x2 – x)(x2 – 2x + 3) 3x2 times everything = (3x2)(x2) + (3x2)(-2x) + (3x2)(3) x times everything + (-x)(x2) + (-x)(-2x) + (-x)(3) = 3x4 – 6x3 + 9x2 – x3 + 2x2 – 3x = 3x4 – 7x3 + 11x2 – 3x. Next Slide

  10. Example (x2 + 2x – 1)(2x2 – 3x + 2) = (x2)(2x2) + (x2)(-3x) + (x2)(2) + (2x)(2x2) + (2x)(-3x) + (2x)(2) + (-1)(2x2) + (-1)(-3x) + (-1)(2) = 2x4 – 3x3 + 2x2 + 4x3 – 6x2 + 4x – 2x2 + 3x – 2 = 2x4 + x3 – 6x2 + 7x – 2 Next Slide

  11. Practice Problems • (x2 + 3)(x – 2) • (x2 + 2x – 1)(x + 7) • (x2 – x + 2)(x2 + 2x – 3) Answers: 1) x3 – 2x2 + 3x – 6 2) x3 + 9x2 + 13x – 7 3) x4 + x3 – 3x2 + 7x – 6 End

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