(2) Dividing both sides of the equation in (1) by the divisor, 3, gives: =
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1 Dividing Polynomials Prepared by: Sa diyya Hendrickson Name: Date: Let s begin by recalling the process of long division for numbers. Consider the following fraction: Recall that fractions are just division problems, so we proceed with the process of long division small We can express our result in one of the following ways: (1) 2683 = 3(894) + 1 (2) Dividing both sides of the equation in (1) by the divisor, 3, gives: = Now, let s consider a fraction, where the numerator and denominator are polynomials. P (x) D(x) = 2x3 + 5x 2 6x + 7 x 2 2x 5 The good news is that the strategies for dividing polynomials are almost identical to those used for numbers! Before exploring an example, we first make note of the Division Algorithm for polynomials. Making Math Possible 1 of 6 c Sa diyya Hendrickson
2 Dividing Polynomials Division Algorithm: If P (x) and D(x) are polynomials, with D(x) = 0, then there exist unique polynomials Q(x) and R(x), where R(x) is either 0 or of degree less than the degree of D(x), such that P (x) R(x) = Q(x) + D(x) D(x) or equivalently, P (x) = D(x) Q(x) + R(x) When the remainder equals zero, we say that D(x) is a factor of the polynomial, P (x). Notice that this is identical to the definition of a factor in the context of integers! Definition: zero/root of a polynomial c R is a zero/root of a polynomial P (x) if P (c) = 0. (e.g. x = 2 is a root of p(x) = x 2 5x + 6.) Exercise 1: Verify that t = is a root of p(t) = t 2 + 2t 1 Sample solution: By definition of a root, we must verify that p( 1 + 2) = 0. p( 1 + 2) = ( 1 + 2) 2 + 2( 1 + 2) 1 = ( 1 + 2)( 1 + 2) = ( 2) = = 0, as required. It is important to note that some polynomials do not have any zeros. For example, because x 2 0 for all x R, p(x) = x always produces values that are greater than or equal to. However, if a polynomial has least one root, there is an interesting and useful relationship between the root and a factor of the polynomial. This relationship is formally stated in the Factor Theorem, given on the next page. Making Math Possible 2 of 6 c Sa diyya Hendrickson
3 Factor Theorem Factor Theorem: Let P (x) denote a polynomial. Then, c is a zero of P (x) if and only if x c is a factor of P (x). The Factor Theorem gives us the following results: 1. If x = c is a zero of P (x), then x c is a factor of P (x). 2. If x c is a factor of P (x), then x = c is a zero of P (x). Exercise 2: Consider the polynomial p(x) = 2x 2 + x 6. Verify that x = 3 2 are roots. What can we then conclude and why? and x = 2 Sample solution: We must verify that p 3 2 = 0 = p( 2) p = = = = = 6 6 = and p ( 2) = 2 ( 2) 2 + ( 2) 6 = 2 (4) 2 6 = 8 8 = 0 Because x = 3 2 and x = 2 are roots, we conclude (by the Factor Theorem) that x 3 2 and x ( 2) = x + 2 are factors. Notice that if we factor by trial and error, we have: 2x 2 + x 6 = (2x 3)(x + 2) 2 = 2x 3 2 = 2 x 3 2 By factoring out 2, we can see the factors! (x + 2) (x + 2) Making Math Possible 3 of 6 c Sa diyya Hendrickson
4 Polynomial Long Division Exercise 3: Suppose we are given the following information about a polynomial: Degree = 2 Roots: x = 1 ± 2 Leading Coefficient = 3 Determine the expanded and factored forms of the polynomial. Sample solution: Because the polynomial is of degree 2, it can be expressed as: p(x) = ax 2 + bx + c, where a, b, c R Furthermore, since x = 1 ± 2 are roots, the Factor Theorem gives that x ( 1 + 2) and x ( 1 2) are factors. Thus, the factored form must be: p(x) = a (x ( 1 + 2))(x ( 1 2)) To achieve a leading coefficient of 3, we require that a = 3 (Why?). For expanded form: p(x) = 3(x ( 1 + 2))(x ( 1 2)) = 3(x 2 ( 1 2)x ( 1 + 2)x + ( 1 + 2)( 1 2)) = 3(x 2 ( )x + ( 1) 2 ( 2) 2 ) = 3(x 2 ( 2)x + 1 2) = 3(x 2 + 2x 1) = 3x 2 + 6x 3 There are two popular methods for dividing polynomials: long division and synthetic division. Synthetic division is an approach that tends to be easier on the eyes and more compact, but it has its shortcomings. In particular, synthetic division can only be used when D(x) is linear. On the other hand, polynomial long division has no such limitations! So, in this handout, we will explore the more robust method of polynomial long division. Let s begin with the following exercise: Exercise 4: Find the quotient and remainder of x 4 16 x + 2 Option 1: Notice that x 4 16 = x (2 2) 4 2 P.O.E = (x 2 ) = (x 2 + 4)(x 2 4) factoring D.O.S. = (x 2 + 4)(x ) = (x 2 + 4)(x + 2)(x 2) factoring D.O.S. Now, we can see that x + 2 is a factor of x Thus, the remainder is zero. Furthermore, the quotient is the product of the remaining factors: (x 2 + 4)(x 2) = x 3 2x 2 + 4x 8. Making Math Possible 4 of 6 c Sa diyya Hendrickson
5 Polynomial Long Division Exercise 4 (Continued) Option 2: To find the quotient and remainder, we consider polynomial long division: The quotient is the cubic polynomial, x 3 2x 2 + 4x 8 and the remainder is zero. Option 3: Notice that x = ±2 are roots of p(x) = x 4 16, since: p(±2) = (±2) 2 16 = = 0 Thus, by the Factor Theorem, x 2 and x ( 2) = x+2 are factors. Furthermore, because x + 2 is a factor, we know that the remainder is zero. Moreover, (x 2)(x + 2) = x 2 4 must be a quadratic factor of p(x). To determine the remaining factor, consider long division: The the remainder is zero and the quotient is the product of the factors: x 2 and x This product gives the cubic polynomial, x 3 2x 2 + 4x 8. Making Math Possible 5 of 6 c Sa diyya Hendrickson
6 Polynomial Long Division Exercise 5: Express 6x4 2x 3 + x 5 3x 3 + x 1 as Q(x) + R(x) D(x). Sample solution: By long division, we have: Making Math Possible 6 of 6 c Sa diyya Hendrickson
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