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assignment10_sol - MATH 135 Assignment 10 Solutions Problem...

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MATH 135 Fall 2005 Assignment 10 Solutions Problem List Chapter 9: 5, 23, 40, 44, 45, 48, 58, 63, 100, 159 Recommended Problems Exercise 9-5: Find the sum, difference and product of each of the following pairs of polynomials with coefficients in the indicated field. x 3 + 2 x + 2 and x 4 + x 2 + x + 1 in Z 3 [ x ] Solution: If f ( x ) = x 3 + 2 x + 2 and g ( x ) = x 4 + x 2 + x + 1 Z 3 [ x ] then, f ( x ) + g ( x ) = x 4 + x 3 + x 2 f ( x ) - g ( x ) = - x 4 + x 3 - x 2 + x + 1 = 2 x 4 + x 3 + 2 x 2 + x + 1 f ( x ) g ( x ) = ( x 3 + 2 x + 2)( x 4 + x 2 + x + 1) = x 7 + x 5 + x 4 + x 3 + 2 x 5 + 2 x 3 + 2 x 2 + 2 x + 2 x 4 + 2 x 2 + 2 x + 2 = x 7 + 3 x 5 + 3 x 4 + 3 x 3 + 4 x 2 + 4 x + 2 = x 7 + x 2 + x + 2 . Exercise 9-23: Find the quotient and remainder when f ( x ) is divided by g ( x ). f ( x ) = x 3 + 2 x 2 + 2 and g ( x ) = 2 x 2 + x + 1 in Z 3 [ x ] Solution: Using long division (since 2(2) = 1 in Z 3 ), 2 x 2 x 2 + x + 1 x 3 + 2 x 2 + 0 x + 2 x 3 + 2 x 2 + 2 x x + 2 Hence when f ( x ) is divided by g ( x ), the quotient is 2 x with remainder x + 2. Therefore, f ( x ) = (2 x )( g ( x )) + ( x + 2). Exercise 9-40: Solve the following polynomial equations over the indicated finite field, stating the multiplicity of any repeated solutions. x 5 + x 4 + x 3 + 2 x 2 + 2 x + 2 = 0 in Z 3 Solution: One approach would be to immediately make a table to find a root in Z 3 , factor out a linear factor and continue. A second approach would be to notice a possible factorization. We rewrite the equation as x 3 ( x 2 + x + 1) + 2( x 2 + x + 1) = 0 ( x 3 + 2)( x 2 + x + 1) = 0 Since we are looking over Z 3 , then either x 3 + 2 = 0 or x 2 + x + 1 = 0. Now, x = 1 is a root to both these equations in Z 3 so ( x - 1) = ( x + 2) is a factor. In fact, x 2 + x + 1 = ( x + 2) 2 and x 3 + 2 = ( x + 2)( x 2 + x + 1) = ( x + 2) 3 (after a bit of work), so the original equation becomes ( x + 2) 5 = 0. Thus, x = 1 is the only root and has multiplicity 5.
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Exercise 9-44: Find all the rational roots of each of the following rational polynomials. x 4 + 5 6 x 3 - 5 18 x 2 + 1 9 Solution: The first step to solving the equation x 4 + 5 6 x 3 - 5 18 x 2 + 1 9 = 0 is to multiply through by the least common multiple of the denominators (18, in this case) to obtain a polynomial equation with integer coefficients, namely 18 x 4 + 15 x 3 - 5 x 2 + 2 = 0 So by the Rational Roots Theorem, the possibilities for rational roots to this equation are rational numbers of the form p q where q | 18 and p | 2, i.e. the possible rational roots are ± 1 , ± 1 2 , ± 1 3 , ± 1 6 , ± 1 9 , ± 1 18 , ± 2 , ± 2 3 , ± 2 9 Checking for integer roots first, we see that x = - 1 is a root (since it satisfies the equation), so we obtain ( x + 1)(18 x 3 - 3 x 2 - 2 x + 2) = 0 and this new cubic polynomial has the same possibilities for rational roots as the previous quartic polynomial.
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