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# Q4SOL - MAP 103 Prociency Algebra Quiz#4 Fall 2009...

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MAP 103: Proficiency Algebra, Quiz #4, Fall 2009 SOLUTIONS 1. Factorize completely: (a) p ( x ) = 8 x 2 + 8 x + 2 Sol. p ( x ) = 8 x 2 + 8 x + 2 = 2(4 x 2 + 4 x + 1) = 2(2 x + 1)(2 x + 1) = 2(2 x + 1) 2 (b) q ( x ) = ( x 2 - 1) 4 - ( x 2 - 1) 2 Sol. Set A = x 2 - 1 , then q ( x ) = ( x 2 - 1) 4 - ( x 2 - 1) 2 = A 4 - A 2 = A 2 ( A 2 - 1) = A 2 ( A - 1)( A + 1) = ( x 2 - 1) 2 ( x 2 - 1 - 1)( x 2 - 1 + 1) = x 2 ( x 2 - 1) 2 ( x 2 - 2) = x 2 [( x - 1)( x + 1)] 2 ( x - 2)( x + 2) 2. Use the Rational Root Theorem to find all the roots of the polynomial p ( x ) = x 3 - 6 x 2 +11 x - 6 . Sol. By the rational root test, we find the possible roots are r = 6 1 = 1 ± 2 ± 3 ± 6 } . By the remainder theorem, we find p (1) = p (2) = p (3) = 0 . Hence, the roots are r 1 = 1 , r 2 = 2 , r 3 = 3 . 3. Find the quotient and remainder (if any) upon long division: x 6 - 5 x 4 + x 2 x 2 - 1 . Sol. In each case, we need place holders. Thus, we are dividing the following: x 6 + 0 x 5 - 5 x 4 + 0 x 3 + x 2 + 0 x + 0 x 2 + 0 x - 1 . Carefully dividing and keeping track of signs, yields the quotient, q ( x ) = x 4 - 4 x 2 +5 , and the remainder, r ( x ) = 0 .

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4. Combine and simplify: (a) 1 x - 1 + 1 x - 2 Sol. We find a common denominator. It is ( x - 1)( x - 2) . Multiplying appropriately, the numerator becomes ( x - 2) + ( x - 1) = 2 x - 3 . Hence, when combined we have 1 x - 1 + 1 x - 2 = 2 x - 3 ( x - 1)( x - 2) . (b) 1 1 - 1 1+ 1 1 - 1 x +1 Sol. In the first denominator, we find 1
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