PRECALCULUS A TEST #2 POLYNOMIALS AND RATIONAL FUNCTIONS, PRACTICE

Transcription

1 PRECALCULUS A TEST # POLYNOMIALS AND RATIONAL FUNCTIONS, PRACTICE SECTION.3 Polynomial and Synthetic Division 1) Divide using long division: ( 6x x 4x 9) ( 3x ) ) Divide using long division: ( x 3 13x 1) ( x 4) 3) Divide using synthetic division: ( x 3 + 5x 7x 1) ( x + 3) 4) Divide using synthetic division: ( x 4 5x 10x 1) ( x ) 5) Find the other solutions of the equation 3 4x 1x x = if one of the solutions is ½. SECTION.4 Real Zeros of Polynomial Functions 6) Use Descartes s Rule of Signs to determine the possible number of positive and negative real zeros of the function f ( x) = 3x x 3x + x 1. 7) Use the Rational Zero Test to list all possible rational zeros of the 4 3 function f ( x) = 4x + 16x 47x + 9x ) Find all the real zeros of the function 9) Find all the real zeros of the function f x x x x 3 = f x x x x x 4 3 = SECTION.5 Complex Numbers 10) Simplify 7. 11) Simplify ( + i) ( i) + ( + i) 1) Simplify ( + 3i )( 4 5 i). 13) Simplify ( i) 14) Simplify + 4 i. 5 3i ) Solve the equation x + 4x + 15 = 0 using the Quadratic Formula.

2 SECTION.6 The Fundamental Theorem of Algebra 16) Find all the zeros of the function f x x x x x 4 3 = ) Use the answers from Problem #16 to write the factorization of 4 3 x x x x ) Find all the zeros of the function f x x x x x 4 3 = ) Use the answers from Problem #18 to write the factorization of 4 3 x x x + 4x +. 0) Find all the zeros of the function is 4 i. 4 3 f ( x) = 3x + 5x + 57x + x 34 if one of the solutions SECTION.7 Rational Functions 1) If f ( x) =, x + 3 provide as much information as possible about the graph of this function. ) 3) x + 4 If f ( x) =, 3x 1 3 x + 1 If f ( x) =, x 9 provide as much information as possible about the graph of this function. provide as much information as possible about the graph of this function. 4) x 5x + 6 If f ( x) =, function. x + 5x 3 provide as much information as possible about the graph of this 5) 8x 16 Graph f ( x) = on the coordinate plane. x + 6 SECTION.8 Partial Fractions 9x + 3 6) Write the partial fraction decomposition of. x + x x + 7 7) Write the partial fraction decomposition of. x x 6 5x 7 8) Write the partial fraction decomposition of. ( x ) 4x 1 9) Write the partial fraction decomposition of. (x + 1)

3 **********************ANSWERS********************** 1) ) 3) 4) x x + x x ( 6x 4x ) 3 ( x 4x ) x 5x 3 3x 6x + 11x 4x 9 15x 4 ( 15x 10x) x x 13 x ( 4x 16x) x 5 3x 6x 9 ( 6x 4) 5 + 4x + 3 x x 4 3x 1 ( 3x 1) x + x x x ) Since there are only three terms left, use the Quadratic Formula ± = = & = = ) f x x x x x = sign changes 3 or 1 positive real zeros f x x x x x ( ) = sign changes or 0 positive real zeros

4 7) Factors of p = ± 1, ±, ± 3, ± 6, ± 9, ± 18 Factors of q = ± 1, ±, ± Possible rational zeros = ± 1, ±, ± 3, ± 6, ± 9, ± 18, ±, ±, ±, ±, ±, ± ) 3 Graph f ( x) = x 5x + 5x 1. The graph appears to cross the x-axis at 1. Divide the original polynomial by 1 using synthetic division ) The remaining polynomial only has three terms, so use the Quadratic Formula. 4 ± 1 4 ± 3 ± 3 Zeros are 1, ± Graph f ( x) = x + 7x 9x + 34x + 0. The graph appears to cross the x-axis at 5. Divide the original polynomial by 5 using synthetic division The remaining polynomial has more than three terms, so you can t use the Quadratic Formula yet. The graph also appears to cross the x-axis at ½, so divide the new polynomial by ½ using synthetic division The remaining polynomial only has three terms, so use the Quadratic Formula. 4 ± 48 10) 7 = 36 1 = 6i 1 Stop here, as -48 is imaginary Zeros are -5 and - 11) + 3i 6 8i + 1+ i = + 3i i + 1+ i = 3 + 1i 1) + 3i 4 5i = 8 10i + 1i 15i = 8 + i + 15 = 3 + i 13) 3 5i = 3 5i 3 5i = 9 15i 15i + 5i = 9 30i 5 = 16 30i 14) + 4i 5 + 3i i + 0i + 1i i 1 + 6i 1+ 13i = = = = 5 3i 5 + 3i 5 9i

5 15) 16) ± ± ± ± ± = = = = i 6 i Graph f ( x) = x + x + 6x + 18x 7. The graph appears to cross the x-axis at 1. Divide the original polynomial by 1 using synthetic division The remaining polynomial has more than three terms, so you can t use the Quadratic Formula yet. The graph also appears to cross the x-axis at 3, so divide the new polynomial by 3 using synthetic division The remaining polynomial only has three terms, so use the Quadratic Formula. 0 ± 36 0 ± 6i 0 + 6i 6i 0 6i 6i = = 3 i & = = 3i Zeros are 1, 3, 3i, and 3i 17) Change zeros into factors x 1 x 1 0 Factor = x 1 = = ( ) x = 3 x + 3 = 0 Factor = ( x + 3) = 3 3 = 0 Factor = ( 3 ) x = 3i x + 3i = 0 Factor = ( x + 3i ) = ( 1)( + 3)( 3 )( + 3 ) x i x i x i 4 3 x x x x x x x i x i

6 18) 4 3 Graph f ( x) = x x x + 4x +. The graph appears to cross the x-axis at 1. Divide the original polynomial by 1 using synthetic division The remaining polynomial has more than three terms, so you can t use the Quadratic Formula yet. The graph also appears to cross the x-axis at ½, so divide the new polynomial by ½ using synthetic division The remaining polynomial only has three terms, so use the Quadratic Formula. 4 ± 16 4 ± 4i 4 + 4i 4 4i 1 = 1 + i & = 1 i Zeros are 1,, 1 + i, 1 i ) Change zeros into factors x = 1 x + 1 = 0 Factor = x x = x = 1 x + 1 = 0 Factor = ( x + 1) x 1 i x 1 i 0 Factor = x 1 i x = 1 i x 1+ i = 0 Factor = x 1+ i = + = ( ) 4 3 x x x + x + = ( x + )( x + )( x + i)( x i)

7 0) If 4 i is a zero, then 4 + i is also a zero. Convert these zeros to factors. 4 3 x x x x x x 4 3 ( 3x 4x 51x ) 3 x + 6x + x 3 ( x 8x 17x) 3x + x ( x i)( x i) x = 4 i x i = 0 Factor = x i x = 4 + i x + 4 i = 0 Factor = x + 4 i Multiply these two factors x + 4x + ix + 4x i ix 4i i = x + 8x = x + 8x + 17 Divide the original polynomial by x + 8x x 16x 34 ( x 16x 34) 0 The remaining polynomial has three terms, so use the Quadratic Formula. 1± 5 1± = = = = & = = Zeros are 4 i, 4 + i, 1, 3 1) To find y-intercept, set x = 0 y = y = y-intercept = 0, To find x-intercept, set numerator = 0 Numerator has no variable No x-intercept To find vertical asymptote, set denominator = 0 x + 3 = 0 x = 3 Degree of numerator = 0, degree of denominator = 1 If degree of denominator is greater than degree of numerator, the horizontal asymptote is the x-axis or y = 0

8 ) To find y-intercept, set x = 0 y = y = y = 4 y-intercept = 0, To find x-intercept, set numerator = 0 x + 4 = 0 x = 4 x = x-intercept =, 0 1 To find vertical asymptote, set denominator = 0 3x 1 = 0 3x = 1 x = 3 Degree of numerator = 1, degree of denominator = 1 If degree of denominator is equal to degree of denominator, horizontal asymptote = leading coefficient of numerator over leading coefficient of denominator y = 3 3) 4) To find y-intercept, set x = 0 y = y = y = y-intercept = 0, To find x-intercept, set numerator = 0 x + 1 = 0 x = 1 x = 1 x-intercept = 1, 0 To find vertical asymptote, set denominator = 0 x 9 0 x 3 x 3 0 = + = x = 3 & x = 3 Degree of numerator = 3, degree of denominator = If degree of numerator is greater than degree of denominator, there is no horizontal asymptote To find y-intercept, set x = 0 y = y = y = y-intercept = ( 0, ) To find x-intercept, set numerator = 0 x 5x 6 0 x x 3 + = x = & x = 3 x-intercepts =, 0 & 3, 0 To find vertical asymptote, set denominator = 0 x 5x 3 0 x 1 x = + = 1 x = & x = 3 If degree of denominator is equal to degree of denominator, horizontal asymptote = leading coefficient of numerator over leading coefficient of denominator y = 1

9 5) Horizontal Asymptote: y = 4 Vertical Asymptote: x = ) 9x + 3 9x + 3 9x + 3 A B = = + Multiply each term by the LCD x + x x x + 1 x x + 1 x x + 1 9x + 3 A B x( x + 1) = x( x + 1) + x( x + 1) 9x + 3 = A ( x + 1) + B x x x + 1 x x + 1 Set x = = A B 1 6 = B 6 = B Set x = = A B 0 3 = A 9x = + x + x x x + 1 7) x + 7 x + 7 x + 7 A B = = + Multiply each term by the LCD x x x x x x x x x + 7 A B + = ( x )( x 3) ( x )( x 3) ( x )( x 3) ( x + )( x 3) x + x 3 x + 7 = A ( x 3) + B ( x + ) Set x = = A ( 3 3) + B ( 3 + ) 10 = 5B = B Set x = + 7 = A ( 3) + B ( + ) 5 = 5A 1 = A x = x x x x

10 8) 9) 5x 7 A B = + Multiply each term by the LCD ( x ) x ( x ) 5x 7 A B ( x ) = ( x ) + ( x ) 5x 7 = A ( x ) + B x ( x ) ( x ) Set x = 5 7 = A + B 3 = B Set x = 3 & B = = A = A = A 5x = + ( x ) x ( x ) 4x 1 A B = + Multiply each term by the LCD ( x + 1) x + 1 ( x + 1) 4x 1 A B ( x + 1) = ( x + 1) + ( x + 1) 4x 1 = A ( x + 1) + B + 1 x ( x ) ( x ) Set x = 4 1 = A B 3 = B Set x = 1 & B = = A = 3A 3 6 = 3A = A 4x 1 3 = + ( x + 1) x + 1 (x + 1)