University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

345

Pages:

4
Author:

Ruby Bates

Question 1 Make a perfect square trinomial out of the binomial by figuring out the constant that needs to be added, then write it down and factor it. 𝑥 2 − 3 𝑥 x 2 −3x Answer: The coefficient of 𝑥 x-term in 𝑥 2 − 3 𝑥 x 2 −3x is − 3 −3. Half of − 3 −3 is: − 3 2 2 −3 Square it: ( − 3 2 ) 2 = 9 4 . ( 2 −3 ) 2 = 4 9 . Add 9 4 4 9 to complete the square: 𝑥 2 − 3 𝑥 + 9 4 . x 2 −3x+ 4 9 . Factor: ( 𝑥 − 3 2 ) 2 . (x− 2 3 ) 2 . Question 2 Find a polynomial of the specified degree that has the given zeros. Degree 4, zeros − 2 , 0 , 2 , 4 −2,0,2,4. Answer: Use the Factor Theorem: For zeros − 2 , 0 , 2 , 4 −2,0,2,4, the factors are: ( 𝑥 + 2 ) , 𝑥 , ( 𝑥 − 2 ) , ( 𝑥 − 4 ) . (x+2),x,(x−2),(x−4). Multiply the factors to form the polynomial: 𝑃 ( 𝑥 ) = 𝑥 ( 𝑥 + 2 ) ( 𝑥 − 2 ) ( 𝑥 − 4 ) . P(x)=x(x+2)(x−2)(x−4). Expand ( 𝑥 + 2 ) ( 𝑥 − 2 ) (x+2)(x−2): ( 𝑥 + 2 ) ( 𝑥 − 2 ) = 𝑥 2 − 4. (x+2)(x−2)=x 2 −4. Expand ( 𝑥 2 − 4 ) ( 𝑥 − 4 ) (x 2 −4)(x−4): ( 𝑥 2 − 4 ) ( 𝑥 − 4 ) = 𝑥 3 − 4 𝑥 − 4 𝑥 2 + 16 = 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16. (x 2 −4)(x−4)=x 3 −4x−4x 2 +16=x 3 −4x 2 −4x+16. Multiply by 𝑥 x: 𝑃 ( 𝑥 ) = 𝑥 ( 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16 ) = 𝑥 4 − 4 𝑥 3 − 4 𝑥 2 + 16 𝑥 . P(x)=x(x 3 −4x 2 −4x+16)=x 4 −4x 3 −4x 2 +16x. Final Answer: 𝑃 ( 𝑥 ) = 𝑥 4 − 4 𝑥 3 − 4 𝑥 2 + 16 𝑥 . P(x)=x 4 −4x 3 −4x 2 +16x. Question 3 To add polynomials, we add _____ terms. Example: ( 3 𝑥 2 + 4 𝑥 + 6 ) + ( 5 𝑥 2 + 2 𝑥 + 1 ) (3x 2 +4x+6)+(5x 2 +2x+1). Answer: To add polynomials, we add similar terms. Combine similar terms: ( 3 𝑥 2 + 4 𝑥 + 6 ) + ( 5 𝑥 2 + 2 𝑥 + 1 ) = 3 𝑥 2 + 5 𝑥 2 + 4 𝑥 + 2 𝑥 + 6 + 1. (3x 2 +4x+6)+(5x 2 +2x+1)=3x 2 +5x 2 +4x+2x+6+1. Simplify: = 8 𝑥 2 + 6 𝑥 + 7. =8x 2 +6x+7. Question 4 To subtract polynomials, we subtract _____ terms. Example: ( 2 𝑥 3 + 9 𝑥 2 + 𝑥 + 10 ) − ( 𝑥 3 + 𝑥 2 + 6 𝑥 + 8 ) (2x 3 +9x 2 +x+10)−(x 3 +x 2 +6x+8). Answer: To subtract polynomials, we subtract like terms. Subtract 𝑥 3 , 𝑥 2 , 𝑥 , x 3 ,x 2 ,x, and constants: ( 2 𝑥 3 + 9 𝑥 2 + 𝑥 + 10 ) − ( 𝑥 3 + 𝑥 2 + 6 𝑥 + 8 ) . (2x 3 +9x 2 +x+10)−(x 3 +x 2 +6x+8). Perform subtraction: = ( 2 𝑥 3 − 𝑥 3 ) + ( 9 𝑥 2 − 𝑥 2 ) + ( 𝑥 − 6 𝑥 ) + ( 10 − 8 ) . =(2x 3 −x 3 )+(9x 2 −x 2 )+(x−6x)+(10−8). Simplify: = 𝑥 3 + 8 𝑥 2 − 5 𝑥 + 2. =x 3 +8x 2 −5x+2. Final Answer: 𝑥 3 + 8 𝑥 2 − 5 𝑥 + 2. x 3 +8x 2 −5x+2. Question 5 The polynomial 𝑃 ( 𝑥 ) P(x) of degree 4 has: A root of multiplicity 2 at 𝑥 = 4 x=4, A root of multiplicity 1 at 𝑥 = 0 x=0, A root of multiplicity 1 at 𝑥 = − 2 x=−2, Passes through the point ( 5 , 24.5 ) (5,24.5). Find a formula for 𝑃 ( 𝑥 ) P(x). Answer: Roots: 𝑃 ( 𝑥 ) = 𝐴 ( 𝑥 − 4 ) 2 𝑥 ( 𝑥 + 2 ) . P(x)=A(x−4) 2 x(x+2). Expand: 𝑃 ( 𝑥 ) = 𝐴 ( 𝑥 2 − 8 𝑥 + 16 ) ( 𝑥 2 + 2 𝑥 ) . P(x)=A(x 2 −8x+16)(x 2 +2x). Expand further: 𝑃 ( 𝑥 ) = 𝐴 ( 𝑥 4 − 6 𝑥 3 + 16 𝑥 2 − 16 𝑥 ) . P(x)=A(x 4 −6x 3 +16x 2 −16x). Use the point ( 5 , 24.5 ) (5,24.5) to find 𝐴 A: Substitute 𝑥 = 5 x=5 and 𝑃 ( 5 ) = 24.5 P(5)=24.5: 24.5 = 𝐴 ( 5 − 4 ) 2 ( 5 ) ( 5 + 2 ) . 24.5=A(5−4) 2 (5)(5+2). Simplify: 24.5 = 𝐴 ( 1 2 ) ( 5 ) ( 7 ) . 24.5=A(1 2 )(5)(7). 24.5 = 35 𝐴 . 24.5=35A. Solve for 𝐴 A: 𝐴 = 24.5 35 = 7 10 . A= 35 24.5 = 10 7 . Final Answer: 𝑃 ( 𝑥 ) = 7 10 ( 𝑥 − 4 ) 2 𝑥 ( 𝑥 + 2 ) . P(x)= 10 7 (x−4) 2 x(x+2).