Question 1 (1 point) Let p () = 24 -8x2+ 16 Find the multiplicity at x= -2 (Hint: type in the number of x = -2 answers of the polynomial above, use synthetic division) O1 2 3 0 4Question 2 (1 point) Identify the increase and decrease intervals for the given polynomial graph: 2 -2 -2 -4 -6 -8\fQuestion 3 (1 point) Use Synthetic division to determine f(-2) given f(I) 5x- 2x- + 3x + 5. Enter only the number as answer and no spaces.Question 4 [1 point} ERROR ANALYSIS: In which step was the error made when solving for the roots AND what are the correct solutions? Step 1: 4x3 + 4x2 25x 25 = 0 Step 2: 4x3(x + 1) 25(x + 1) = 0 Step 3: (4x2 - 25)(x + 1) = D Step4: 36:; x: 1 a) Step 2 is wrong; Solutions are @ = 21-1 O b) Step 3 is wrong; Solutions are@ = 19, -1 (multiplicity = 2) O c) Step 4 is wrong; Solutions are = NOT (multiplicity = 2), -1 O d) Step 2 is wrong; Solutions are = 2,-1 Oe) No Error Made Of) Step 4 is wrong; Solutions are @ = ,-1Question 5 (1 point) Identify the end behavior of the function in the graph. Select two (2) of the options below. 120 100 80 60 40 20 -2 -1 0 2 3 -20 -40 -60 .80 -100 Jas x - -00, y - -00 Jas x -+ -00, y -+ 00 Jas x -7 00, y - Do Jas x -+ 00, y - -00Question 6 (1 point) State if the given binomial is a factor of the given polynomial. (463 + 1462 -656- 30) + (b+6) Yes O NoQuestion 7 (1 point) Find all roots. 2x' - 2x- + x- 1=0 O iV2 iV2 1. 2 2 O iV3 iV3 -3. O iV6 iV6 -3, 2 2 iV2 iV2 -2, 2 2Question 8 (1 point) Solve by finding the factored form. Then, state the roots. 2x3 - x2 -32x + 16 =0 O a) (x+8)(x-2)(x+1)=0 ; Roots are x = -8, -2, 2, -1 b) (x-4)(x+4)(2x-1)=0 ; Roots are x = 4, -4, 1/2 O c) (x+4)(x-4)(x-2)=0 ; Roots are x = -4, 4, 2 O d) (x-8)(x+4)(x-4)=0 ; Roots are x = 8, -4, 4 O e) (x+4)(x-4)(2x+1)=0 ; Roots are x = -4, 4, -1/2Question 9 (1 point) What is one factor of the polynomial below? fx =x2 + 7x+ 12 (x+12) (x+3) O (x+6) O (x+2)Question 10 (1 point) Main Content Describe the end behavior of the function below. Ax) =x' - 2x2 -4 Ox + 00 f(z) + -00 0-> - 00 f(@) -> -00 f(x) + 00 0 - - 00 f(2) -> -00 Ox + 0 f(z) - -00 0-> - 00 f(2) -+ 00 Ox + 00 f(x) -+ 00 -> -00 f(@) -> 00