Columbus Secondary School

Home of the Wildcats

'20-'21 Algebra II Students:
here's your supply list for class.
I'm looking forward to August! Enjoy the remainder of your summer.

• binder with loose leaf paper OR spiral notebook with pockets and enough paper to last all school year-for MATH only! (A 1-inch binder with a zippered pencil pouch works best.)
• pencils, pens (dark blue or black for regular classroom purposes),
• large eraser
• colored pencils or an assortment of colored pens for graphing
• graph paper
• protractor
• box of facial tissues and container of disinfecting wipes

Week Six Assignment (Friday, May 15-Thursday, May 21)

Web-Based

Algebra II Khan Academy
 Sign up using code B4E55JC8. 
You’ll review operations on polynomials and working with complex numbers.

Non Web-Based
1. Factor the polynomial 6a²-3a-18.
2. Solve x⁴-14x²+45=0.
3. One factor of x³+x²-14x-24 is x-4. Find the remaining factors.

SOLUTIONS to Week Six

1. 3(2a²-a-6)=3(2a+3)(a-2)
2. (x²-9)(x²-5)     x=±3 and x=±√5
3. Use synthetic division to factor out (x-4). The remaining polynomial is x²+5x+6.
The factors of this remaining trinomial are (x+3) and (x+2).

Week Five Assignment (Friday, May 8-Thursday, May 14)

Web-Based

Algebra II Khan Academy
 Sign up using code B4E55JC8. 
You’ll review operations on polynomials and working with complex numbers.

Non Web-Based
Use synthetic division in each of the following problems.

1. Find f(-5) and f(2) for the polynomial f(x)= 3x⁴+ x³ -2x² +x + 12

2. Determine if (x+4) is a factor of the polynomial 2x³+7x²-53x-28.

3. Is (x-4) a factor to the polynomial 2x³+7x²-53x-28?

SOLUTIONS to Week Five

1. When using synthetic division, the solution is the remainder.
f(-5)=1707       f(2)=62

2. When using synthetic division to test determine a factor, the remainder must be zero.
Don't forget to first set the tested factor x+4 equal to zero. Thus, use -4 as the divisor.
The remainder is 168, so (x+4) is not a factor.

3. Use the same procedure as in problem 2, but this time the divisor used in synthetic division will be four. The remainder is zero. Thus, (x-4) is indeed a factor of the polynomial.

Week Four Assignment (Friday, May 1-Thursday, May 7)

Web-Based

Algebra II Khan Academy
 Sign up using code B4E55JC8. 
You’ll review operations on polynomials and working with complex numbers.

Non Web-Based
1. Write a polynomial function of least degree with integral coefficients that have the given zeros.
a) -3, -5, 1                        b) -1, √3, -√3.                c) -1, 1, i√6, -I√6

2. State the possible number of positive real zeros, negative real zeros, and imaginary zeros of the function,
                                        f(x)=7x⁴+3x³-2x²-x+1

Week Three Assignment (Friday, April 24-Thursday, April 30)

Web-Based

Algebra II Khan Academy
 Sign up using code B4E55JC8. 
You’ll review operations on polynomials and working with complex numbers.

Non Web-Based
Solve each equation by factoring:
1. x² - x -12 = 0

2.  3x² -7 x +2 = 0

3.  x² - 2x -15 = 0

4.  2x² +5 x -3 = 0

SOLUTIONS to Week Three
1. (x-4)(x+3)=0         x=4 and x=-3

2. (3x-1)( x-2)=0       x= 1/3 and x=2

3. (x-5)(x+3)=0        x=5 and x=-3

4. (2x-1)(x+3)=0       x= 1/2 and x=-3

Algebra II Khan Academy
 Sign up using code B4E55JC8. 
You’ll review operations on polynomials and working with complex numbers.

1. Simplify 
   a.The square root of –81 (hint: the result is an imaginary number).
   b.The square root of –48 (again, an imaginary number).
2.  Add, subtract, multiply and divide (3-i) and (4+2i)

 Solutions to Week 2
1. a) √-81 = 9i.      b) √-48 = √16∙√3∙I = 4i√3.  

2. Addition: 7 + I.   Subtract: -1-3i.    Multiply: 14 + 2i.   Divide: (10-10i)/20 simplifies to (1-i)/2 .

Algebra II Khan Academy
 Sign up using code B4E55JC8.
You’ll review operations on polynomials and working with complex numbers.

1. Add, subtract and multiply complex numbers (8 – 6i) and (-12+2i). Be sure to completely simplify each problem.
2. Simplify i to the power of a) 5, b) 12, and c) 19.
3. Rationalize the denominator and simplify the expression: (7+2i)/(4- 5i).

SOLUTIONS to Week One
1. (8-6i)+(-12+2i)= -4-4i
    (8-6i)-(-12+2i)= 20-8i
   (8-6i)(-12+2i)= -96+16i+72i-12i² = -96 + 88i -12(-1) = -96 + 88i + 12 = -84+ 88i
  (8-6i)/(-12+2i)- Rationalize by multiplying by (-12 - 2i)/(-12-2i) = (-108+56i)/148, which simplifies to (-27+14i)/37.
2. a) i⁵ = i     b) i¹²= 1.   c) I¹⁹=-I
3. Multiply (7+2i)/(4- 5i) by (4 + 5i)/(4 + 5i) and simplify. The result is (18 + 43i)/41.