Khan Academy Algebra II

x + 2 - (1/x-2) (-Notice the numerator is missing a 1st degree. Let's add it as 0x (between the x^2 and -5))

(x^2-5)/(x-2) https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/e/quad-by-linear-remainders?modal=1

x^3 + 5x^2 + 3x

(x^4 + 5x^3 + 3x^2)/x

x^4 + 6 + 2/x

(x^5 + 6x + 2)/x

10x^6

Find the missing factor 'C' that makes the following equality true. 40x^9 = (C)(4x^3)

c = 4 (Plug x = 3 in for x and set equation = to 0)

Find the value of 'c' so that the polynomial p(x) is divisible by (x-3) p(x) = -x^3 + cx^2 - 4x + 3

8

Find the value of c so that the polynomial p(x) is divisible by (x+2). p(x) = 4x^3 + cx^2 + x + 2 c =_____

Both of them

Ibuki factored 12x^7 as (4x^3)(3x^4) Melodie factored 12x^7 as (2x^6)(6x) Which of them factored 12x^7 correctly?

b, c (Plug 1 in for x in each equation; the equations that equal 0 are the ones that are divisible by 0)

Select all polynomials that are divisible by (x - 1): a) A(x) = 3x^3 + 2x^2 - x b) B(x) = 5x^3 - 4x^2 - x c) C(x)= 2x^3 - 3x^2 + 2x - 1 d) D(x) = x^3 + 2x^2 + 3x + 2

a, c

Select all polynomials that have (x + 2) as a factor a) A(x) = x^3 - 3x^2 - 10x b) B(x) = x^3 + 5x^2 + 4x c) C(x) = x^3 - 2x^2 - 13x - 10 d) D(x) = x^3 - 6x^2 + 11x - 6

8x^5

Solve: (2x^2)(4x^3) =

2x^3+4x^2-x

(2x^4 + 4x^3 - x^2)/x

2x^3 + 5 + 4/x

(2x^4 + 5x + 4)/x

3x - (10/x)

(3x^2 - 10)/x

3x^4-1

(3x^5 - x)/x

6x^4 - 2x^3- 1/x (1. (6x^5- 2x^4 - 1)/x = (6x^5)/x - (2x^4)/x - 1/x)

(6x^5 - 2x^4 - 1)/x https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:poly-div-by-x/e/poly-by-x-remainders?modal=1

x + 3

(x^2 + 6x + 9)/(x + 3) https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/e/quad-by-linear-no-remainders?modal=1

x-1

(x^2 - 2x + 1)/(x -1)

x + 2 (1. Factoring the numerator: x^2 - 3x - 10 = (x + 2)(x - 5) 2. Cancel common factors: (x + 2)(x - 5)/(x - 5) -cancel (x-5) in numerator and denominator 3. Left with x + 2)

(x^2 - 3x - 10)/(x - 5)

x - 4 - (6/x-5)

(x^2 - 9x + 14)/(x - 5) https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/e/quad-by-linear-remainders?modal=1

Andrei and Amit

Andrei, Amit and Andrew were each asked to factor the term 20x^6 as the produce of two monomials. Their responses are shown below: Andrei- 20x^6 = (2x)(10x^5) Amit- 20x^6 = (4x^3)(5x^3) Andrew- 20x^6 = (20x^2)(x^3)

(3x)(x-4)(x - 1)^2

Factor completely: (3x^2 - 12x)(x^2 - 2x + 1) https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:factor-high-deg/e/factor-higher-degree-polynomials?modal=1

(x + 3)(x - 2)(2x)(x + 2)

Factor completely: (x^2 + x - 6)(2x^2 + 4x)

2x^2(x + 5)(x - 3) ---------------------- 1. 2x^2(x^2 + 2x - 15) 2. 2x^2 (x + 5)(x - 3)

Factor completely: 2x^4 + 4x^3 - 30x^2

3x^3 (x + 5)(x - 5)

Factor completely: 3x^5 - 75x^3

(7x^3)(x - 2)(x - 1)

Factor completely: 7x^5 - 21x^4 + 14x^3 https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:factor-high-deg/e/factor-higher-degree-polynomials?modal=1

(x -8)(x^2 - 2) ---------------------- 1. Notice the expression has four terms. In cases like this, we will try to factor the expression by grouping. 2. Before we factor, we need to find the common factors for the two pairs of terms. The first pair is x^3 and -8x^2. Their greatest common factor is x^2. The 2nd pair is -2x and 16. Their greatest common factor is 2. Since the 1st term is negative, let's factor out -2. 3. Factoring by grouping: x^2(x) + x^2(-8) - 2(x) - 2(-8) =x^2(x - 8) - 2(x-8) 4. Factor out the common factor (x - 8) = (x - 8)(x^2 - 2)

Factor completely: x^3 - 8x^2 - 2x + 16

(x^4 + 3)(x - 1)

Factor completely: x^5 - x^4 + 3x - 3

9m(m^2 - 3)

Factor the polynomial by its greatest common monomial factor. 9m^3- 27m

200 (1. 200! = 200 x 199! 2. 200!/199! = 200 x 199!/199! 3. = 200 (both 199!'s in the numerator and denominator cancel each other out))

What is 200!/199!

6

What is 3!? https://www.khanacademy.org/partner-content/pixar/crowds/crowds2/e/calculating-factorials

c (-To factor a monomial means to express it as a product of two or more monomials. -To factor an integer completely, we write it as a product of primes; for ex: we know that 30 = 2 x 3 x 5)

Which of the following is the *complete* factorization of 6x^2? a) (6)(x)(x) b) (2)(3)(x^2) c) (2)(3)(x)(x)

x + 7 + (3/ x - 1)

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:quad-div-by-linear/e/quad-by-linear-remainders?modal=1 (x^2+6x-4)/x-1

120 (5! = 5 x 4 x 3 x 2 x 1)

https://www.khanacademy.org/partner-content/pixar/crowds/crowds2/e/calculating-factorials We have 5 different robots and we want to arrange them all in a line. How many different ways are there to arrange them?

x - 4

x^2 - 16/x + 4