Polynomials Synthetic Division

• 2x² + 2x - 4  R(-12)

• 2x² - 2x + 6  R(-23)

• 2x² - 2x + 8  R(-20)

• 2x² - 4x + 1  R0

Tags

CCSS.HSA.APR.D.6

Use synthetic division

• x - 7

• x² - 7

• x + 7

• x² + 3x - 54

Tags

CCSS.HSA.APR.D.6

• Yes, (x-2) is a factor.

  There is a remainder.

• No, (x-2) is not a factor.

  The remainder is zero.

• Yes, (x-2) is a factor.

  The remainder is zero.

• No, (x-2) is not a factor.

  There is a remainder. 

Answer explanation

Two ways to solve:

Synthetic division using x = 2

OR

Remainder Theorem: (recommended) Evaluate for x=2 to find the remainder!

Tags

CCSS.HSA.APR.B.2

What should be the correct order of the coefficients when setting up synthetic division for (3x-4x3+6x4+1)÷(x+3) ?

• 3   0   -4   6   1

• 6   -4   3   1   0

• 6   -4   0   3   1

• -6   4   0   -3   -1

Tags

CCSS.HSA.APR.D.6

Use synthetic division to divide the polynomial (2x^3 + 4x^2 - 3x + 1) by the binomial (x - 1).

• 2x^2 + 6x - 3

• 2x^2 + 4x + 3

• 2x^2 + 2x + 3

• 2x^2 + 6x + 3

Tags

CCSS.HSA.APR.D.6

Divide using synthetic division

2x³ - 5x² + 3x + 7 ÷ x - 2

• 2x³ - x² + x + 9

• 2x² - x + 1

• 2x² - x + 1 + 9/(x - 2)

• 2x² - 9x - 15 - 23/(x - 2)

Tags

CCSS.HSA.APR.D.6

Divide using synthetic division
(2x³ + 4x² - 5) by (x + 3).

• 2x² + 2x - 4  R(-12)

• 2x² - 2x + 6  R(-23)

• 2x² - 2x + 8  R(-20)

• 2x² - 4x + 1  R0

Tags

CCSS.HSA.APR.D.6

Use synthetic division to divide the polynomial (2x^4 - 5x^3 + 3x^2 + 4x - 1) by the binomial (x + 1).

• 2x^3 - 7x^2 + 10x - 1

• 2x^3 - 7x^2 + 10x + 6

• 2x^3 - 7x^2 + 10x + 1

• 2x^3 - 7x^2 + 10x - 6

Tags

CCSS.HSA.APR.D.6

Tags

CCSS.HSA.APR.D.6

Use synthetic division to divide the polynomial (5x^3 + 2x^2 - 4x + 1) by the binomial (x + 3).

• 5x^2 - 13x + 35 - 104/(x - 3)

• 5x^2 - 13x + 35 + 104/(x + 3)

• 5x^2 - 13x + 35 - 104/(x + 2)

• 5x^2 - 13x + 35 - 104/(x + 3)

Tags

CCSS.HSA.APR.D.6