Solve Polynomials

What are the roots of the following factored polynomial?
(x+1)(x-1)(2x+1)=0

• x= -1/2, -1, 1

• x= 1/2, -1, 1

• x = - ± 1

• x= 1/2, 1

Tags

CCSS.HSF-IF.C.7C

Factor:  27x³ + 8

• (3x + 2)(9x² - 6x + 4)

• (9x + 2)(3x² + 18x + 4)

• (3x²+4)(9x + 2)

• (27x + 8)(x² - 216x + 64)

Tags

CCSS.HSA.APR.C.4

Solve the polynomial equation by factoring: x^4+13x^2=-36

Tags

CCSS.HSA-REI.B.4B

• a = 0

• a = 2

• a = -2, 0

• a = 2, 0

Tags

CCSS.HSA-REI.B.4B

• a = 0 (repeated)

• a = 1 (repeated)

• a = 1 (repeated), 0 (repeated)

• a = -1 (repeated), 0 (repeated)

Tags

CCSS.HSF-IF.C.7C

Solve: -6x-1+5x²=8x²

• x = 5 ±√67 /6

• x = 1, 2/7

• x = 1 ±√ 22

• x = 3 ±√6 / -3

Tags

CCSS.HSA-REI.B.4B

Solve x⁴ - 256 = 0

• x = 4, x = - 4

• x = 4, x = 4i

• x = 4, x -4, x = 4i

• x = 4, x -4, x = 4i,         x = -4i

Tags

CCSS.HSA-REI.B.4B

Find the quotient

Mathematical Equivalence

ON

Tags

CCSS.HSA.APR.D.6

Answer explanation

The roots of the polynomial equation can be found by factoring it as (x-1)(x+2)(x-3)=0, giving x=1, -2, 3 as the solutions.

Tags

CCSS.HSA-REI.B.4B

Answer explanation

The roots of the polynomial equation can be found by factoring the equation as (x-1)(x+2)(x-3)=0. Therefore, x = 1, -2, 3.

Tags

CCSS.HSA-REI.B.4B