Solving polynomials by factoring help

Question

1)) 9x^3 - 3x^2 - 3x + 1 = 0 

answer: 3x^2(3x-1) 1(x-3) 

Am I correct? 

I need help on this one too 

2) 3x^5 + 18x^4 - 21x^3 = 0 

3) -x^4 + 2x^3 + 8x^2 = 0

 

 

Answer 1

1) The polynomial 9x³ - 3x² - 3x + 1 = 0

(9x³ - 3x²) + (- 3x + 1) = 0

3x²(3x - 1) - 1 (3x - 1) = 0

(3x - 1)(3x² - 1) = 0

(3x - 1)[(√3x)² - 1²] = 0

Factoring of 9x³ - 3x² - 3x + 1 = (3x - 1)[√3x - 1)(√3x + 1).

(3x - 1)[√3x - 1)(√3x + 1) = 0

Apply zero product properrty.

3x - 1 = 0, √3x - 1 = 0 and √3x + 1 = 0

Solve 3x - 1 = 0

3x = 1

x = 1/3

Solve √3x - 1 = 0

√3x = 1

x = 1/√3

Solve √3x + 1 = 0

√3x = - 1

x = - 1/√3

Solutions are x = 1/3, 1/√3 and -1/√3.

Answer 2

2) The polynomial 3x⁵ + 18x⁴ - 21x³ = 0

x³(3x² + 18x - 21) = 0

x³(3x² + 21x - 3x - 21) = 0

x³[3x(x + 7) - 3(x + 7)] = 0

x³[(x + 7)(3x - 3)] = 0

x(x)(x)(x + 7)(3x - 3) = 0

Apply zero product property.

x = 0 , x = 0, x = 0, x + 7 = 0 and 3x - 3 = 0

Solutions are x = 0, 0 , 0, - 7 and x = 1.

Answer 3

3) The polynomial - x⁴ + 2x³ + 8x² = 0

- x²(x² - 2x - 8) = 0

- x²(x² - 4x + 2x - 8) = 0

- x²[x(x - 4) + 2(x - 4)] = 0

- x²[(x - 4)(x + 2)] = 0

(- x)(x)(x - 4)(x + 2) = 0

Apply zero product property.

- x = 0 , x = 0 , x - 4 = 0 and x + 2 = 0

Solutions are x = 0, 0, 4 and - 2.