What are the solutions to the equation 2x^3-7x^2+32x-112=0 ?

The solutions to the equation 2x^3-7x^2+32x-112=0 are x= 7/2 ,x=4i,x=-4i

Find the zeros of 2x^3-7x^2+32x-112=0

The zeros of 2x^3-7x^2+32x-112=0 are x= 7/2 ,x=4i,x=-4i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{3} - 7 x^{2} + 32 x - 112 = 0

x = \frac{7}{2}, x = 4 i, x = - 4 i

Solution steps

2 x^{3} - 7 x^{2} + 32 x - 112 = 0

Factor

2 x^{3} - 7 x^{2} + 32 x - 112 : (2 x - 7) (x^{2} + 16)

(2 x - 7) (x^{2} + 16) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 7 = 0 or x^{2} + 16 = 0

Solve

2 x - 7 = 0 : x = \frac{7}{2}

Solve

x^{2} + 16 = 0 : x = 4 i, x = - 4 i

The solutions are

x = \frac{7}{2}, x = 4 i, x = - 4 i

F, t = (F^{3} + G) a

y^{3} + 2 y^{2} = 9 y + 18

- x^{3} + 4 x^{2} - x - 6 = 0

x^{4} - 6 x^{3} - 8 x^{2} + 96 x - 128 = 0

10 = 500 (x)^{3}

What are the solutions to the equation 2x^3-7x^2+32x-112=0 ?
The solutions to the equation 2x^3-7x^2+32x-112=0 are x= 7/2 ,x=4i,x=-4i
Find the zeros of 2x^3-7x^2+32x-112=0
The zeros of 2x^3-7x^2+32x-112=0 are x= 7/2 ,x=4i,x=-4i