What are the solutions to the equation r^3-3r^2+7r-5=0 ?

The solutions to the equation r^3-3r^2+7r-5=0 are r=1,r=1+2i,r=1-2i

Find the zeros of r^3-3r^2+7r-5=0

The zeros of r^3-3r^2+7r-5=0 are r=1,r=1+2i,r=1-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{3} - 3 r^{2} + 7 r - 5 = 0

r = 1, r = 1 + 2 i, r = 1 - 2 i

Solution steps

r^{3} - 3 r^{2} + 7 r - 5 = 0

Factor

r^{3} - 3 r^{2} + 7 r - 5 : (r - 1) (r^{2} - 2 r + 5)

(r - 1) (r^{2} - 2 r + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 1 = 0 or r^{2} - 2 r + 5 = 0

Solve

r - 1 = 0 : r = 1

Solve

r^{2} - 2 r + 5 = 0 : r = 1 + 2 i, r = 1 - 2 i

The solutions are

r = 1, r = 1 + 2 i, r = 1 - 2 i

(x + 1)^{2} (x - 1) = 0

- 4 h - 5 h^{2} + h^{3} - 4 = 0

c^{3} - 5 c^{2} + 7 c - 9 = 100

25 x^{3} - 50 x^{2} - 4 x + 8 = 0

3 - (x - 1)^{4} = 0

What are the solutions to the equation r^3-3r^2+7r-5=0 ?
The solutions to the equation r^3-3r^2+7r-5=0 are r=1,r=1+2i,r=1-2i
Find the zeros of r^3-3r^2+7r-5=0
The zeros of r^3-3r^2+7r-5=0 are r=1,r=1+2i,r=1-2i