What are the solutions to the equation r^3-7r^2+16r-12=0 ?

The solutions to the equation r^3-7r^2+16r-12=0 are r=2,r=3

Find the zeros of r^3-7r^2+16r-12=0

The zeros of r^3-7r^2+16r-12=0 are r=2,r=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

r^{3} - 7 r^{2} + 16 r - 12 = 0

r = 2, r = 3

Solution steps

r^{3} - 7 r^{2} + 16 r - 12 = 0

Factor

r^{3} - 7 r^{2} + 16 r - 12 : (r - 2)^{2} (r - 3)

(r - 2)^{2} (r - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r - 2 = 0 or r - 3 = 0

Solve

r - 2 = 0 : r = 2

Solve

r - 3 = 0 : r = 3

The solutions are

r = 2, r = 3

x^{5} - 3 x^{4} - 15 x^{3} + 45 x^{2} - 16 x + 48 = 0

x^{4} - x^{2} = 3 x^{3} - 3 x

m^{4} - 2 m^{2} - 8 = 0

(x + 3)^{4} + 2 (x + 3)^{2} - 8 = 0

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

What are the solutions to the equation r^3-7r^2+16r-12=0 ?
The solutions to the equation r^3-7r^2+16r-12=0 are r=2,r=3
Find the zeros of r^3-7r^2+16r-12=0
The zeros of r^3-7r^2+16r-12=0 are r=2,r=3