What are the solutions to the equation k^3-3k^2-16k+48=0 ?

The solutions to the equation k^3-3k^2-16k+48=0 are k=3,k=-4,k=4

Find the zeros of k^3-3k^2-16k+48=0

The zeros of k^3-3k^2-16k+48=0 are k=3,k=-4,k=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

k^{3} - 3 k^{2} - 16 k + 48 = 0

k = 3, k = - 4, k = 4

Solution steps

k^{3} - 3 k^{2} - 16 k + 48 = 0

Factor

k^{3} - 3 k^{2} - 16 k + 48 : (k - 3) (k + 4) (k - 4)

(k - 3) (k + 4) (k - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k - 3 = 0 or k + 4 = 0 or k - 4 = 0

Solve

k - 3 = 0 : k = 3

Solve

k + 4 = 0 : k = - 4

Solve

k - 4 = 0 : k = 4

The solutions are

k = 3, k = - 4, k = 4

3 x^{3} + 7 x^{2} = 6 x

y^{4} - 37 y^{2} = - 36

5 - x^{3} = - 15

x^{3} + 7 x^{2} - 16 x - 112 = 0

x^{2} + x - 43 = 3 x^{4}

What are the solutions to the equation k^3-3k^2-16k+48=0 ?
The solutions to the equation k^3-3k^2-16k+48=0 are k=3,k=-4,k=4
Find the zeros of k^3-3k^2-16k+48=0
The zeros of k^3-3k^2-16k+48=0 are k=3,k=-4,k=4