What are the solutions to the equation 4x(3-x)(4x+1)^2=0 ?

The solutions to the equation 4x(3-x)(4x+1)^2=0 are x=0,x=3,x=-1/4

Find the zeros of 4x(3-x)(4x+1)^2=0

The zeros of 4x(3-x)(4x+1)^2=0 are x=0,x=3,x=-1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x (3 - x) (4 x + 1)^{2} = 0

x = 0, x = 3, x = - \frac{1}{4}

Decimal Notation

x = 0, x = 3, x = - 0.25

Solution steps

4 x (3 - x) (4 x + 1)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 3 - x = 0 or 4 x + 1 = 0

Solve

3 - x = 0 : x = 3

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

The solutions are

x = 0, x = 3, x = - \frac{1}{4}

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

20000 = 17100 \cdot (1 + x)^{3}

x^{4} - 6 x - 4 = 0

8 x^{6} + 9 x^{3} + 1 = 0

2 x^{4} - 11 x^{2} + 5 = 0

What are the solutions to the equation 4x(3-x)(4x+1)^2=0 ?
The solutions to the equation 4x(3-x)(4x+1)^2=0 are x=0,x=3,x=-1/4
Find the zeros of 4x(3-x)(4x+1)^2=0
The zeros of 4x(3-x)(4x+1)^2=0 are x=0,x=3,x=-1/4