What are the solutions to the equation 12x^4+7x^3+49x^2+28x+4=0 ?

The solutions to the equation 12x^4+7x^3+49x^2+28x+4=0 are x=-1/3 ,x=-1/4 ,x=2i,x=-2i

Find the zeros of 12x^4+7x^3+49x^2+28x+4=0

The zeros of 12x^4+7x^3+49x^2+28x+4=0 are x=-1/3 ,x=-1/4 ,x=2i,x=-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 x^{4} + 7 x^{3} + 49 x^{2} + 28 x + 4 = 0

x = - \frac{1}{3}, x = - \frac{1}{4}, x = 2 i, x = - 2 i

Solution steps

12 x^{4} + 7 x^{3} + 49 x^{2} + 28 x + 4 = 0

Factor

12 x^{4} + 7 x^{3} + 49 x^{2} + 28 x + 4 : (3 x + 1) (4 x + 1) (x^{2} + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 1 = 0 or 4 x + 1 = 0 or x^{2} + 4 = 0

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

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

Solve

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

The solutions are

x = - \frac{1}{3}, x = - \frac{1}{4}, x = 2 i, x = - 2 i

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

x (x + 2) (x + 3) = - 1

3 x^{4} - 16 x^{3} + 24 x^{2} - 9 = 0

y, x^{3} \cdot y = 9

x^{3} - 2 x^{2} - 6 x - 8

What are the solutions to the equation 12x^4+7x^3+49x^2+28x+4=0 ?
The solutions to the equation 12x^4+7x^3+49x^2+28x+4=0 are x=-1/3 ,x=-1/4 ,x=2i,x=-2i
Find the zeros of 12x^4+7x^3+49x^2+28x+4=0
The zeros of 12x^4+7x^3+49x^2+28x+4=0 are x=-1/3 ,x=-1/4 ,x=2i,x=-2i