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

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

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

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

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

x = - 1, x = \frac{1}{2}

Decimal Notation

x = - 1, x = 0.5

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 2 x - 1 = 0

Solve

x + 1 = 0 : x = - 1

Solve

2 x - 1 = 0 : x = \frac{1}{2}

The solutions are

x = - 1, x = \frac{1}{2}

(1 + x)^{3} = 4.82

x^{4} - 22 x^{2} + 105 = 0

p^{4} + 40 = 14 p^{2}

3 x^{4} - (15 x)^{2} = 23

1180 = 1000 (1 + x)^{6}

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