What are the solutions to the equation 16x^4-32x^3+24x^2-8x+1=0 ?

The solutions to the equation 16x^4-32x^3+24x^2-8x+1=0 are x= 1/2

Find the zeros of 16x^4-32x^3+24x^2-8x+1=0

The zeros of 16x^4-32x^3+24x^2-8x+1=0 are x= 1/2

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16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1 = 0

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1 = 0

Factor

16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1 : (2 x - 1)^{4}

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 1 = 0

Solve

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

The solution is

x = \frac{1}{2}

x^{3} = 6 x^{2} + 3 x + 10

5 = k^{6}

z^{4} = 3 + 2 i

x^{4} + 6 x^{3} + 18 x^{2} + 22 x + 17 = 0

0 = x^{3} + 8

What are the solutions to the equation 16x^4-32x^3+24x^2-8x+1=0 ?
The solutions to the equation 16x^4-32x^3+24x^2-8x+1=0 are x= 1/2
Find the zeros of 16x^4-32x^3+24x^2-8x+1=0
The zeros of 16x^4-32x^3+24x^2-8x+1=0 are x= 1/2