What are the solutions to the equation (x-2)^4-13x^2+52x-16=0 ?

The solutions to the equation (x-2)^4-13x^2+52x-16=0 are x=0,x=-1,x=4,x=5

Find the zeros of (x-2)^4-13x^2+52x-16=0

The zeros of (x-2)^4-13x^2+52x-16=0 are x=0,x=-1,x=4,x=5

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(x - 2)^{4} - 13 x^{2} + 52 x - 16 = 0

x = 0, x = - 1, x = 4, x = 5

Solution steps

(x - 2)^{4} - 13 x^{2} + 52 x - 16 = 0

Expand

(x - 2)^{4} - 13 x^{2} + 52 x - 16 : x^{4} - 8 x^{3} + 11 x^{2} + 20 x

x^{4} - 8 x^{3} + 11 x^{2} + 20 x = 0

Factor

x^{4} - 8 x^{3} + 11 x^{2} + 20 x : x (x + 1) (x - 4) (x - 5)

x (x + 1) (x - 4) (x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0 or x - 4 = 0 or x - 5 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 4 = 0 : x = 4

Solve

x - 5 = 0 : x = 5

The solutions are

x = 0, x = - 1, x = 4, x = 5

2 x^{3} - 5 x - 9 = 0

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

- 1 - 9 n^{3} = - 136

- x^{3} + 6 x^{2} - 7 x = 4

2 x^{4} - 12 x^{2} = 0

What are the solutions to the equation (x-2)^4-13x^2+52x-16=0 ?
The solutions to the equation (x-2)^4-13x^2+52x-16=0 are x=0,x=-1,x=4,x=5
Find the zeros of (x-2)^4-13x^2+52x-16=0
The zeros of (x-2)^4-13x^2+52x-16=0 are x=0,x=-1,x=4,x=5