What are the solutions to the equation (3-x)^5+(x-5)^5=-32 ?

The solutions to the equation (3-x)^5+(x-5)^5=-32 are x=3,x=5,x=4+sqrt(3)i,x=4-sqrt(3)i

Find the zeros of (3-x)^5+(x-5)^5=-32

The zeros of (3-x)^5+(x-5)^5=-32 are x=3,x=5,x=4+sqrt(3)i,x=4-sqrt(3)i

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(3 - x)^{5} + (x - 5)^{5} = - 32

x = 3, x = 5, x = 4 + 3 i, x = 4 - 3 i

Solution steps

(3 - x)^{5} + (x - 5)^{5} = - 32

Expand

(3 - x)^{5} + (x - 5)^{5} : - 10 x^{4} + 160 x^{3} - 980 x^{2} + 2720 x - 2882

- 10 x^{4} + 160 x^{3} - 980 x^{2} + 2720 x - 2882 = - 32

Move

32

to the left side

- 10 x^{4} + 160 x^{3} - 980 x^{2} + 2720 x - 2850 = 0

Factor

- 10 x^{4} + 160 x^{3} - 980 x^{2} + 2720 x - 2850 : - 10 (x - 3) (x - 5) (x^{2} - 8 x + 19)

- 10 (x - 3) (x - 5) (x^{2} - 8 x + 19) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x - 5 = 0 or x^{2} - 8 x + 19 = 0

Solve

x - 3 = 0 : x = 3

Solve

x - 5 = 0 : x = 5

Solve

x^{2} - 8 x + 19 = 0 : x = 4 + 3 i, x = 4 - 3 i

The solutions are

x = 3, x = 5, x = 4 + 3 i, x = 4 - 3 i

c, c^{1} c^{2} = r^{1} + r^{2}

x^{7} = 120

x^{4} - 2 = 0

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

x^{7} - x^{5} = - 11

What are the solutions to the equation (3-x)^5+(x-5)^5=-32 ?
The solutions to the equation (3-x)^5+(x-5)^5=-32 are x=3,x=5,x=4+sqrt(3)i,x=4-sqrt(3)i
Find the zeros of (3-x)^5+(x-5)^5=-32
The zeros of (3-x)^5+(x-5)^5=-32 are x=3,x=5,x=4+sqrt(3)i,x=4-sqrt(3)i