What are the solutions to the equation (x^2+6x)(x+3)^2=400 ?

The solutions to the equation (x^2+6x)(x+3)^2=400 are x=2,x=-8,x=-3+4i,x=-3-4i

Find the zeros of (x^2+6x)(x+3)^2=400

The zeros of (x^2+6x)(x+3)^2=400 are x=2,x=-8,x=-3+4i,x=-3-4i

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(x^{2} + 6 x) (x + 3)^{2} = 400

x = 2, x = - 8, x = - 3 + 4 i, x = - 3 - 4 i

Solution steps

(x^{2} + 6 x) (x + 3)^{2} = 400

Expand

(x^{2} + 6 x) (x + 3)^{2} : x^{4} + 12 x^{3} + 45 x^{2} + 54 x

x^{4} + 12 x^{3} + 45 x^{2} + 54 x = 400

Move

400

to the left side

x^{4} + 12 x^{3} + 45 x^{2} + 54 x - 400 = 0

Factor

x^{4} + 12 x^{3} + 45 x^{2} + 54 x - 400 : (x - 2) (x + 8) (x^{2} + 6 x + 25)

(x - 2) (x + 8) (x^{2} + 6 x + 25) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 8 = 0 or x^{2} + 6 x + 25 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 8 = 0 : x = - 8

Solve

x^{2} + 6 x + 25 = 0 : x = - 3 + 4 i, x = - 3 - 4 i

The solutions are

x = 2, x = - 8, x = - 3 + 4 i, x = - 3 - 4 i

x^{4} - 14 x^{2} + 48 = 0

x^{3} + 6 x^{2} + 11 x - 6 = 0

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

t^{4} - 16 = 0

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

What are the solutions to the equation (x^2+6x)(x+3)^2=400 ?
The solutions to the equation (x^2+6x)(x+3)^2=400 are x=2,x=-8,x=-3+4i,x=-3-4i
Find the zeros of (x^2+6x)(x+3)^2=400
The zeros of (x^2+6x)(x+3)^2=400 are x=2,x=-8,x=-3+4i,x=-3-4i