What are the solutions to the equation (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0 ?

The solutions to the equation (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0 are x=-6,x=-2,x=-4

Find the zeros of (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0

The zeros of (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0 are x=-6,x=-2,x=-4

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(x + 2)^{2} \cdot (x + 6)^{3} + (x + 2) \cdot (x + 6)^{4} = 0

x = - 6, x = - 2, x = - 4

Solution steps

(x + 2)^{2} (x + 6)^{3} + (x + 2) (x + 6)^{4} = 0

Factor

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

2 (x + 6)^{3} (x + 2) (x + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 6 = 0 or x + 2 = 0 or x + 4 = 0

Solve

x + 6 = 0 : x = - 6

Solve

x + 2 = 0 : x = - 2

Solve

x + 4 = 0 : x = - 4

The solutions are

x = - 6, x = - 2, x = - 4

x^{6} = 5

(3 x - 2) (x^{1}) - (2 x^{3}) (2 - x) = 182 x

(x^{2} - 3 x)^{2} - 16 (x^{2} - 3 x) - 36 = 0

x^{3} - x = 1

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

What are the solutions to the equation (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0 ?
The solutions to the equation (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0 are x=-6,x=-2,x=-4
Find the zeros of (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0
The zeros of (x+2)^2*(x+6)^3+(x+2)*(x+6)^4=0 are x=-6,x=-2,x=-4