What are the solutions to the equation (x-2)^4(x+1)(x+4)=0 ?

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

Find the zeros of (x-2)^4(x+1)(x+4)=0

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

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

x = 2, x = - 1, x = - 4

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 1 = 0 or x + 4 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

Solve

x + 4 = 0 : x = - 4

The solutions are

x = 2, x = - 1, x = - 4

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

x, a = (2 x^{5} y^{3})^{3}

a^{3} + 12 a - 6.5 = 0

8 (1.2 - x^{2})^{2} = 0.5 x^{2}

76 = \frac{1}{2} (h (h^{2} + 3))

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