What are the solutions to the equation (1-x)(x+2)^2(x+3)=0 ?

The solutions to the equation (1-x)(x+2)^2(x+3)=0 are x=1,x=-2,x=-3

Find the zeros of (1-x)(x+2)^2(x+3)=0

The zeros of (1-x)(x+2)^2(x+3)=0 are x=1,x=-2,x=-3

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(1 - x) (x + 2)^{2} (x + 3) = 0

x = 1, x = - 2, x = - 3

Solution steps

(1 - x) (x + 2)^{2} (x + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

1 - x = 0 or x + 2 = 0 or x + 3 = 0

Solve

1 - x = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x + 3 = 0 : x = - 3

The solutions are

x = 1, x = - 2, x = - 3

(x) = (x^{3} + x) (x^{3} + x^{4})

y, x = (y + 1)^{3} - 2

3 x (4 x - 5) (x^{2} + 25) = 0

0 = x^{3} - 11 x^{2} + 37 x - 42

x, y^{2} - 12 x^{3} = 8 y

What are the solutions to the equation (1-x)(x+2)^2(x+3)=0 ?
The solutions to the equation (1-x)(x+2)^2(x+3)=0 are x=1,x=-2,x=-3
Find the zeros of (1-x)(x+2)^2(x+3)=0
The zeros of (1-x)(x+2)^2(x+3)=0 are x=1,x=-2,x=-3