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) \cdot 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^{3} + 5 x^{2} - 2 x = 0

- λ^{3} + 6 λ^{2} - 11 λ + 6 = 0

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

x^{4} + 2 x^{2} = 24

y^{3} + y = 2

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