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

The solutions to the equation (x+3)(x^2+2x+6)+(x+3)(x-4)=0 are x=-3,x=-1,x=-2

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

The zeros of (x+3)(x^2+2x+6)+(x+3)(x-4)=0 are x=-3,x=-1,x=-2

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

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

Solution steps

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

Factor

(x + 3) (x^{2} + 2 x + 6) + (x + 3) (x - 4) : (x + 3) (x + 1) (x + 2)

(x + 3) (x + 1) (x + 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 3 = 0 : x = - 3

Solve

x + 1 = 0 : x = - 1

Solve

x + 2 = 0 : x = - 2

The solutions are

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

x, x^{3} = y + 1

6 x^{3} - 7 x^{2} - 5 x = 0

729 = s^{3}

x^{6} - 7 x^{3} + 6 = 0

16 x^{3} = (8 x)

What are the solutions to the equation (x+3)(x^2+2x+6)+(x+3)(x-4)=0 ?
The solutions to the equation (x+3)(x^2+2x+6)+(x+3)(x-4)=0 are x=-3,x=-1,x=-2
Find the zeros of (x+3)(x^2+2x+6)+(x+3)(x-4)=0
The zeros of (x+3)(x^2+2x+6)+(x+3)(x-4)=0 are x=-3,x=-1,x=-2