What are the solutions to the equation (x+3)(x-1)(x-2)=96 ?

The solutions to the equation (x+3)(x-1)(x-2)=96 are x=5,x=-5/2+i(sqrt(47))/2 ,x=-5/2-i(sqrt(47))/2

Find the zeros of (x+3)(x-1)(x-2)=96

The zeros of (x+3)(x-1)(x-2)=96 are x=5,x=-5/2+i(sqrt(47))/2 ,x=-5/2-i(sqrt(47))/2

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(x + 3) (x - 1) (x - 2) = 96

x = 5, x = - \frac{5}{2} + i \frac{47}{2}, x = - \frac{5}{2} - i \frac{47}{2}

Solution steps

(x + 3) (x - 1) (x - 2) = 96

Expand

(x + 3) (x - 1) (x - 2) : x^{3} - 7 x + 6

x^{3} - 7 x + 6 = 96

Move

96

to the left side

x^{3} - 7 x - 90 = 0

Factor

x^{3} - 7 x - 90 : (x - 5) (x^{2} + 5 x + 18)

(x - 5) (x^{2} + 5 x + 18) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 5 = 0 or x^{2} + 5 x + 18 = 0

Solve

x - 5 = 0 : x = 5

Solve

x^{2} + 5 x + 18 = 0 : x = - \frac{5}{2} + i \frac{47}{2}, x = - \frac{5}{2} - i \frac{47}{2}

The solutions are

x = 5, x = - \frac{5}{2} + i \frac{47}{2}, x = - \frac{5}{2} - i \frac{47}{2}

a^{3} = 0

3 (t^{2} - 9)^{2} + 22 (t^{2} - 9) = - 7

3 (x + 7)^{3} = 648

4 x^{3} - 20 x^{2} = 0

0 = x^{4} - 4 x^{3} + 7

What are the solutions to the equation (x+3)(x-1)(x-2)=96 ?
The solutions to the equation (x+3)(x-1)(x-2)=96 are x=5,x=-5/2+i(sqrt(47))/2 ,x=-5/2-i(sqrt(47))/2
Find the zeros of (x+3)(x-1)(x-2)=96
The zeros of (x+3)(x-1)(x-2)=96 are x=5,x=-5/2+i(sqrt(47))/2 ,x=-5/2-i(sqrt(47))/2