What are the solutions to the equation x^2(x-3)=16(x-3) ?

The solutions to the equation x^2(x-3)=16(x-3) are x=3,x=-4,x=4

Find the zeros of x^2(x-3)=16(x-3)

The zeros of x^2(x-3)=16(x-3) are x=3,x=-4,x=4

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x^{2} (x - 3) = 16 (x - 3)

x = 3, x = - 4, x = 4

Solution steps

x^{2} (x - 3) = 16 (x - 3)

Move

16 (x - 3)

to the left side

x^{2} (x - 3) - 16 (x - 3) = 0

Factor

x^{2} (x - 3) - 16 (x - 3) : (x - 3) (x + 4) (x - 4)

(x - 3) (x + 4) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 3 = 0 or x + 4 = 0 or x - 4 = 0

Solve

x - 3 = 0 : x = 3

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = 3, x = - 4, x = 4

3 x^{3} + 1 = x^{2} + 3 x

x^{3} + 4 x^{2} - 5 x = - 3 x^{2} - 5 x + 15

x^{3} - 10 x^{2} + 24 = 0

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

(x^{2} - 4) (x^{2} + 8 x + 16) = 0

What are the solutions to the equation x^2(x-3)=16(x-3) ?
The solutions to the equation x^2(x-3)=16(x-3) are x=3,x=-4,x=4
Find the zeros of x^2(x-3)=16(x-3)
The zeros of x^2(x-3)=16(x-3) are x=3,x=-4,x=4