What are the solutions to the equation p^3-7p^2-4p=-28 ?

The solutions to the equation p^3-7p^2-4p=-28 are p=7,p=-2,p=2

Find the zeros of p^3-7p^2-4p=-28

The zeros of p^3-7p^2-4p=-28 are p=7,p=-2,p=2

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p^{3} - 7 p^{2} - 4 p = - 28

p = 7, p = - 2, p = 2

Solution steps

p^{3} - 7 p^{2} - 4 p = - 28

Move

28

to the left side

p^{3} - 7 p^{2} - 4 p + 28 = 0

Factor

p^{3} - 7 p^{2} - 4 p + 28 : (p - 7) (p + 2) (p - 2)

(p - 7) (p + 2) (p - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

p - 7 = 0 or p + 2 = 0 or p - 2 = 0

Solve

p - 7 = 0 : p = 7

Solve

p + 2 = 0 : p = - 2

Solve

p - 2 = 0 : p = 2

The solutions are

p = 7, p = - 2, p = 2

x = 524 x^{3} - 2 x^{2} - 70 x - 75

x^{3} + 3 = 0

x^{4} = 100

12 = 4 c [x^{7}]

x^{4} - 36 x^{2} = 0

What are the solutions to the equation p^3-7p^2-4p=-28 ?
The solutions to the equation p^3-7p^2-4p=-28 are p=7,p=-2,p=2
Find the zeros of p^3-7p^2-4p=-28
The zeros of p^3-7p^2-4p=-28 are p=7,p=-2,p=2