What are the solutions to the equation a^3-44a+48=0 ?

The solutions to the equation a^3-44a+48=0 are a=6,a=-3+sqrt(17),a=-3-sqrt(17)

Find the zeros of a^3-44a+48=0

The zeros of a^3-44a+48=0 are a=6,a=-3+sqrt(17),a=-3-sqrt(17)

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a^{3} - 44 a + 48 = 0

a = 6, a = - 3 + 17, a = - 3 - 17

Decimal Notation

a = 6, a = 1.12310 \dots, a = - 7.12310 \dots

Solution steps

a^{3} - 44 a + 48 = 0

Factor

a^{3} - 44 a + 48 : (a - 6) (a^{2} + 6 a - 8)

(a - 6) (a^{2} + 6 a - 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a - 6 = 0 or a^{2} + 6 a - 8 = 0

Solve

a - 6 = 0 : a = 6

Solve

a^{2} + 6 a - 8 = 0 : a = - 3 + 17, a = - 3 - 17

The solutions are

a = 6, a = - 3 + 17, a = - 3 - 17

x^{3} + x = 0

x + x^{3} = 0

(x + 1)^{3} = 64

x^{3} + 4 x - 1 = 0

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

What are the solutions to the equation a^3-44a+48=0 ?
The solutions to the equation a^3-44a+48=0 are a=6,a=-3+sqrt(17),a=-3-sqrt(17)
Find the zeros of a^3-44a+48=0
The zeros of a^3-44a+48=0 are a=6,a=-3+sqrt(17),a=-3-sqrt(17)