What are the solutions to the equation 3a^3-147a=-2a^2+98 ?

The solutions to the equation 3a^3-147a=-2a^2+98 are a=-2/3 ,a=-7,a=7

Find the zeros of 3a^3-147a=-2a^2+98

The zeros of 3a^3-147a=-2a^2+98 are a=-2/3 ,a=-7,a=7

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3 a^{3} - 147 a = - 2 a^{2} + 98

a = - \frac{2}{3}, a = - 7, a = 7

Solution steps

3 a^{3} - 147 a = - 2 a^{2} + 98

Move

98

to the left side

3 a^{3} - 147 a - 98 = - 2 a^{2}

Move

2 a^{2}

to the left side

3 a^{3} - 147 a - 98 + 2 a^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

3 a^{3} + 2 a^{2} - 147 a - 98 = 0

Factor

3 a^{3} + 2 a^{2} - 147 a - 98 : (3 a + 2) (a + 7) (a - 7)

(3 a + 2) (a + 7) (a - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 a + 2 = 0 or a + 7 = 0 or a - 7 = 0

Solve

3 a + 2 = 0 : a = - \frac{2}{3}

Solve

a + 7 = 0 : a = - 7

Solve

a - 7 = 0 : a = 7

The solutions are

a = - \frac{2}{3}, a = - 7, a = 7

695 = 5 x^{3}

z^{3} + 1 - i = 0

2 x^{5} + 5 x = 2215

12 x^{3} - 26 x^{2} - 12 x = 0

25 x^{4} + 9 x^{2} - 16 = 0

What are the solutions to the equation 3a^3-147a=-2a^2+98 ?
The solutions to the equation 3a^3-147a=-2a^2+98 are a=-2/3 ,a=-7,a=7
Find the zeros of 3a^3-147a=-2a^2+98
The zeros of 3a^3-147a=-2a^2+98 are a=-2/3 ,a=-7,a=7