What are the solutions to the equation solvefor a,(a^3+27)=(a+3)(a^2+ma+9) ?

The solutions to the equation solvefor a,(a^3+27)=(a+3)(a^2+ma+9) are a=0,a=-3;\quad m\ne-3

Find the zeros of solvefor a,(a^3+27)=(a+3)(a^2+ma+9)

The zeros of solvefor a,(a^3+27)=(a+3)(a^2+ma+9) are a=0,a=-3;\quad m\ne-3

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a, (a^{3} + 27) = (a + 3) (a^{2} + ma + 9)

a = 0, a = - 3; m \neq = - 3

Solution steps

(a^{3} + 27) = (a + 3) (a^{2} + ma + 9)

Expand

(a + 3) (a^{2} + ma + 9) : a^{3} + m a^{2} + 9 a + 3 a^{2} + 3 ma + 27

a^{3} + 27 = a^{3} + m a^{2} + 9 a + 3 a^{2} + 3 ma + 27

Switch sides

a^{3} + m a^{2} + 9 a + 3 a^{2} + 3 ma + 27 = a^{3} + 27

Move

27

to the left side

a^{3} + m a^{2} + 3 a^{2} + 9 a + 3 ma = a^{3}

Move

a^{3}

to the left side

m a^{2} + 3 a^{2} + 3 ma + 9 a = 0

Write in the standard form

a x^{2} + b x + c = 0

(m + 3) a^{2} + (3 m + 9) a = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( 3 m + 9 ) \pm ( 3 m + 9 ) ^{2} - 4 ( m + 3 ) \cdot 0}{2 ( m + 3 )}

Simplify

(3 m + 9)^{2} - 4 (m + 3) \cdot 0 : 3 m + 9

a_{1, 2} = \frac{- ( 3 m + 9 ) \pm ( 3 m + 9 )}{2 ( m + 3 )}; m \neq = - 3

Separate the solutions

a_{1} = \frac{- ( 3 m + 9 ) + 3 m + 9}{2 ( m + 3 )}, a_{2} = \frac{- ( 3 m + 9 ) - ( 3 m + 9 )}{2 ( m + 3 )}

a = \frac{- ( 3 m + 9 ) + 3 m + 9}{2 ( m + 3 )} : 0

a = \frac{- ( 3 m + 9 ) - ( 3 m + 9 )}{2 ( m + 3 )} : - 3

The solutions to the quadratic equation are:

a = 0, a = - 3; m \neq = - 3

6 x^{2} - 84 x + 228 = 0

x^{2} - 6 x + 12

x^{2} + 6 x - 4

v^{2} = - 4

v^{2} = - 5

What are the solutions to the equation solvefor a,(a^3+27)=(a+3)(a^2+ma+9) ?
The solutions to the equation solvefor a,(a^3+27)=(a+3)(a^2+ma+9) are a=0,a=-3;\quad m\ne-3
Find the zeros of solvefor a,(a^3+27)=(a+3)(a^2+ma+9)
The zeros of solvefor a,(a^3+27)=(a+3)(a^2+ma+9) are a=0,a=-3;\quad m\ne-3