Solve the equations:
$4 x^{3} + 16 x^{2} - 9 x - 36 = 0$ , the sum of two of the roots being zero.

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Solution

${4 x}^{3} + {16 x}^{2} - 9 x - 36 = 0$
Let the roots be $a, - a, b$
$a - a + b = - \frac{16}{4} = - 4 \Rightarrow b = - 4 a (- a) + (- a) b + a b = \frac{- 9}{4} a^{2} = \frac{9}{4} \Rightarrow a = \pm \frac{3}{2}$
So the roots of the equation are $\frac{3}{2}, - \frac{3}{2}, - 4$

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