What are the real zeroes of x³ + 6x² - 9x - 54?

Solution:

Given, the equation is x³ + 6x² - 9x - 54.

We have to find the real zeroes of the given equation.

To find the real zeroes of the given equation,

x³ + 6x² - 9x - 54 = 0

On factorizing,

x³ + 6x² - 9x - 54 = x²(x + 6) - 9(x + 6)

= (x² - 9)(x + 6)

Now, (x² - 9)(x + 6) = 0

x² - 9 = 0

x² = 9

x = ±3

Similarly, x + 6 = 0

x = -6

Therefore, the roots of the equation are +3, -3 and -6.

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What are the real zeroes of x³ + 6x² - 9x - 54?

Summary:

The real zeroes of x³ + 6x² - 9x - 54 are +3, -3 and -6.