What are the real zeros of x³ + 4x² - 9x - 36?

Solution:

Given expression is :

x³ + 4x² - 9x - 36

Rearranging the given expression, we have:

x³ - 9x + 4x² - 36

x(x² - 9) + 4(x² - 9)

(x² - 9)(x + 4)

We know from algebraic identity:

a² - b² = (a - b)(a + b)

Hence,

x ² - 9 = (x)² - (3)² = (x + 3)(x - 3)

Therefore

(x² - 9)(x + 4) = (x + 3)(x - 3)(x + 4)

The real zeros of the given equation are:

x + 3 = 0 ⇒ x = -3

x - 3 = 0 ⇒ x = 3

x + 4 = 0 ⇒ x = -4

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What are the real zeros of x³ + 4x² - 9x - 36?

Summary:

The real zeros of x³ + 4x² - 9x - 36 are x = -3, x = 3 and x = -4