Question

What are the real zeros of x^3+4x^2-9x-36

Answer 1

Answer:

$x = -4\\x = -3\\x = 3$

Step-by-step explanation:

To find the real zeros you must match the function to zero and factor the expression.

We have a polynomial of degree 3.

We try to group the terms to perform the factorization

$x ^ 3 + 4x ^ 2-9x-36 = 0\\\\x ^ 2(x + 4) - 9(x + 4) = 0$

Now we take (x + 4) as a common factor

$(x + 4)(x ^ 2 -9) = 0$

If we have an expression of the form $(a ^ 2-b ^ 2)$ we know that this expression is equivalent to:

$(a ^ 2-b ^ 2) = (a + b)(a-b)$

In this case

$a = x\\b = 3$

So:

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

Finally:

$x ^ 3 + 4x ^ 2-9x-36 = (x + 4)(x + 3)(x-3) = 0$

The solutions are:

$x = -4\\x = -3\\x = 3$