Question

Which are real zeroes of this function? x3 + 2x2 – 9x – 18.

A. 1, 2, 3
B. 3, -3, 2
C. 2, 3, -6
D. 1, -1, 18
E. -2, 3, -3

Answer 1

The real zeros of function are (-2, 3, -3)

Explanation:

It was E on my computer but always double check to match!

Answer 2

Option E

The real zeros of function are (-2, 3, -3)

Solution:

Given function is:
`x^3 + 2x^2 -9x - 18`

To find the zeroes of the function, we just need to evaluate the values and check if the function/expression tends to 0.

Let
`f(x) = x^3 + 2x^2 -9x - 18`

On grouping the terms we get,

`f(x) = (x^3 + 2x^2) - (9x +18)`

Taking
`x^2` as common term from first two terms and "9" as common term from last two terms,

`f(x) = x^2(x + 2) - 9(x + 2)`

Taking (x + 2) as common term,

`f(x) = (x^2 - 9)(x + 2)`

`f(x) = (x^2 - 3^2)(x + 2)`

We know that,

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

Applying the above identity,

`f(x) = (x + 3)(x - 3)(x + 2)`

We equate the function to 0 , to find the zeros of polynomial function

(x + 3)(x - 3)(x + 2) = 0

Therefore zeros are:

x + 3 = 0

x = -3

x - 3 = 0

x = 3

x + 2 = 0

x = -2

Thus zeros are (-2, 3, -3) Option E is correct