Question

given ( d(x) = -x^3 - 3x^2 + 4x ), how many real zeros does ( d(x) ) have?
options:
no real zeros
exactly three real zeros
exactly two real zeros
exactly one real zero

Explanation:

Step1: Factor the polynomial

First, we factor the given polynomial \( d(x)=-x^{3}-3x^{2} + 4x\). We can factor out a common factor of \(-x\) first:
\[
d(x)=-x(x^{2}+3x - 4)
\]
Then, we factor the quadratic expression \(x^{2}+3x - 4\). We need to find two numbers that multiply to \(- 4\) and add up to \(3\). The numbers are \(4\) and \(-1\), so:
\[
x^{2}+3x - 4=(x + 4)(x-1)
\]
So the fully factored form of \(d(x)\) is:
\[
d(x)=-x(x + 4)(x - 1)
\]

Step2: Find the real zeros

To find the real zeros, we set \(d(x) = 0\):
\[
-x(x + 4)(x - 1)=0
\]
Using the zero - product property (if \(ab = 0\), then either \(a = 0\), \(b = 0\) or both), we have:

• If \(-x=0\), then \(x = 0\).
• If \(x + 4=0\), then \(x=-4\).
• If \(x - 1=0\), then \(x = 1\).

So the real zeros of \(d(x)\) are \(x=-4\), \(x = 0\) and \(x = 1\).

Answer:

Exactly three real zeros