To find the zeros of the polynomial [tex]\( P(x) = x^3 + 3x^2 - 4 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which [tex]\( P(x) = 0 \)[/tex].
Step-by-step Solution:
1. Start with the polynomial equation set to zero:
[tex]\[
x^3 + 3x^2 - 4 = 0
\][/tex]
2. To solve this polynomial equation, we look for values of [tex]\( x \)[/tex] that satisfy this equation. These values are known as the "zeros" or "roots" of the polynomial.
3. Let's assume that [tex]\( x = a \)[/tex] is a root of the polynomial. Then, [tex]\( a \)[/tex] satisfies:
[tex]\[
a^3 + 3a^2 - 4 = 0
\][/tex]
4. By inspecting possible rational roots (potential rational roots could be factors of the constant term divided by factors of the leading coefficient), we can try different values such as [tex]\(\pm 1, \pm 2, \pm 4\)[/tex] to see if any of them satisfy the polynomial equation.
5. By checking, we find that [tex]\( x = -2 \)[/tex] is a root, i.e.,
[tex]\[
(-2)^3 + 3(-2)^2 - 4 = -8 + 12 - 4 = 0
\][/tex]
Therefore, [tex]\( x = -2 \)[/tex] is a zero of the polynomial.
6. Next, we use synthetic or polynomial division to divide [tex]\( x^3 + 3x^2 - 4 \)[/tex] by [tex]\( (x + 2) \)[/tex] to find the quotient polynomial.
Performing the division, the quotient polynomial is [tex]\( x^2 + x - 2 \)[/tex].
7. Now, we need to find the zeros of the quadratic polynomial [tex]\( x^2 + x - 2 \)[/tex].
8. Solve the quadratic equation:
[tex]\[
x^2 + x - 2 = 0
\][/tex]
We can factor this quadratic equation:
[tex]\[
(x - 1)(x + 2) = 0
\][/tex]
9. From the factorization, the solutions to the quadratic equation are:
[tex]\[
x - 1 = 0 \quad \text{or} \quad x + 2 = 0
\][/tex]
This gives us:
[tex]\[
x = 1 \quad \text{or} \quad x = -2
\][/tex]
10. Therefore, the zeros of the polynomial [tex]\( P(x) = x^3 + 3x^2 - 4 \)[/tex] are:
[tex]\[
x = -2, \quad x = 1
\][/tex]
In conclusion, the polynomial [tex]\( P(x) = x^3 + 3x^2 - 4 \)[/tex] has two distinct real zeros:
[tex]\[
x = -2 \quad \text{and} \quad x = 1
\][/tex]
