Answer :
To find the real zeros of the function [tex]\( f(x) = x^3 + 4x^2 + x - 6 \)[/tex], we aim to determine the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].
The polynomial in question is a cubic polynomial. The general form of a cubic polynomial [tex]\( ax^3 + bx^2 + cx + d \)[/tex] can have up to three real roots, which could be distinct or have multiplicity. Let's follow a systematic approach to find these roots.
### Step 1: Check for Rational Roots
Using the Rational Root Theorem, we consider possible rational roots of the polynomial [tex]\( ax^3 + bx^2 + cx + d \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], [tex]\( c = 1 \)[/tex], and [tex]\( d = -6 \)[/tex]. The potential rational roots are given by the factors of the constant term [tex]\( d \)[/tex] (in this case, [tex]\(-6\)[/tex]) divided by the factors of the leading coefficient [tex]\( a \)[/tex] (in this case, [tex]\( 1 \)[/tex]). Therefore, the potential rational roots are [tex]\(\pm 1, \pm 2, \pm 3, \pm 6\)[/tex].
### Step 2: Testing Rational Roots
Let's test each of these possible roots by substituting them into the polynomial [tex]\( f(x) \)[/tex] to see if [tex]\( f(x) = 0 \)[/tex].
1. [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^3 + 4(1)^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is a root.
2. [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^3 + 4(-1)^2 + (-1) - 6 = -1 + 4 - 1 - 6 = -4 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
3. [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^3 + 4(2)^2 + 2 - 6 = 8 + 16 + 2 - 6 = 20 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
4. [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6 = -8 + 16 - 2 - 6 = 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is a root.
5. [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 3^3 + 4(3)^2 + 3 - 6 = 27 + 36 + 3 - 6 = 60 \][/tex]
So, [tex]\( x = 3 \)[/tex] is not a root.
6. [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^3 + 4(-3)^2 + (-3) - 6 = -27 + 36 - 3 - 6 = 0 \][/tex]
So, [tex]\( x = -3 \)[/tex] is a root.
### Step 3: List the Real Zeros
The roots we found to make [tex]\( f(x) = 0 \)[/tex] are:
[tex]\[ x = -3, x = -2, \text{ and } x = 1 \][/tex]
### Conclusion
Therefore, the function [tex]\( f(x) = x^3 + 4x^2 + x - 6 \)[/tex] has exactly three real zeros:
[tex]\[ x = -3, x = -2, \text{ and } x = 1 \][/tex]
The polynomial in question is a cubic polynomial. The general form of a cubic polynomial [tex]\( ax^3 + bx^2 + cx + d \)[/tex] can have up to three real roots, which could be distinct or have multiplicity. Let's follow a systematic approach to find these roots.
### Step 1: Check for Rational Roots
Using the Rational Root Theorem, we consider possible rational roots of the polynomial [tex]\( ax^3 + bx^2 + cx + d \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], [tex]\( c = 1 \)[/tex], and [tex]\( d = -6 \)[/tex]. The potential rational roots are given by the factors of the constant term [tex]\( d \)[/tex] (in this case, [tex]\(-6\)[/tex]) divided by the factors of the leading coefficient [tex]\( a \)[/tex] (in this case, [tex]\( 1 \)[/tex]). Therefore, the potential rational roots are [tex]\(\pm 1, \pm 2, \pm 3, \pm 6\)[/tex].
### Step 2: Testing Rational Roots
Let's test each of these possible roots by substituting them into the polynomial [tex]\( f(x) \)[/tex] to see if [tex]\( f(x) = 0 \)[/tex].
1. [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^3 + 4(1)^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex] is a root.
2. [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^3 + 4(-1)^2 + (-1) - 6 = -1 + 4 - 1 - 6 = -4 \][/tex]
So, [tex]\( x = -1 \)[/tex] is not a root.
3. [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^3 + 4(2)^2 + 2 - 6 = 8 + 16 + 2 - 6 = 20 \][/tex]
So, [tex]\( x = 2 \)[/tex] is not a root.
4. [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6 = -8 + 16 - 2 - 6 = 0 \][/tex]
So, [tex]\( x = -2 \)[/tex] is a root.
5. [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 3^3 + 4(3)^2 + 3 - 6 = 27 + 36 + 3 - 6 = 60 \][/tex]
So, [tex]\( x = 3 \)[/tex] is not a root.
6. [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^3 + 4(-3)^2 + (-3) - 6 = -27 + 36 - 3 - 6 = 0 \][/tex]
So, [tex]\( x = -3 \)[/tex] is a root.
### Step 3: List the Real Zeros
The roots we found to make [tex]\( f(x) = 0 \)[/tex] are:
[tex]\[ x = -3, x = -2, \text{ and } x = 1 \][/tex]
### Conclusion
Therefore, the function [tex]\( f(x) = x^3 + 4x^2 + x - 6 \)[/tex] has exactly three real zeros:
[tex]\[ x = -3, x = -2, \text{ and } x = 1 \][/tex]
