What are the factors of x 3+ 4x 2– 11x – 30?
(x + 2), (x – 3) and (x + 5)
Understanding Polynomial Factorization
The question asks us to find the factors of the polynomial \( P(x) = x^3 + 4x^2 - 11x - 30 \). Finding the factors of a polynomial means expressing it as a product of simpler polynomials, typically linear factors like \( (x-a) \), \( (x-b) \), etc. For a cubic polynomial like this, we expect to find three linear factors.
Using the Factor Theorem to Find Polynomial Factors
A very useful tool for finding factors of a polynomial is the Factor Theorem. The theorem states:
- If \( (x-a) \) is a factor of a polynomial \( P(x) \), then \( P(a) = 0 \).
- Conversely, if \( P(a) = 0 \) for some value \( a \), then \( (x-a) \) is a factor of \( P(x) \).
In this problem, the options provide sets of potential linear factors. We can test these potential factors by using the Factor Theorem. If \( (x-a) \) is a factor, then substituting \( x=a \) into the polynomial \( P(x) \) should result in zero.
Testing the Potential Factors
Let's test the factors provided in one of the options. The correct option suggests the factors are \( (x+2) \), \( (x-3) \), and \( (x+5) \). According to the Factor Theorem, if these are indeed the factors, then the polynomial \( P(x) \) must be zero when \( x = -2 \) (from \( x+2=0 \)), \( x = 3 \) (from \( x-3=0 \)), and \( x = -5 \) (from \( x+5=0 \)).
Let's evaluate \( P(x) \) at these values:
Testing \( x = -2 \) for Factor \( (x+2) \)
Substitute \( x = -2 \) into \( P(x) = x^3 + 4x^2 - 11x - 30 \):
\( P(-2) = (-2)^3 + 4(-2)^2 - 11(-2) - 30 \)
\( P(-2) = -8 + 4(4) - (-22) - 30 \)
\( P(-2) = -8 + 16 + 22 - 30 \)
\( P(-2) = 8 + 22 - 30 \)
\( P(-2) = 30 - 30 \)
\( P(-2) = 0 \)
Since \( P(-2) = 0 \), \( (x - (-2)) = (x+2) \) is a factor of the polynomial \( x^3 + 4x^2 - 11x - 30 \).
Testing \( x = 3 \) for Factor \( (x-3) \)
Substitute \( x = 3 \) into \( P(x) = x^3 + 4x^2 - 11x - 30 \):
\( P(3) = (3)^3 + 4(3)^2 - 11(3) - 30 \)
\( P(3) = 27 + 4(9) - 33 - 30 \)
\( P(3) = 27 + 36 - 33 - 30 \)
\( P(3) = 63 - 33 - 30 \)
\( P(3) = 30 - 30 \)
\( P(3) = 0 \)
Since \( P(3) = 0 \), \( (x - 3) \) is a factor of the polynomial \( x^3 + 4x^2 - 11x - 30 \).
Testing \( x = -5 \) for Factor \( (x+5) \)
Substitute \( x = -5 \) into \( P(x) = x^3 + 4x^2 - 11x - 30 \):
\( P(-5) = (-5)^3 + 4(-5)^2 - 11(-5) - 30 \)
\( P(-5) = -125 + 4(25) - (-55) - 30 \)
\( P(-5) = -125 + 100 + 55 - 30 \)
\( P(-5) = -25 + 55 - 30 \)
\( P(-5) = 30 - 30 \)
\( P(-5) = 0 \)
Since \( P(-5) = 0 \), \( (x - (-5)) = (x+5) \) is a factor of the polynomial \( x^3 + 4x^2 - 11x - 30 \).
Conclusion
Since substituting the roots \( x=-2 \), \( x=3 \), and \( x=-5 \) into the polynomial \( P(x) \) all result in zero, according to the Factor Theorem, the corresponding expressions \( (x+2) \), \( (x-3) \), and \( (x+5) \) are the factors of the polynomial \( x^3 + 4x^2 - 11x - 30 \).
Thus, the factorization is \( x^3 + 4x^2 - 11x - 30 = (x+2)(x-3)(x+5) \). You can expand this product to verify it matches the original polynomial.
The factors are \( (x+2) \), \( (x-3) \) and \( (x+5) \).
Revision Table: Key Polynomial Concepts
| Concept | Description | Relevance to this Problem |
|---|---|---|
| Polynomial | An expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. | The problem is about factoring a cubic polynomial. |
| Factor of a Polynomial | A polynomial that divides another polynomial evenly (with no remainder). | We are looking for linear factors that divide the given cubic polynomial. |
| Factor Theorem | A polynomial \( P(x) \) has a factor \( (x-a) \) if and only if \( P(a) = 0 \). | This theorem is the primary method used to verify the given potential factors. |
| Root of a Polynomial | A value \( x=a \) for which \( P(a) = 0 \). Roots are closely related to factors. If \( x=a \) is a root, then \( (x-a) \) is a factor. | We tested the roots \( -2 \), \( 3 \), and \( -5 \) to check if they make \( P(x) = 0 \). |
Additional Information: Other Factorization Methods
Besides the Factor Theorem, there are other ways to factor polynomials, especially after finding one factor using the theorem:
- Polynomial Long Division: Once you find one factor, say \( (x-a) \), you can divide the original polynomial \( P(x) \) by \( (x-a) \) using polynomial long division. The result will be a polynomial of lower degree (in this case, a quadratic), which can then be factored further.
- Synthetic Division: A shorthand method for polynomial division when dividing by a linear factor \( (x-a) \). It's generally faster than long division for this specific case.
- Grouping: Sometimes, polynomials can be factored by grouping terms, although this is less common for general cubic polynomials like the one in the question unless they have a specific structure.
For this problem, finding one factor using the Factor Theorem would allow you to use synthetic or long division to find the resulting quadratic, and then factor the quadratic. However, testing all three proposed roots directly using the Factor Theorem is efficient when options are provided.
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