4x 3 + 12x 2 - x - 3 is divisible by
Both (2x + 1) and (2x - 1)
Understanding Polynomial Divisibility
The question asks us to determine if the polynomial \(4x^3 + 12x^2 - x - 3\) is divisible by the linear factors \((2x + 1)\), \((2x - 1)\), both, or neither. To check for divisibility by a linear factor \((ax - b)\), we can use the Factor Theorem. The Factor Theorem states that a polynomial \(P(x)\) is divisible by \((ax - b)\) if and only if \(P(\frac{b}{a}) = 0\).
Applying the Factor Theorem to Check Divisibility
Let our polynomial be \(P(x) = 4x^3 + 12x^2 - x - 3\).
Checking Divisibility by \((2x + 1)\)
The factor is \((2x + 1)\). This is in the form \((ax - b)\) where \(a = 2\) and \(b = -1\). According to the Factor Theorem, \(P(x)\) is divisible by \((2x + 1)\) if \(P(\frac{-1}{2}) = 0\).
Let's evaluate \(P(-\frac{1}{2})\):
\(P(-\frac{1}{2}) = 4(-\frac{1}{2})^3 + 12(-\frac{1}{2})^2 - (-\frac{1}{2}) - 3\)
\(P(-\frac{1}{2}) = 4(-\frac{1}{8}) + 12(\frac{1}{4}) + \frac{1}{2} - 3\)
\(P(-\frac{1}{2}) = -\frac{4}{8} + \frac{12}{4} + \frac{1}{2} - 3\)
\(P(-\frac{1}{2}) = -\frac{1}{2} + 3 + \frac{1}{2} - 3\)
\(P(-\frac{1}{2}) = (-\frac{1}{2} + \frac{1}{2}) + (3 - 3)\)
\(P(-\frac{1}{2}) = 0 + 0\)
\(P(-\frac{1}{2}) = 0\)
Since \(P(-\frac{1}{2}) = 0\), the polynomial \(4x^3 + 12x^2 - x - 3\) is divisible by \((2x + 1)\).
Checking Divisibility by \((2x - 1)\)
The factor is \((2x - 1)\). This is in the form \((ax - b)\) where \(a = 2\) and \(b = 1\). According to the Factor Theorem, \(P(x)\) is divisible by \((2x - 1)\) if \(P(\frac{1}{2}) = 0\).
Let's evaluate \(P(\frac{1}{2})\):
\(P(\frac{1}{2}) = 4(\frac{1}{2})^3 + 12(\frac{1}{2})^2 - (\frac{1}{2}) - 3\)
\(P(\frac{1}{2}) = 4(\frac{1}{8}) + 12(\frac{1}{4}) - \frac{1}{2} - 3\)
\(P(\frac{1}{2}) = \frac{4}{8} + \frac{12}{4} - \frac{1}{2} - 3\)
\(P(\frac{1}{2}) = \frac{1}{2} + 3 - \frac{1}{2} - 3\)
\(P(\frac{1}{2}) = (\frac{1}{2} - \frac{1}{2}) + (3 - 3)\)
\(P(\frac{1}{2}) = 0 + 0\)
\(P(\frac{1}{2}) = 0\)
Since \(P(\frac{1}{2}) = 0\), the polynomial \(4x^3 + 12x^2 - x - 3\) is divisible by \((2x - 1)\).
Conclusion on Divisibility
We found that \(P(-\frac{1}{2}) = 0\) and \(P(\frac{1}{2}) = 0\). This means the polynomial \(4x^3 + 12x^2 - x - 3\) is divisible by both \((2x + 1)\) and \((2x - 1)\).
Revision Table: Key Concepts
| Concept | Description | Application Here |
|---|---|---|
| Polynomial | An expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. | \(P(x) = 4x^3 + 12x^2 - x - 3\) is the polynomial. |
| Divisibility of Polynomials | A polynomial \(P(x)\) is divisible by a polynomial \(D(x)\) if dividing \(P(x)\) by \(D(x)\) results in a remainder of zero. | Checking if \(P(x)\) divided by \((2x+1)\) or \((2x-1)\) gives a remainder of zero. |
| Factor Theorem | For a polynomial \(P(x)\) and a number \(c\), \(x - c\) is a factor of \(P(x)\) if and only if \(P(c) = 0\). For a factor \((ax-b)\), it's a factor if \(P(\frac{b}{a})=0\). | Used to check divisibility by evaluating \(P(x)\) at the roots of the linear factors (\(-\frac{1}{2}\) and \(\frac{1}{2}\)). |
| Root of a Polynomial | A value of the variable for which the polynomial evaluates to zero. If \(P(c) = 0\), then \(c\) is a root. | \(-\frac{1}{2}\) is the root of \((2x+1)\) and \(\frac{1}{2}\) is the root of \((2x-1)\). |
Additional Information on Polynomial Factors
Since both \((2x + 1)\) and \((2x - 1)\) are factors of the polynomial \(4x^3 + 12x^2 - x - 3\), their product is also a factor (unless they are repeated factors). The product \((2x + 1)(2x - 1)\) is a difference of squares, which equals \((2x)^2 - (1)^2 = 4x^2 - 1\). Therefore, the polynomial is also divisible by \(4x^2 - 1\). We could perform polynomial long division to find the other factor. Dividing \(4x^3 + 12x^2 - x - 3\) by \(4x^2 - 1\) gives \(x + 3\). So, the polynomial can be factored as \((4x^2 - 1)(x + 3)\) or \((2x+1)(2x-1)(x+3)\).
This confirms that both \((2x + 1)\) and \((2x - 1)\) are indeed factors and the polynomial is divisible by both.
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