Which one of the following is a factor of the polynomial (x - 1)(x - 2)(x - 4) - 90 ?
x - 7
Finding a Factor of the Polynomial
The question asks us to identify which of the given options is a factor of the polynomial \((x - 1)(x - 2)(x - 4) - 90\). A factor of a polynomial \(P(x)\) is an expression, say \((x - a)\), that divides \(P(x)\) evenly, leaving no remainder. According to the Factor Theorem, \((x - a)\) is a factor of a polynomial \(P(x)\) if and only if \(P(a) = 0\).
We can use the Factor Theorem to test each option. Each option is in the form \((x - a)\), so we need to evaluate the polynomial at the value of \(a\) for each option and see which one results in 0.
Applying the Factor Theorem to the Options
Let the given polynomial be \(P(x) = (x - 1)(x - 2)(x - 4) - 90\).
We will evaluate \(P(x)\) for the potential roots suggested by the options:
- Option 1: \(x + 14\). This factor corresponds to a root \(x = -14\). \(P(-14) = (-14 - 1)(-14 - 2)(-14 - 4) - 90\) \(P(-14) = (-15)(-16)(-18) - 90\) \(P(-14) = (240)(-18) - 90\) \(P(-14) = -4320 - 90\) \(P(-14) = -4410\) Since \(P(-14) \neq 0\), \((x + 14)\) is not a factor of the polynomial.
- Option 2: \(x - 14\). This factor corresponds to a root \(x = 14\). \(P(14) = (14 - 1)(14 - 2)(14 - 4) - 90\) \(P(14) = (13)(12)(10) - 90\) \(P(14) = (156)(10) - 90\) \(P(14) = 1560 - 90\) \(P(14) = 1470\) Since \(P(14) \neq 0\), \((x - 14)\) is not a factor of the polynomial.
- Option 3: \(x - 6\). This factor corresponds to a root \(x = 6\). \(P(6) = (6 - 1)(6 - 2)(6 - 4) - 90\) \(P(6) = (5)(4)(2) - 90\) \(P(6) = (20)(2) - 90\) \(P(6) = 40 - 90\) \(P(6) = -50\) Since \(P(6) \neq 0\), \((x - 6)\) is not a factor of the polynomial.
- Option 4: \(x - 7\). This factor corresponds to a root \(x = 7\). \(P(7) = (7 - 1)(7 - 2)(7 - 4) - 90\) \(P(7) = (6)(5)(3) - 90\) \(P(7) = (30)(3) - 90\) \(P(7) = 90 - 90\) \(P(7) = 0\) Since \(P(7) = 0\), according to the Factor Theorem, \((x - 7)\) is a factor of the polynomial \((x - 1)(x - 2)(x - 4) - 90\).
The evaluation shows that only \(x = 7\) makes the polynomial equal to zero. Therefore, \((x - 7)\) is a factor.
Alternatively: Expanding and Factoring the Polynomial
We can also expand the polynomial and attempt to factor it:
\(P(x) = (x - 1)(x - 2)(x - 4) - 90\)
First, multiply the first two factors:
\((x - 1)(x - 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2\)
Now, multiply this result by \((x - 4)\):
\((x^2 - 3x + 2)(x - 4) = x(x^2 - 3x + 2) - 4(x^2 - 3x + 2)\)
\(= x^3 - 3x^2 + 2x - 4x^2 + 12x - 8\)
Combine like terms:
\(= x^3 - 7x^2 + 14x - 8\)
Now, include the \(-90\):
\(P(x) = x^3 - 7x^2 + 14x - 8 - 90\)
\(P(x) = x^3 - 7x^2 + 14x - 98\)
We can try dividing this polynomial by \((x - 7)\) using synthetic division:
| \(1\) | \(-7\) | \(14\) | \(-98\) | |
|---|---|---|---|---|
| \(7\) | \(7\) | \(0\) | \(98\) | |
| \(1\) | \(0\) | \(14\) | \(0\) |
The remainder is \(0\), and the resulting quadratic is \(x^2 + 0x + 14 = x^2 + 14\). Thus, the polynomial can be factored as \((x - 7)(x^2 + 14)\). This confirms that \((x - 7)\) is indeed a factor of the polynomial.
Summary of Results
Using the Factor Theorem, we evaluated the polynomial at the roots corresponding to each option. We found that evaluating the polynomial at \(x=7\) results in 0, proving that \((x-7)\) is a factor.
| Option | Potential Factor | Root (\(a\)) | Value of \(P(a)\) | Is it a Factor? |
|---|---|---|---|---|
| 1 | \(x+14\) | \(-14\) | \(-4410\) | No |
| 2 | \(x-14\) | \(14\) | \(1470\) | No |
| 3 | \(x-6\) | \(6\) | \(-50\) | No |
| 4 | \(x-7\) | \(7\) | \(0\) | Yes |
Revision Table: Key Concepts for Polynomial Factors
| Concept | Explanation | Relevance to 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 given expression \((x - 1)(x - 2)(x - 4) - 90\) is a polynomial. |
| Factor of a Polynomial | A polynomial \(D(x)\) is a factor of \(P(x)\) if \(P(x)\) can be written as \(P(x) = D(x) \cdot Q(x)\) for some polynomial \(Q(x)\). If \(D(x)\) is a linear factor \((x-a)\), then \(a\) is a root of \(P(x)\). | We are looking for a linear factor \((x-a)\) of the given polynomial. |
| Factor Theorem | A polynomial \(P(x)\) has a factor \((x-a)\) if and only if \(P(a) = 0\). This means that \(a\) is a root of the polynomial. | This theorem is the primary method used to solve the problem by testing each option. |
| Root of a Polynomial | A value \(a\) for which \(P(a) = 0\). The roots are the values of \(x\) where the polynomial crosses the x-axis on a graph. | Finding a factor \((x-a)\) is equivalent to finding a root \(a\). |
Additional Information: Polynomial Factorization Techniques
Besides the Factor Theorem, there are other methods to factor polynomials, especially once one factor is found:
- Polynomial Long Division: If you know that \((x-a)\) is a factor of \(P(x)\), you can divide \(P(x)\) by \((x-a)\) using long division. The quotient will be another polynomial of a lower degree, which can then potentially be factored further.
- Synthetic Division: This is a simplified method of polynomial division specifically used when dividing by a linear factor of the form \((x-a)\) or \((x+a)\). It is quicker than long division for this specific case and helps find the quotient efficiently, as demonstrated in the alternative solution method.
- Grouping: For some polynomials with four terms, terms can be grouped and common factors pulled out to reveal a common binomial factor. This method is typically used for specific polynomial structures.
- Recognizing Patterns: Special polynomial forms like difference of squares (\(a^2 - b^2 = (a-b)(a+b)\)), sum/difference of cubes (\(a^3 \pm b^3\)), or perfect square trinomials can be factored directly using formulas.
In this problem, finding one linear factor using the Factor Theorem was the most straightforward approach given the multiple-choice options.
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