For what value of k can the expression x 3+ kx 2– 7x + 6 be resolved into three linear factors?
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Finding the Value of k for Linear Factors
The problem asks for the value of \(k\) such that the expression \(x^3 + kx^2 - 7x + 6\) can be resolved into three linear factors. A cubic polynomial with real coefficients can be factored into three linear factors if and only if it has three roots (counting multiplicity).
Let the polynomial be \(P(x) = x^3 + kx^2 - 7x + 6\). If this polynomial can be factored into three linear factors, say \((x-r_1)(x-r_2)(x-r_3)\), then \(r_1, r_2, r_3\) are the roots of the polynomial equation \(P(x)=0\).
Relationship Between Roots and Coefficients (Vieta's Formulas)
For a cubic polynomial \(ax^3 + bx^2 + cx + d\), the relationships between the roots (\(r_1, r_2, r_3\)) and the coefficients are:
- Sum of roots: \(r_1 + r_2 + r_3 = -\frac{b}{a}\)
- Sum of products of roots taken two at a time: \(r_1r_2 + r_1r_3 + r_2r_3 = \frac{c}{a}\)
- Product of roots: \(r_1r_2r_3 = -\frac{d}{a}\)
In our polynomial \(P(x) = x^3 + kx^2 - 7x + 6\), we have \(a=1\), \(b=k\), \(c=-7\), and \(d=6\). Comparing the coefficients with the formulas:
- \(-k = r_1 + r_2 + r_3\)
- \(-7 = r_1r_2 + r_1r_3 + r_2r_3\)
- \(-6 = r_1r_2r_3\)
From the last equation, the product of the three roots must be \(-6\). If we assume the roots are integers (or rational numbers), they must be divisors of the constant term 6. The integer divisors of 6 are \(\pm 1, \pm 2, \pm 3, \pm 6\).
Finding Potential Roots and the Value of k
We need to find three numbers whose product is \(-6\) and whose sum of pairwise products is \(-7\). Let's test combinations of integer divisors of 6.
Consider the set of roots \(\{1, 2, -3\}\).
- Product of roots: \(1 \times 2 \times (-3) = -6\). This matches the condition \(r_1r_2r_3 = -6\).
- Sum of products of roots taken two at a time: \((1 \times 2) + (1 \times -3) + (2 \times -3) = 2 - 3 - 6 = -7\). This matches the condition \(r_1r_2 + r_1r_3 + r_2r_3 = -7\).
Since these roots satisfy the conditions for the coefficient of \(x\) and the constant term, they must be the roots of the polynomial for some value of \(k\).
Now, let's find \(k\) using the sum of these roots:
- \(-k = r_1 + r_2 + r_3\)
- \(-k = 1 + 2 + (-3)\)
- \(-k = 3 - 3\)
- \(-k = 0\)
- \(k = 0\)
So, if \(k=0\), the polynomial \(x^3 - 7x + 6\) has the roots 1, 2, and -3, and thus can be factored into \((x-1)(x-2)(x+3)\).
Verification Using the Factor Theorem
Alternatively, we can use the Factor Theorem. If a polynomial \(P(x)\) has a linear factor \((x-r)\), then \(r\) is a root, meaning \(P(r)=0\). If the polynomial has integer roots, they must be divisors of the constant term 6.
Let's test if \(x=1\) is a root for some value of \(k\). Substitute \(x=1\) into \(P(x)\):
\(P(1) = (1)^3 + k(1)^2 - 7(1) + 6\)
\(P(1) = 1 + k - 7 + 6\)
\(P(1) = k\)
For \(x=1\) to be a root, \(P(1)\) must be 0. So, \(k=0\).
Now, let's see if the polynomial \(x^3 - 7x + 6\) (when \(k=0\)) can be factored into three linear factors. Since \(k=0\), \(x=1\) is a root, so \((x-1)\) is a factor. We can divide \(x^3 - 7x + 6\) by \((x-1)\).
Using synthetic division:
| 1 | 0 | -7 | 6 | |
|---|---|---|---|---|
| 1 | 1 | 1 | -6 | |
| 1 | 1 | -6 | 0 |
The quotient is \(x^2 + x - 6\). So, \(x^3 - 7x + 6 = (x-1)(x^2 + x - 6)\).
Now, we factor the quadratic \(x^2 + x - 6\). We look for two numbers that multiply to -6 and add to 1. These numbers are 3 and -2.
\(x^2 + x - 6 = (x+3)(x-2)\)
Therefore, for \(k=0\), the polynomial is \(x^3 - 7x + 6 = (x-1)(x+3)(x-2)\), which are three linear factors.
Thus, the value of \(k\) for which the expression can be resolved into three linear factors is 0.
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Linear Factor | A polynomial of degree 1, e.g., \((x-r)\). | The problem asks for resolution into three linear factors. |
| Root of a Polynomial | A value \(r\) such that \(P(r)=0\). If \((x-r)\) is a factor, \(r\) is a root. | Finding roots helps in finding linear factors. |
| Factor Theorem | If \(P(r)=0\), then \((x-r)\) is a factor of \(P(x)\). | Used to test potential roots and find \(k\). |
| Vieta's Formulas | Relate coefficients of a polynomial to sums and products of its roots. | Used to find relationships between \(k\), \(-7\), \(6\) and the roots. |
Additional Information: Polynomial Roots and Factoring
For a cubic polynomial \(P(x)\) with real coefficients:
- It always has at least one real root.
- The roots can be three real roots (distinct or repeated), or one real root and two complex conjugate roots.
- If the polynomial can be resolved into three linear factors over real numbers, it must have three real roots. The problem statement "three linear factors" typically implies factors over the same field as the coefficients (real numbers in this case).
- The Rational Root Theorem can help identify potential rational roots \(\frac{p}{q}\), where \(p\) divides the constant term and \(q\) divides the leading coefficient. In our case (\(x^3 + kx^2 - 7x + 6\)), the leading coefficient is 1, so any rational root must be an integer divisor of 6.
- Once a root \(r\) is found (either by testing divisors or by Vieta's formulas), synthetic division or polynomial division can be used to divide \(P(x)\) by \((x-r)\). The result for a cubic is a quadratic polynomial. Factoring this quadratic (using the quadratic formula, factoring by grouping, or trial and error) yields the remaining two linear factors or reveals if they are complex.
In this problem, finding a value of \(k\) that makes one of the integer divisors of 6 a root proved to be an efficient way to determine \(k\). The value \(k=0\) allows \(x=1\) to be a root, and the remaining quadratic factors into two more linear terms.
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