Let p and q be non-zero integers. Consider the polynomial A(x) = x 2+ px + q. It is given that (x − m) and (x − km) are simple factors of A(x), where m is a non-zero integer and k is positive integer, k ≥ 2. Which one of the following is correct?
(k + 1) 2q = kp 2
Understanding Factors and Roots of Polynomials
A factor of a polynomial \(A(x)\) is an expression \((x - r)\) such that \(A(r) = 0\). If \((x - r)\) is a factor, then \(r\) is a root of the polynomial. The problem states that \((x - m)\) and \((x - km)\) are simple factors of the polynomial \(A(x) = x^2 + px + q\). This means that \(m\) and \(km\) are the roots of the polynomial \(A(x)\). A 'simple' factor implies that each root has a multiplicity of 1.
The given polynomial is a quadratic polynomial of the form \(ax^2 + bx + c\), where in this case, \(a=1\), \(b=p\), and \(c=q\). For a quadratic polynomial, there are well-known relationships between its roots and coefficients, often referred to as Vieta's formulas.
Applying Vieta's Formulas for Quadratic Polynomials
Let the roots of a quadratic equation \(ax^2 + bx + c = 0\) be \(\alpha\) and \(\beta\). Vieta's formulas state the following relationships:
- Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
- Product of roots: \(\alpha \beta = \frac{c}{a}\)
In our problem, the polynomial is \(A(x) = x^2 + px + q\), and the roots are \(m\) and \(km\). Comparing this to the standard form, we have \(a=1\), \(b=p\), and \(c=q\). The roots are \(\alpha = m\) and \(\beta = km\).
Using Vieta's formulas, we can write two equations based on the roots \(m\) and \(km\) and the coefficients \(p\) and \(q\):
- Sum of roots: \(m + km = -\frac{p}{1} = -p\)
- Product of roots: \(m \times km = \frac{q}{1} = q\)
Deriving the Relationship between p, q, and k
From the sum of roots equation, we have:
\(m(1 + k) = -p\)
Since \(m\) is a non-zero integer and \(k\) is a positive integer (\(k \ge 2\)), \(1+k \ne 0\). We can express \(m\) in terms of \(p\) and \(k\):
\(m = \frac{-p}{1 + k}\) (Equation 1)
From the product of roots equation, we have:
\(km^2 = q\) (Equation 2)
Now, we can substitute the expression for \(m\) from Equation 1 into Equation 2 to eliminate \(m\) and find a relationship between \(p\), \(q\), and \(k\):
\(k \left(\frac{-p}{1 + k}\right)^2 = q\)
\(k \left(\frac{(-p)^2}{(1 + k)^2}\right) = q\)
\(k \left(\frac{p^2}{(1 + k)^2}\right) = q\)
To clear the denominator, multiply both sides of the equation by \((1 + k)^2\):
\(k p^2 = q (1 + k)^2\)
Rearranging the terms to match the options, we get:
\((k + 1)^2 q = k p^2\)
This is the required relationship between \(p\), \(q\), and \(k\). We can compare this derived relationship with the given options.
Analyzing the Options
Let's compare our derived relationship \((k + 1)^2 q = k p^2\) with the given options:
- Option 1: \((k + 1)^2 p^2 = kq\) (Incorrect)
- Option 2: \((k + 1)^2 q = kp^2\) (Matches our result)
- Option 3: \(k^2 q = (k + 1) p^2\) (Incorrect)
- Option 4: \(k^2 p^2 = (k + 1)^2 q\) (Incorrect, this is \(k^2 p^2 = (k+1)^2 q\), not \(k p^2 = (k+1)^2 q\))
The derived relationship \((k + 1)^2 q = k p^2\) exactly matches Option 2.
| Step | Description | Formula/Equation |
|---|---|---|
| 1 | Identify roots from factors | Roots are \(m\) and \(km\) |
| 2 | Apply Vieta's formula for sum of roots | \(m + km = -p\) |
| 3 | Apply Vieta's formula for product of roots | \(m \times km = q\) |
| 4 | Solve sum equation for \(m\) | \(m = \frac{-p}{1+k}\) |
| 5 | Substitute \(m\) into product equation | \(k \left(\frac{-p}{1+k}\right)^2 = q\) |
| 6 | Simplify and rearrange | \((k+1)^2 q = kp^2\) |
Conclusion on the Polynomial Relationship
Based on the property that \(m\) and \(km\) are the roots of the polynomial \(A(x) = x^2 + px + q\), we used Vieta's formulas to establish a system of equations involving \(m, p, q,\) and \(k\). By eliminating \(m\) from these equations, we successfully derived the relationship that must hold true for the non-zero integers \(p\) and \(q\), the non-zero integer \(m\), and the positive integer \(k \ge 2\). The resulting relationship is \((k + 1)^2 q = k p^2\), which corresponds to the second option.
Revision Table: Polynomial Factors and Roots
| Concept | Explanation | Relevance Here |
|---|---|---|
| Factor \((x-r)\) | If \((x-r)\) is a factor of \(A(x)\), then \(A(r)=0\). | \((x-m)\) and \((x-km)\) are factors, so \(m\) and \(km\) are roots. |
| Root of Polynomial | A value \(r\) for which \(A(r)=0\). Roots are also called zeros. | \(m\) and \(km\) are the roots of \(x^2+px+q\). |
| Vieta's Formulas (Quadratic) | Relates coefficients of a polynomial to sums and products of its roots. For \(ax^2+bx+c=0\), roots \(\alpha, \beta\): \(\alpha+\beta=-b/a\), \(\alpha\beta=c/a\). | Used to set up equations: \(m+km = -p\) and \(m \times km = q\). |
| Simple Factor | A factor \((x-r)\) where the root \(r\) has multiplicity 1. | Indicates \(m\) and \(km\) are distinct roots (since \(k \ge 2\) and \(m \ne 0\)). |
Additional Information: Vieta's Formulas Generalization
Vieta's formulas are powerful tools that apply to polynomials of any degree, not just quadratics. For a general polynomial of degree \(n\), \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\), with roots \(r_1, r_2, \dots, r_n\), the formulas relate the coefficients \(a_i\) to elementary symmetric polynomials of the roots:
- Sum of roots: \(\sum r_i = r_1 + r_2 + \dots + r_n = -\frac{a_{n-1}}{a_n}\)
- Sum of products of roots taken two at a time: \(\sum_{i<j} r_i r_j = r_1 r_2 + r_1 r_3 + \dots + r_{n-1} r_n = \frac{a_{n-2}}{a_n}\)
- Sum of products of roots taken three at a time: \(\sum_{i<j<k} r_i r_j r_k = -\frac{a_{n-3}}{a_n}\)
- ...
- Product of roots: \(r_1 r_2 \dots r_n = (-1)^n \frac{a_0}{a_n}\)
For the quadratic \(x^2 + px + q = 0\) (\(a_2=1, a_1=p, a_0=q\)), these reduce to:
- Sum of roots (\(n=2\), \((-1)^1 a_1/a_2\)): \(r_1 + r_2 = -p/1 = -p\)
- Product of roots (\(n=2\), \((-1)^2 a_0/a_2\)): \(r_1 r_2 = q/1 = q\)
These are exactly the formulas we used in solving the problem, confirming their application to the given quadratic polynomial.
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