Problem 13 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \((x+a)\) is a factor of both the quadratic polynomials \(x^{2}+p x+q\) and \(x^{2}+l x+m\), where \(p, q, l\) and \(m\) are constants, then which one of the following is correct? (a) \(a=(m-q) /(l-p)(l \neq p)\) (b) \(a=(m+q) /(l+p)(l \neq-p)\) (c) \(l=(m-q) /(a-p)(a \neq p)\) (d) \(p=(m-q) /(a-l)(a \neq l)\)

Step-by-Step Solution

Answer

Correct answer is (a): \(a = \frac{m-q}{l-p}(l \neq p)\).

Step 1: Understand the Factor Theorem

According to the factor theorem, if \((x + a)\) is a factor of a polynomial \(f(x)\), then \(f(-a) = 0\). Here, \((x + a)\) is a factor of both \(x^2 + px + q\) and \(x^2 + lx + m\). Thus, we will use \(f(-a) = 0\) on both polynomials.

Step 2: Apply the Factor Theorem to the First Polynomial

For the polynomial \(x^2 + px + q\), if \((x + a)\) is a factor, then substituting \((-a)\), we get:\[ (-a)^2 + p(-a) + q = 0 \] \[ a^2 - pa + q = 0 \]

Step 3: Apply the Factor Theorem to the Second Polynomial

For the polynomial \(x^2 + lx + m\), if \((x + a)\) is a factor, then substituting \((-a)\), we get:\[ (-a)^2 + l(-a) + m = 0 \] \[ a^2 - la + m = 0 \]

Step 4: Set the Two Equations Equal to Each Other

Since both expressions \(a^2 - pa + q = 0\) and \(a^2 - la + m = 0\) must be true, the equations can be set equal to each other:\[ a^2 - pa + q = a^2 - la + m \]

Step 5: Simplify and Solve for Constants

Cancel out \(a^2\) from both sides and simplify the equation:\[ -pa + q = -la + m \]Simplify further to get:\[ pa - la = m - q \]Thus, \[ (p - l)a = m - q \]

Step 6: Solve for 'a'

Rearrange the equation from Step 5 to solve for \(a\):\[ a = \frac{m - q}{l - p} \]Make sure \(l eq p\) is stated to avoid division by zero.

Key Concepts

Polynomial Factorization

Polynomial factorization is the process of expressing a polynomial as a product of its factors. These factors are usually polynomials of lower degrees and can sometimes be linear terms like \((x + a)\). Factorization is essential because it helps simplify polynomials, making them easier to solve or evaluate.

• One of the key theorems used in polynomial factorization is the extbf{Factor Theorem}. It states that a polynomial \(f(x)\) has a factor \((x + a)\) if and only if \(f(-a) = 0\).
• If you know a number \(a\) such that \(f(a) = 0\), you can claim that \((x - a)\) is a factor of \(f(x)\).
• This theorem is extremely useful when dealing with polynomials of any degree, as it provides a systematic way to detect linear factors.

Polynomial factorization is not just about finding factors automatically; it's about understanding the structure of the polynomial and using properties like the Factor Theorem to find these factors efficiently.

Quadratic Polynomials

Quadratic polynomials are polynomials of degree 2, and their general form is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. Quadratics are fundamental because they appear frequently in various mathematical contexts, including equations of motion and optimizations.

• When factoring quadratics, one commonly looks for two numbers that multiply to give the product \(ac\) and add to give \(b\).
• A quadratic can sometimes be factored into linear factors: \((x + r)(x + s)\), where \(r\) and \(s\) satisfy the conditions mentioned above.
• Quadratics may not always be factorable using integers or rationals. In such cases, the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can be used to find the roots or solutions.

In the particular problem at hand, recognizing that \((x + a)\) is a factor of both polynomials means that substituting \(-a\) into each polynomial will result in zero.

Polynomial Equations

Polynomial equations involve expressions equal to zero and can have different degrees, including linear and quadratic forms. Solving polynomial equations requires techniques like factorization, substitution, and utilizing fundamental theorems.

• One method to solve polynomial equations is to "set them equal to zero," using the roots of the polynomial to find solutions.
• For polynomials like \(x^2 + px + q = 0\), factoring into a product of lower-degree polynomials—or linear terms when possible—allows us to solve for \(x\).
• In this exercise, we are equating two polynomial expressions, leading to simplification and solving for unknown constants by setting the equations equal, canceling terms, and simplifying.

Effective solving of polynomial equations not only involves algebraic manipulation but also understanding the relationships between coefficients and roots, especially when they must also satisfy the Factor Theorem.