Understanding HCF and Polynomials

The Highest Common Factor (HCF) of two polynomials is the polynomial of the highest possible degree that divides both polynomials. In this question, we are given that \((x - k)\) is the HCF of the two polynomials \(x^2 + ax + b\) and \(x^2 + cx + d\).

If \((x - k)\) is the HCF, it means \((x - k)\) is a factor of both polynomials. According to the Factor Theorem, if \((x - k)\) is a factor of a polynomial \(P(x)\), then \(P(k)\) must be equal to zero.

We can apply this theorem to both given polynomials.

Applying the Factor Theorem

Let the first polynomial be \(P_1(x) = x^2 + ax + b\). Since \((x - k)\) is a factor, \(P_1(k) = 0\). Substituting \(x = k\), we get:

\(k^2 + ak + b = 0\) (Equation 1)

Let the second polynomial be \(P_2(x) = x^2 + cx + d\). Since \((x - k)\) is also a factor, \(P_2(k) = 0\). Substituting \(x = k\), we get:

\(k^2 + ck + d = 0\) (Equation 2)

Solving for k

We now have a system of two equations involving \(k\):

1. \(k^2 + ak + b = 0\)

2. \(k^2 + ck + d = 0\)

We want to find the value of \(k\). We can eliminate the \(k^2\) term by subtracting Equation 2 from Equation 1:

\((k^2 + ak + b) - (k^2 + ck + d) = 0 - 0\)

\(k^2 + ak + b - k^2 - ck - d = 0\)

The \(k^2\) terms cancel out:

\(ak + b - ck - d = 0\)

Now, group the terms with \(k\) and the constant terms:

\(ak - ck = d - b\)

Factor out \(k\) from the terms on the left side:

\(k(a - c) = d - b\)

To find \(k\), divide both sides by \((a - c)\), assuming \(a \neq c\). If \(a=c\), then the polynomials \(x^2+ax+b\) and \(x^2+cx+d\) would have the same coefficient for \(x\), and the subtraction would result in \(b-d=0\), implying \(b=d\). In this case, the polynomials would be identical, and their HCF would be themselves, not necessarily \((x-k)\) unless they are factorable as \((x-k)(x-p)\). However, the problem implies \(a \neq c\) or \(b \neq d\) for a unique value of \(k\) in this form.

Assuming \(a \neq c\):

\(k = \frac{{d - b}}{{a - c}}\)

This is the value of \(k\) in terms of \(a\), \(b\), \(c\), and \(d\).

Step-by-Step Derivation

1. Identify that \((x - k)\) is a factor of both \(x^2 + ax + b\) and \(x^2 + cx + d\).
2. Use the Factor Theorem: If \((x - k)\) is a factor of \(P(x)\), then \(P(k) = 0\).
3. Apply the theorem to \(x^2 + ax + b\): \(k^2 + ak + b = 0\).
4. Apply the theorem to \(x^2 + cx + d\): \(k^2 + ck + d = 0\).
5. Subtract the second equation from the first: \((k^2 + ak + b) - (k^2 + ck + d) = 0\).
6. Simplify the equation: \(ak + b - ck - d = 0\).
7. Rearrange the terms: \(ak - ck = d - b\).
8. Factor out \(k\): \(k(a - c) = d - b\).
9. Solve for \(k\): \(k = \frac{{d - b}}{{a - c}}\) (assuming \(a \neq c\)).

Comparing with Options

Let's compare our derived value of \(k\) with the given options:

Our result, \(k = \frac{{d - b}}{{a - c}}\), matches Option 2.

Revision Table: Key Concepts for HCF of Polynomials

Additional Information: Common Roots and Polynomials

When two polynomials share a common factor like \((x - k)\), it means they share a common root, which is \(k\). If they share *only* \((x - k)\) as a common factor (up to a constant multiple), then \((x - k)\) is their HCF.

If \(k\) is a common root of two polynomials \(P(x)\) and \(Q(x)\), then \(P(k) = 0\) and \(Q(k) = 0\). This implies that \(k\) satisfies both equations simultaneously. The method used above, subtracting the two equations, is a standard technique to find common roots of two polynomial equations, especially when the highest degree term is the same (like \(k^2\) here).

The general idea is that if \(k\) is a common root of \(P(x)=0\) and \(Q(x)=0\), then \(k\) is also a root of any linear combination \(m P(x) + n Q(x) = 0\) for constants \(m\) and \(n\). By choosing \(m=1\) and \(n=-1\), we get \(P(x) - Q(x) = 0\), which eliminates the \(x^2\) term and gives a linear equation in \(x\) (or \(k\) in our case) that is easy to solve, provided the linear term coefficient is non-zero \((a-c \neq 0)\).