Finding the Highest Common Factor (HCF) of Polynomials

The problem asks for the Highest Common Factor (HCF) of two given polynomials:

• Polynomial 1: \(\text{acx}^3 + \text{bcx}^2 + \text{adx}^2 + \text{acdx} + \text{bdx} + \text{bcd}\)
• Polynomial 2: \(\text{adx}^3 + \text{acx}^2 + \text{bdx}^2 + \text{bcx} + \text{acdx} + \text{bcd}\)

We are also given the conditions that HCF(c, d) = 1 and c \(\neq\) d. This implies c and d are coprime integers and are not equal.

Understanding HCF of Polynomials

The HCF of polynomials is the polynomial of the highest degree that divides each of the given polynomials without leaving a remainder. To find the HCF, we typically factorize each polynomial and identify the common factors.

Factoring the Polynomials

Let the first polynomial be \(P_1(x) = \text{acx}^3 + \text{bcx}^2 + \text{adx}^2 + \text{acdx} + \text{bdx} + \text{bcd}\).

Let the second polynomial be \(P_2(x) = \text{adx}^3 + \text{acx}^2 + \text{bdx}^2 + \text{bcx} + \text{acdx} + \text{bcd}\).

We can attempt to factor these polynomials by grouping terms or by testing potential factors. Looking at the options, potential linear factors are of the form \((ax+b)\), \((cx+d)\), \((bx+c)\), or \((ax+d)\).

Testing Potential Factors using the Root Theorem

If \((ax+b)\) is a factor of a polynomial, then \(x = -b/a\) must be a root of the polynomial (i.e., substituting \(x = -b/a\) makes the polynomial equal to zero).

Let's test if \((ax+b)\) is a factor of \(P_1(x)\) by substituting \(x = -b/a\):

\(\text{ac}(-b/a)^3 + \text{bc}(-b/a)^2 + \text{ad}(-b/a)^2 + \text{acd}(-b/a) + \text{bd}(-b/a) + \text{bcd}\)

\(= \text{ac}(-b^3/a^3) + \text{bc}(b^2/a^2) + \text{ad}(b^2/a^2) + \text{acd}(-b/a) + \text{bd}(-b/a) + \text{bcd}\)

\(= -\frac{\text{b}^3\text{c}}{\text{a}^2} + \frac{\text{b}^3\text{c}}{\text{a}^2} + \frac{\text{b}^2\text{d}}{\text{a}} - \frac{\text{abcd}}{\text{a}} - \frac{\text{b}^2\text{d}}{\text{a}} + \text{bcd}\)

\(= -\frac{\text{b}^3\text{c}}{\text{a}^2} + \frac{\text{b}^3\text{c}}{\text{a}^2} + \frac{\text{b}^2\text{d}}{\text{a}} - \text{bcd} - \frac{\text{b}^2\text{d}}{\text{a}} + \text{bcd}\)

\(= 0 + 0 + \frac{\text{b}^2\text{d}}{\text{a}} - \text{bcd} - \frac{\text{b}^2\text{d}}{\text{a}} + \text{bcd} = 0\)

Since \(P_1(-b/a) = 0\), \((x - (-b/a)) = (x + b/a)\), which is proportional to \((ax+b)\), is a factor of \(P_1(x)\). Therefore, \((ax+b)\) is a factor of \(P_1(x)\).

Now, let's test if \((ax+b)\) is a factor of \(P_2(x)\) by substituting \(x = -b/a\):

\(\text{ad}(-b/a)^3 + \text{ac}(-b/a)^2 + \text{bd}(-b/a)^2 + \text{bc}(-b/a) + \text{acd}(-b/a) + \text{bcd}\)

\(= \text{ad}(-b^3/a^3) + \text{ac}(b^2/a^2) + \text{bd}(b^2/a^2) + \text{bc}(-b/a) + \text{acd}(-b/a) + \text{bcd}\)

\(= -\frac{\text{b}^3\text{d}}{\text{a}^2} + \frac{\text{b}^2\text{c}}{\text{a}} + \frac{\text{b}^3\text{d}}{\text{a}^2} - \frac{\text{b}^2\text{c}}{\text{a}} - \frac{\text{abcd}}{\text{a}} + \text{bcd}\)

\(= -\frac{\text{b}^3\text{d}}{\text{a}^2} + \frac{\text{b}^2\text{c}}{\text{a}} + \frac{\text{b}^3\text{d}}{\text{a}^2} - \frac{\text{b}^2\text{c}}{\text{a}} - \text{bcd} + \text{bcd}\)

\(= 0 + 0 + 0 - 0 - \text{bcd} + \text{bcd} = 0\)

Since \(P_2(-b/a) = 0\), \((ax+b)\) is also a factor of \(P_2(x)\).

Since \((ax+b)\) is a common factor of both polynomials, it is a candidate for the HCF.

Complete Factorization of the Polynomials

Let's factor \(P_1(x)\) by grouping terms to explicitly show the \((ax+b)\) factor:

\(P_1(x) = \text{acx}^3 + \text{bcx}^2 + \text{adx}^2 + \text{acdx} + \text{bdx} + \text{bcd}\)

Group terms: \((\text{acx}^3 + \text{adx}^2) + (\text{bcx}^2 + \text{bdx}) + (\text{acdx} + \text{bcd})\)

\(= \text{ax}^2(\text{cx} + \text{d}) + \text{bx}(\text{cx} + \text{d}) + \text{cd}(\text{ax} + \text{b})\)

Group the first two terms: \((\text{ax}^2 + \text{bx})(\text{cx} + \text{d}) + \text{cd}(\text{ax} + \text{b})\)

\(= \text{x}(\text{ax} + \text{b})(\text{cx} + \text{d}) + \text{cd}(\text{ax} + \text{b})\)

Factor out the common term \((\text{ax} + \text{b})\):

\(P_1(x) = (\text{ax} + \text{b}) [\text{x}(\text{cx} + \text{d}) + \text{cd}]\)

\(P_1(x) = (\text{ax} + \text{b}) (\text{cx}^2 + \text{dx} + \text{cd})\)

Now let's factor \(P_2(x)\) by grouping terms to explicitly show the \((ax+b)\) factor:

\(P_2(x) = \text{adx}^3 + \text{acx}^2 + \text{bdx}^2 + \text{bcx} + \text{acdx} + \text{bcd}\)

Group terms: \((\text{adx}^3 + \text{bdx}^2) + (\text{acx}^2 + \text{bcx}) + (\text{acdx} + \text{bcd})\)

\(= \text{dx}^2(\text{adx} + \text{b}) + \text{cx}(\text{acx} + \text{b}) + \text{cd}(\text{ax} + \text{b})\) - This grouping was incorrect in thought process. Let's regroup based on \((ax+b)\).

Let's rearrange \(P_2(x)\): \(\text{adx}^3 + \text{acx}^2 + \text{bdx}^2 + \text{bcx} + \text{acdx} + \text{bcd}\)

Group terms differently: \((\text{adx}^3 + \text{acx}^2) + (\text{bdx}^2 + \text{bcx}) + (\text{acdx} + \text{bcd})\)

\(= \text{ax}^2(\text{dx} + \text{c}) + \text{bx}(\text{dx} + \text{c}) + \text{cd}(\text{ax} + \text{b})\)

Group the first two terms: \((\text{ax}^2 + \text{bx})(\text{dx} + \text{c}) + \text{cd}(\text{ax} + \text{b})\)

\(= \text{x}(\text{ax} + \text{b})(\text{dx} + \text{c}) + \text{cd}(\text{ax} + \text{b})\)

Factor out the common term \((\text{ax} + \text{b})\):

\(P_2(x) = (\text{ax} + \text{b}) [\text{x}(\text{dx} + \text{c}) + \text{cd}]\)

\(P_2(x) = (\text{ax} + \text{b}) (\text{dx}^2 + \text{cx} + \text{cd})\)

So the factorizations are:

• \(P_1(x) = (\text{ax} + \text{b}) (\text{cx}^2 + \text{dx} + \text{cd})\)
• \(P_2(x) = (\text{ax} + \text{b}) (\text{dx}^2 + \text{cx} + \text{cd})\)

Determining the Highest Common Factor

We have identified \((\text{ax} + \text{b})\) as a common factor. The HCF is the common factor of highest degree.

We need to check if the remaining quadratic factors \((\text{cx}^2 + \text{dx} + \text{cd})\) and \((\text{dx}^2 + \text{cx} + \text{cd})\) share any common factors, particularly linear factors, given the conditions HCF(c, d) = 1 and c \(\neq\) d.

Suppose the two quadratic factors share a common root \(r\). Then \(cr^2 + dr + cd = 0\) and \(dr^2 + cr + cd = 0\).

Subtracting the two equations gives:

\((cr^2 + dr + cd) - (dr^2 + cr + cd) = 0 - 0\)

\((c-d)r^2 + (d-c)r = 0\)

\((c-d)r^2 - (c-d)r = 0\)

\((c-d)r(r-1) = 0\)

Since \(c \neq d\), \((c-d) \neq 0\). Thus, \(r(r-1) = 0\), which means \(r=0\) or \(r=1\).

• If \(r=0\) is a common root, then substituting \(x=0\) into the quadratic factors should give 0. \(c(0)^2 + d(0) + cd = cd\). For this to be 0, \(cd=0\). This would mean either \(c=0\) or \(d=0\). However, the condition HCF(c, d) = 1 implies that both c and d must be non-zero (and coprime integers). So, \(r=0\) is not a common root under the given conditions.
• If \(r=1\) is a common root, then substituting \(x=1\) into the quadratic factors should give 0. \(c(1)^2 + d(1) + cd = c+d+cd\). For this to be 0, \(c+d+cd=0\). \(d(1)^2 + c(1) + cd = d+c+cd\). This is the same condition. For integers c and d, \(c+d+cd=0\) can be rewritten as \(cd+c+d+1=1\), or \((c+1)(d+1)=1\). Since c and d are integers, \((c+1)\) and \((d+1)\) must be integer factors of 1. The possible integer pairs are (1, 1) or (-1, -1). If \(c+1=1\) and \(d+1=1\), then \(c=0\) and \(d=0\). HCF(0,0) is not 1. If \(c+1=-1\) and \(d+1=-1\), then \(c=-2\) and \(d=-2\). This contradicts the condition \(c \neq d\). Therefore, for coprime integers c and d where c \(\neq\) d, the condition \(c+d+cd=0\) is not met. So, \(r=1\) is not a common root.

This analysis shows that the quadratic factors \((\text{cx}^2 + \text{dx} + \text{cd})\) and \((\text{dx}^2 + \text{cx} + \text{cd})\) do not share common linear factors of the form \((x-r)\) where \(r=0\) or \(r=1\). Under the general conditions provided, these quadratic factors are unlikely to share any common factors.

Thus, the Highest Common Factor of the two polynomials is just the common factor we found through substitution and factorization.

The HCF is \((\text{ax} + \text{b})\).

The common factor is \((\text{ax} + \text{b})\).

Conclusion

The Highest Common Factor (HCF) of the given polynomials is \((\text{ax} + \text{b})\).

Revision Table: HCF of Polynomials

Additional Information: Finding HCF in Algebra

Finding the HCF of polynomials is similar to finding the HCF of integers. The fundamental idea is to break down the expressions into their prime factors and find the common ones. For polynomials, this involves factorization techniques such as:

• Grouping terms
• Using algebraic identities
• Applying the Remainder Theorem or Factor Theorem (related to finding roots)
• Polynomial long division or synthetic division
• For numerical coefficients, finding their HCF (e.g., HCF of \(6x^2\) and \(9x\) is \(3x\)).

In this problem, a combination of testing potential factors and grouping helped reveal the structure leading to the HCF.