Understanding the Problem

We are given two polynomials, \(f(x)\) and \(g(x)\), and their Highest Common Factor (HCF). Our goal is to find the specific values of the constants \(a\) and \(b\) that are part of these polynomials.

• The first polynomial is \(f(x) = (x – 1) (x^2 + 3x + a)\).
• The second polynomial is \(g(x) = (x + 2) (x^2 + 2x + b)\).
• The HCF of \(f(x)\) and \(g(x)\) is given as \(x^2 + x – 2\).

The key concept here is that the HCF of two polynomials must be a factor of both polynomials. If a polynomial \(P(x)\) is a factor of another polynomial \(Q(x)\), then any root of \(P(x)\) must also be a root of \(Q(x)\).

Analyzing the HCF of Polynomials

The given HCF is \(h(x) = x^2 + x – 2\). To understand its factors, we need to factorize this quadratic expression.

We look for two numbers that multiply to -2 and add up to +1. These numbers are +2 and -1.

So, the HCF can be factored as:

\[ h(x) = x^2 + x – 2 = (x + 2)(x – 1) \]

This means the factors of the HCF are \((x + 2)\) and \((x – 1)\). Since \((x + 2)(x – 1)\) is the HCF of \(f(x)\) and \(g(x)\), both \((x + 2)\) and \((x – 1)\) must be common factors of \(f(x)\) and \(g(x)\).

Finding the Value of 'a' in f(x)

The polynomial \(f(x)\) is given as \(f(x) = (x – 1) (x^2 + 3x + a)\). We know that \((x + 2)(x – 1)\) is a factor of \(f(x)\).

The term \((x – 1)\) is already explicitly a factor of \(f(x)\).

For \((x + 2)(x – 1)\) to be a factor of \(f(x)\), the term \((x + 2)\) must be a factor of the remaining part of \(f(x)\), which is \((x^2 + 3x + a)\).

If \((x + 2)\) is a factor of \((x^2 + 3x + a)\), then \(x = -2\) must be a root of the quadratic expression \(x^2 + 3x + a\). This means that when we substitute \(x = -2\) into \(x^2 + 3x + a\), the expression must evaluate to zero.

Substitute \(x = -2\):

\[ (-2)^2 + 3(-2) + a = 0 \]

\[ 4 – 6 + a = 0 \]

\[ -2 + a = 0 \]

Solving for \(a\):

\[ a = 2 \]

Finding the Value of 'b' in g(x)

The polynomial \(g(x)\) is given as \(g(x) = (x + 2) (x^2 + 2x + b)\). We know that \((x + 2)(x – 1)\) is a factor of \(g(x)\).

The term \((x + 2)\) is already explicitly a factor of \(g(x)\).

For \((x + 2)(x – 1)\) to be a factor of \(g(x)\), the term \((x – 1)\) must be a factor of the remaining part of \(g(x)\), which is \((x^2 + 2x + b)\).

If \((x – 1)\) is a factor of \((x^2 + 2x + b)\), then \(x = 1\) must be a root of the quadratic expression \(x^2 + 2x + b\). This means that when we substitute \(x = 1\) into \(x^2 + 2x + b\), the expression must evaluate to zero.

Substitute \(x = 1\):

\[ (1)^2 + 2(1) + b = 0 \]

\[ 1 + 2 + b = 0 \]

\[ 3 + b = 0 \]

Solving for \(b\):

\[ b = -3 \]

Verification

Let's check if these values of \(a\) and \(b\) yield the correct HCF.

If \(a = 2\), then the second factor of \(f(x)\) is \(x^2 + 3x + 2\). Factoring this gives \((x + 1)(x + 2)\).
So, \(f(x) = (x – 1)(x + 1)(x + 2)\).

If \(b = -3\), then the second factor of \(g(x)\) is \(x^2 + 2x – 3\). Factoring this gives \((x – 1)(x + 3)\).
So, \(g(x) = (x + 2)(x – 1)(x + 3)\).

Now, let's find the HCF of \(f(x)\) and \(g(x)\):

• Factors of \(f(x)\) are \((x – 1)\), \((x + 1)\), and \((x + 2)\).
• Factors of \(g(x)\) are \((x + 2)\), \((x – 1)\), and \((x + 3)\).

The common factors are \((x – 1)\) and \((x + 2)\).

The HCF is the product of the common factors with the lowest power (which is 1 for both in this case).

HCF\((f(x), g(x))\) = \((x – 1)(x + 2) = x^2 + 2x – x – 2 = x^2 + x – 2\).

This matches the given HCF, so our values for \(a\) and \(b\) are correct.

Summary of Results

We found the values of \(a\) and \(b\) to be:

\[ a = 2 \]

\[ b = -3 \]

The values are \(a = 2\) and \(b = -3\) respectively.

Revision Table: Key Concepts in Polynomial HCF

Additional Information: LCM of Polynomials

While this problem focused on HCF, it's useful to know about the Least Common Multiple (LCM) of polynomials as well.

The LCM of two polynomials is the lowest degree polynomial that is a multiple of both given polynomials.

There is a relationship between HCF and LCM for two polynomials, say P(x) and Q(x):

\[ P(x) \times Q(x) = \text{HCF}(P(x), Q(x)) \times \text{LCM}(P(x), Q(x)) \]

Using the factored forms we found:

\(f(x) = (x – 1)(x + 1)(x + 2)\)
\(g(x) = (x + 2)(x – 1)(x + 3)\)
HCF\((f(x), g(x))\) = \((x – 1)(x + 2)\)

To find the LCM, we take all unique factors from both polynomials and use the highest power of each factor present in either polynomial. In this case, all factors have a power of 1.

Unique factors are \((x – 1)\), \((x + 1)\), \((x + 2)\), and \((x + 3)\).

LCM\((f(x), g(x))\) = \((x – 1)(x + 1)(x + 2)(x + 3)\).

This relationship between HCF and LCM is a fundamental concept in polynomial algebra problems.