If (x + k) is the HCF of x 2+ 5x + 6 and x 2+ 8x + 15, then what is the value of k?
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Finding the Value of k using Polynomial HCF
The question asks us to find the value of \(k\) given that \((x + k)\) is the Highest Common Factor (HCF) of two polynomial expressions: \(x^2 + 5x + 6\) and \(x^2 + 8x + 15\). To find the HCF of polynomials, we first need to factorize each polynomial into its irreducible factors.
Step 1: Factorize the First Polynomial
The first polynomial is \(x^2 + 5x + 6\). This is a quadratic trinomial of the form \(ax^2 + bx + c\), where \(a=1\), \(b=5\), and \(c=6\). To factorize it, we look for two numbers that multiply to \(c\) (which is 6) and add up to \(b\) (which is 5).
Let the two numbers be \(m\) and \(n\). We need:
- \(m \times n = 6\)
- \(m + n = 5\)
Let's list pairs of integers that multiply to 6:
- 1 and 6 (Sum = 7)
- 2 and 3 (Sum = 5)
- -1 and -6 (Sum = -7)
- -2 and -3 (Sum = -5)
The pair that adds up to 5 is 2 and 3. So we can rewrite the middle term (\(5x\)) as \(2x + 3x\):
\(x^2 + 5x + 6 = x^2 + 2x + 3x + 6\)
Now, we can group the terms and factor by grouping:
\((x^2 + 2x) + (3x + 6)\)
\(x(x + 2) + 3(x + 2)\)
Factor out the common binomial factor \((x + 2)\):
\((x + 2)(x + 3)\)
So, the factorization of \(x^2 + 5x + 6\) is \((x + 2)(x + 3)\).
Step 2: Factorize the Second Polynomial
The second polynomial is \(x^2 + 8x + 15\). This is also a quadratic trinomial where \(a=1\), \(b=8\), and \(c=15\). We look for two numbers that multiply to \(c\) (which is 15) and add up to \(b\) (which is 8).
Let the two numbers be \(p\) and \(q\). We need:
- \(p \times q = 15\)
- \(p + q = 8\)
Let's list pairs of integers that multiply to 15:
- 1 and 15 (Sum = 16)
- 3 and 5 (Sum = 8)
- -1 and -15 (Sum = -16)
- -3 and -5 (Sum = -8)
The pair that adds up to 8 is 3 and 5. So we rewrite the middle term (\(8x\)) as \(3x + 5x\):
\(x^2 + 8x + 15 = x^2 + 3x + 5x + 15\)
Now, we group the terms and factor by grouping:
\((x^2 + 3x) + (5x + 15)\)
\(x(x + 3) + 5(x + 3)\)
Factor out the common binomial factor \((x + 3)\):
\((x + 3)(x + 5)\)
So, the factorization of \(x^2 + 8x + 15\) is \((x + 3)(x + 5)\).
Step 3: Find the HCF of the Polynomials
The HCF of two or more polynomials is the product of the common factors with the lowest power. In this case, we have the factorizations:
- \(x^2 + 5x + 6 = (x + 2)(x + 3)\)
- \(x^2 + 8x + 15 = (x + 3)(x + 5)\)
The common factor between these two factorizations is \((x + 3)\). Both polynomials have \((x + 3)\) raised to the power of 1. Therefore, the HCF is \((x + 3)\).
| Polynomial | Factors |
|---|---|
| \(x^2 + 5x + 6\) | \((x + 2), (x + 3)\) |
| \(x^2 + 8x + 15\) | \((x + 3), (x + 5)\) |
| Common Factor (HCF) | \((x + 3)\) |
Step 4: Determine the Value of k
We are given that the HCF of the two polynomials is \((x + k)\). From our factorization and HCF calculation, we found that the HCF is \((x + 3)\).
Therefore, we can equate the given HCF with the calculated HCF:
\(x + k = x + 3\)
To find the value of \(k\), we can subtract \(x\) from both sides of the equation:
\(x + k - x = x + 3 - x\)
\(k = 3\)
Thus, the value of \(k\) is 3.
Revision Table: Polynomial HCF Calculation
| Step | Action | Polynomial 1 (\(x^2 + 5x + 6\)) | Polynomial 2 (\(x^2 + 8x + 15\)) |
|---|---|---|---|
| 1 | Factorize | \((x + 2)(x + 3)\) | \((x + 3)(x + 5)\) |
| 2 | Identify Common Factors | Common factor is \((x + 3)\) | Common factor is \((x + 3)\) |
| 3 | Determine HCF | The HCF is the common factor, which is \((x + 3)\) | |
| 4 | Equate HCF with \((x + k)\) | \((x + k) = (x + 3)\) | |
| 5 | Solve for \(k\) | Comparing both sides, \(k = 3\) | |
Additional Information: Understanding Polynomial HCF and Factorization
The Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD), of two or more polynomials is the polynomial of the highest possible degree that is a factor of all the given polynomials. Just like finding the HCF of numbers involves prime factorization, finding the HCF of polynomials involves polynomial factorization.
Polynomial Factorization is the process of expressing a polynomial as a product of irreducible polynomials. Irreducible polynomials are polynomials that cannot be factored further into non-constant polynomials with coefficients in a given set (like rational numbers or real numbers). For quadratic trinomials of the form \(x^2 + bx + c\) where the leading coefficient is 1, factorization often involves finding two numbers that multiply to \(c\) and add to \(b\).
If \((x - r)\) is a factor of a polynomial \(P(x)\), then \(r\) is a root of the polynomial, meaning \(P(r) = 0\). In our problem, since \((x + 3)\) is the HCF, it is a factor of both \(x^2 + 5x + 6\) and \(x^2 + 8x + 15\). This means that \(x = -3\) should be a root for both polynomials:
- For \(x^2 + 5x + 6\): Substitute \(x = -3\). \((-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0\). This confirms \((x + 3)\) is a factor.
- For \(x^2 + 8x + 15\): Substitute \(x = -3\). \((-3)^2 + 8(-3) + 15 = 9 - 24 + 15 = 0\). This confirms \((x + 3)\) is a factor.
This property also verifies our factorization and HCF calculation. The HCF represents the common "building blocks" of the polynomials when expressed as products of simpler polynomials.
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