What is the LCM of the polynomials x 3+ 3x 2+ 3x + 1, x 3+ 5x 2+ 5x + 4 and x 2+ 5x + 4?
(x + 1) 3(x + 4)(x 2+ x + 1)
LCM Calculation for Polynomials
This problem requires finding the Least Common Multiple (LCM) of three given polynomials. The polynomials are:
- \(P_1(x) = x^3 + 3x^2 + 3x + 1\)
- \(P_2(x) = x^3 + 5x^2 + 5x + 4\)
- \(P_3(x) = x^2 + 5x + 4\)
The LCM is the polynomial of the lowest degree that is a multiple of all the given polynomials.
Polynomial Factorization Steps
To determine the LCM, the first crucial step is to factorize each polynomial completely into its simplest factors.
Factorizing \(x^3 + 3x^2 + 3x + 1\)
The polynomial \(P_1(x) = x^3 + 3x^2 + 3x + 1\) is recognizable as the expansion of a binomial cube. It fits the form \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\). By setting \(a=x\) and \(b=1\), we get:
\((x+1)^3 = x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3 = x^3 + 3x^2 + 3x + 1\).
Thus, the factorization is \(P_1(x) = (x+1)^3\).
Factorizing \(x^3 + 5x^2 + 5x + 4\)
To factorize \(P_2(x) = x^3 + 5x^2 + 5x + 4\), we can look for rational roots. According to the Rational Root Theorem, possible rational roots are factors of the constant term (4) divided by factors of the leading coefficient (1). Potential roots include \(\pm1, \pm2, \pm4\).
Let's test \(x=-4\): \(P_2(-4) = (-4)^3 + 5(-4)^2 + 5(-4) + 4\) \(P_2(-4) = -64 + 5(16) - 20 + 4\) \(P_2(-4) = -64 + 80 - 20 + 4 = 0\).
Since \(P_2(-4) = 0\), \((x+4)\) is a factor of \(P_2(x)\).
We can use polynomial division or synthetic division to find the remaining factor. Dividing \(x^3 + 5x^2 + 5x + 4\) by \((x+4)\) gives the quotient \(x^2 + x + 1\).
The quadratic factor \(x^2 + x + 1\) can be checked for further factorization by examining its discriminant, \(\Delta = b^2 - 4ac\). Here, \(a=1, b=1, c=1\).
\(\Delta = 1^2 - 4(1)(1) = 1 - 4 = -3\).
Since the discriminant \(\Delta\) is negative (\(\Delta\) < 0), the quadratic \(x^2 + x + 1\) has no real roots and is considered irreducible over the real numbers.
So, the factorization is \(P_2(x) = (x+4)(x^2 + x + 1)\).
Factorizing \(x^2 + 5x + 4\)
The polynomial \(P_3(x) = x^2 + 5x + 4\) is a simple quadratic expression. We need to find two numbers that multiply to 4 and add up to 5. These numbers are 1 and 4.
Thus, the factorization is \(P_3(x) = (x+1)(x+4)\).
LCM Synthesis Process
We have the factorizations for each polynomial:
- \(P_1(x) = (x+1)^3\)
- \(P_2(x) = (x+4)(x^2 + x + 1)\)
- \(P_3(x) = (x+1)(x+4)\)
To find the LCM, we must identify all unique factors present across these polynomials and take the highest power of each unique factor.
The unique factors identified are \((x+1)\), \((x+4)\), and \((x^2 + x + 1)\).
Now, let's determine the highest power for each factor:
- Highest power of \((x+1)\) is 3 (from \(P_1(x)\)).
- Highest power of \((x+4)\) is 1 (from \(P_2(x)\) and \(P_3(x)\)).
- Highest power of \((x^2 + x + 1)\) is 1 (from \(P_2(x)\)).
The LCM is the product of these factors raised to their highest powers:
LCM = \((x+1)^3 \times (x+4)^1 \times (x^2 + x + 1)^1\)
LCM = \((x+1)^3 (x+4) (x^2 + x + 1)\)
Verification with Provided Options
The calculated LCM is \((x+1)^3 (x+4) (x^2 + x + 1)\). We compare this result with the given options:
- \((x + 1)^3(x + 4)(x^2 + x + 1)\) – This matches our calculated LCM.
- \((x + 4)(x^2 + x + 1)\) – This is only \(P_2(x)\), missing the higher power of \((x+1)\).
- \((x + 1)(x^2 + x + 1)\) – This is incorrect; it lacks the factor \((x+4)\) and uses a lower power of \((x+1)\).
- \((x + 1)^2(x + 4)(x^2 + x + 1)\) – This is incorrect as the power of \((x+1)\) should be 3, not 2.
Based on the comparison, the first option is the correct LCM.
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