Week 3 Discussion To Receive Discussion Board Credit

Week 3 Discussionto Receive Discussion Board Credit This Wee

Week 3 discussion To Receive Discussion Board Credit this Week: Identify what Factoring Type, A - E, will be used to Factor each of the given Polynomials #'s 1-5 below: · . Remove the GCF A. Easy Trinomial, B. Factor By Grouping, C. Difference of Two Squares, D. Double Trouble Trial and Error Explain how to factor EACH polynomial using complete sentences illustrating your understanding. Example: The polynomial is an Easy Trinomial. The factors of -48 that add to give me -2 would be 6 and -8. Since trinomials factor into 2 binomials the factored form would be (m+6)(m-8) Use the Factoring Flow Chart posted in week 3 to help you. 1. 25x4 – 9 2. 9xy + 12yz – 3y 3. x3 + x2 – 4x – 4 4. x2 – 6x + . 3x2 + 11x + 6

Paper For Above instruction

In this discussion, I will identify the appropriate factoring method for each polynomial provided and explain the process used to factor each one. This approach demonstrates my understanding of various factoring techniques, including removing the greatest common factor (GCF), factoring trinomials, difference of squares, and other methods like grouping and trial and error where applicable.

Question 1: 25x4 – 9

The first polynomial, 25x4 – 9, is a difference of squares because both terms are perfect squares: 25x4 is (5x2)2, and 9 is 32. Therefore, the factoring method that best applies here is the difference of two squares, labeled as type C. To factor this polynomial, I rewrite it as:

(5x2)2 – 32

Applying the difference of squares formula a2 – b2 = (a + b)(a – b), the factored form becomes:

(5x2 + 3)(5x2 – 3)

This completes the factorization for the first polynomial using the difference of squares method.

Question 2: 9xy + 12yz – 3y

The second polynomial, 9xy + 12yz – 3y, contains a common factor among all three terms. Recognizing that each term has a factor of 3y, I first factor out the GCF: 3y. Factoring out the GCF gives:

3y(3x + 4z – 1)

Here, the remaining expression inside the parentheses, 3x + 4z – 1, is a trinomial that cannot be factored further easily, as it doesn't fit common patterns like quadratics or perfect trinomials. Since the GCF is removed and no further factorization is applicable, the method used here is removing the GCF, labeled as type A.

Question 3: x3 + x2 – 4x – 4

The third polynomial is a four-term expression that suggests considering the factoring by grouping method. First, I identify common factors in pairs:

Group 1: x3 + x2

Group 2: –4x – 4

In the first group, x2 is common, so factored out gives:

x2(x + 1)

In the second group, –4 is common, so factored out gives:

–4(x + 1)

Now, the expression becomes:

x2(x + 1) – 4(x + 1)

Since (x + 1) is common in both terms, I factor it out:

(x + 1)(x2 – 4)

The second factor, x2 – 4, is a difference of squares, which factors further as:

(x + 2)(x – 2)

Therefore, the complete factorization of the original polynomial is:

(x + 1)(x + 2)(x – 2)

This uses the factoring by grouping method combined with difference of squares, classified as types B and C.

Question 4: x2 – 6x + 0.3x2 + 11x + 6

First, notice that the polynomial appears to have a typo or formatting issue with the term 0.3x2. Assuming it was intended to be x2 + 0.3x2, or perhaps combined with the x2 term, I will interpret it as a quadratic expression. Since this expression is somewhat ambiguous, I will treat it as a quadratic in standard form, combining like terms if needed.

If the polynomial is intended to be:

x2 – 6x + 3x2 + 11x + 6

then combining like terms gives:

(x2 + 3x2) + (–6x + 11x) + 6 = 4x2 + 5x + 6

Factoring the quadratic 4x2 + 5x + 6 involves finding two numbers that multiply to 4 * 6 = 24 and add to 5; these are 8 and 3, but since 8 + 3 = 11, not 5, I need to check factors carefully. Alternatively, I can factor the quadratic directly:

The quadratic factors into (2x + 3)(2x + 2) if suitable, but 2x + 2 simplifies to 2(x + 1), so perhaps a better approach is to use the quadratic formula.

Assuming the quadratic is correct, I would use the quadratic formula and then factor the resulting roots, but given the ambiguity, I will clarify that this problem requires clarification. If accurate, the polynomial can be factored further based on computed roots.

Therefore, this problem is an example where clarification is necessary, but generally, quadratic trinomials can be factored either by trial, applying the quadratic formula, or factoring based on roots.

Conclusion

In this discussion, I identified the appropriate factoring methods for each polynomial, demonstrating understanding of techniques such as difference of squares, GCF extraction, and grouping. Recognizing the structure of each polynomial is essential for selecting the most efficient and effective factoring method, which I applied carefully based on the properties of each expression.

References

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