Week 6 Unit 7 X5 And 8w4 X2 Find The Greatest Common Factor

1 20w6 U7 X5 And 8w4 X2find The Greatest Common Factor Of These

1) Find the greatest common factor of the following algebraic expressions: 20w^6 u^7 x^5 and 8w^4 x^2.

2) Factor the expression: 30u^6 w^9 – 18u w^3 y^7.

3) Factor the difference of squares: 4 – 49x^2.

4) Factor by grouping: 4v^3 – 5v^2 – 24v + 30.

5) Factor the quadratic trinomial: z^2 + 7z + 12.

6) Factor the quadratic: 2x^2 + 3x – 35.

7) Factor completely: 4w^5 – 6w^4 – 18w^3.

8) Solve for V: V^2 – 3v – 28 = 0. Find all solutions (values of V).

9) Solve for W: 2w^2 – 13w + 16 = (w – 6)^2. Find all solutions (values of W).

10) The area of a rectangle is 28 yd^2. The length of the rectangle is 1 yd more than twice the width. Find the dimensions of the rectangle.

Paper For Above instruction

Introduction

Algebraic expressions and equations are fundamental components of mathematics, providing essential tools for solving real-world problems and understanding mathematical relationships. This paper addresses core algebraic concepts such as factoring, greatest common factors (GCF), quadratic equations, and word problem applications. The goal is to develop strategies to simplify expressions, solve equations accurately, and interpret geometric contexts involving algebraic formulas.

Finding the Greatest Common Factor (GCF)

The first problem involves determining the GCF of two algebraic expressions: 20w^6 u^7 x^5 and 8w^4 x^2. To find the GCF, we analyze the coefficients, variables, and their exponents separately.

For the coefficients, 20 and 8, the GCF is 4. For the variable w, exponents are 6 and 4; thus, the GCF is w^4. For u, only the first expression has u^7, while the second has no u term; therefore, u does not appear in the GCF. For x, exponents are 5 and 2; the GCF is x^2.

Combining these, the GCF of the expressions is 4w^4 x^2, which can be factored out from both expressions to simplify related algebraic work or further problem-solving.

Factoring Algebraic Expressions

The subsequent problems involve various factoring techniques. For example, the second problem requires factoring 30u^6 w^9 – 18u w^3 y^7. First, identify common factors in coefficients and variables. The GCF of coefficients, 30 and 18, is 6. For u, the minimum exponent between 6 and 1 is 1, so u appears as u in the GCF. For w, minimum exponent is 3, so w^3. For y, only the second term has y^7, so y does not factor out globally, but we can restate the expression accordingly after factoring out what is common.

Factoring the expression yields 6u w^3 (5u^5 w^6 – 3 y^7), further simplifying the expression and facilitating solving or analyzing the algebraic structure.

The third problem involves recognizing the difference of squares: 4 – 49x^2. Since 4 and 49x^2 are perfect squares, this factors as (2 – 7x)(2 + 7x). Factoring such binomials is a standard skill in algebra and critical for solving equations involving quadratic forms.

The fourth problem asks for factoring by grouping: 4v^3 – 5v^2 – 24v + 30. Group the terms: (4v^3 – 5v^2) + (–24v + 30). Factor each group: v^2(4v – 5) –6(4v – 5); now, factor out the common binomial (4v – 5), resulting in (4v – 5)(v^2 – 6).

Quadratic Factoring and Solving

Quadratic factorizations are critical for solving many algebraic problems. For example, the quadratic z^2 + 7z + 12 factors as (z + 3)(z + 4), since 3 and 4 sum to 7 and multiply to 12. Similarly, 2x^2 + 3x – 35 can be factored as (2x – 5)(x + 7). These factorizations allow for straightforward solutions through zero-product property in equations.

The problem involving factoring completely 4w^5 – 6w^4 – 18w^3 involves factoring out the greatest common factor first, which is 2w^3, resulting in 2w^3(2w^2 – 3w – 9). Further factoring may involve quadratic techniques or quadratic formula if needed.

Solving Quadratic Equations

For the quadratic equation V^2 – 3v – 28 = 0, apply the quadratic formula: V = [3 ± √(9 + 112)]/2, which yields solutions V = (3 ± √121)/2 = (3 ± 11)/2. Therefore, the solutions are V = (3 + 11)/2 = 7, and V = (3 – 11)/2 = –4.

Similarly, solving 2w^2 – 13w + 16 = (w – 6)^2 involves interpreting the quadratic, rewriting or expanding the right side if necessary, and solving for w. Expanding (w – 6)^2 = w^2 – 12w + 36, the equation becomes 2w^2 – 13w + 16 = w^2 – 12w + 36. Subtract w^2 –12w + 36 from both sides to set the equation to zero and find solutions for w.

Application of Algebra in Geometry

The final problem involves a word problem about a rectangle. Given the area is 28 square yards and the length is 1 yard more than twice the width, we define width as w, then length as 2w + 1. The area equation is:

w(2w + 1) = 28.

This simplifies to 2w^2 + w = 28; rearranged as 2w^2 + w – 28 = 0. Applying quadratic formula or factoring helps find the width(s), then substitute back to find the length.

Using quadratic formula: w = [–1 ± √(1 + 224)] / 4 = [–1 ± √225]/4 = [–1 ± 15]/4. Solutions are w = (–1 + 15)/4 = 14/4 = 3.5, and w = (–1 – 15)/4 = –16/4 = –4. Since width cannot be negative in this context, the feasible solution is w = 3.5 yards. The length is then 2(3.5) + 1 = 8 yards, confirming that the dimensions are approximately 3.5 yards in width and 8 yards in length.

Conclusion

This analysis addressed foundational algebraic concepts and techniques including finding GCF, factoring different forms of quadratic and polynomial expressions, solving quadratic equations, and applying algebra to geometric problems. Mastery of these skills supports advanced problem-solving and understanding of mathematics that is essential for academic success and real-world applications.

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