Expanding (1/2 - 2x)^5 And Finding Coefficients Of X^2 And X^3

Introduction

This article delves into the expansion of the binomial expression ${(\frac{1}{2} - 2x)^5}$ up to the term in x³. Binomial expansion is a fundamental concept in algebra, particularly useful in calculus, statistics, and other advanced mathematical fields. Understanding how to expand such expressions allows us to manipulate and analyze complex mathematical models more effectively. Our primary focus here will be on systematically expanding the given binomial and identifying the coefficients of the relevant terms. Following the expansion, we will tackle a related problem involving another expression, ${(1 + ax + 3x^2)(\frac{1}{2} - 2x)^5}$. The goal is to find the coefficient of x³ in this new expression, given that the coefficient of x² is ${\frac{13}{2}}$. This part of the problem combines the understanding of binomial expansion with algebraic manipulation to solve for unknown coefficients. This combination of tasks provides a comprehensive exercise in polynomial algebra, showcasing the application of binomial theorem and coefficient determination. Mastering these techniques is crucial for anyone pursuing further studies in mathematics or related fields.

Expanding (1/2 - 2x)^5 up to the Term in x^3

To begin, we will expand ${(\frac{1}{2} - 2x)^5}$ using the binomial theorem. The binomial theorem provides a formula for expanding expressions of the form ${(a + b)^n}$, where n is a non-negative integer. The general formula is:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

where ${\binom{n}{k}}$ represents the binomial coefficient, also known as "n choose k", calculated as:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

In our case, ${a = \frac{1}{2}}$, ${b = -2x}$, and ${n = 5}$. We need to expand the expression up to the term in x³, so we will calculate the terms for k = 0, 1, 2, and 3.

For k = 0:

$$\binom{5}{0} \left(\frac{1}{2}\right)^{5-0} (-2x)^0 = 1 \cdot \left(\frac{1}{2}\right)^5 \cdot 1 = \frac{1}{32}$$

For k = 1:

$$\binom{5}{1} \left(\frac{1}{2}\right)^{5-1} (-2x)^1 = 5 \cdot \left(\frac{1}{2}\right)^4 \cdot (-2x) = 5 \cdot \frac{1}{16} \cdot (-2x) = -\frac{10}{16}x = -\frac{5}{8}x$$

For k = 2:

$$\binom{5}{2} \left(\frac{1}{2}\right)^{5-2} (-2x)^2 = 10 \cdot \left(\frac{1}{2}\right)^3 \cdot (4x^2) = 10 \cdot \frac{1}{8} \cdot 4x^2 = \frac{40}{8}x^2 = 5x^2$$

For k = 3:

$$\binom{5}{3} \left(\frac{1}{2}\right)^{5-3} (-2x)^3 = 10 \cdot \left(\frac{1}{2}\right)^2 \cdot (-8x^3) = 10 \cdot \frac{1}{4} \cdot (-8x^3) = -\frac{80}{4}x^3 = -20x^3$$

Therefore, the expansion of ${(\frac{1}{2} - 2x)^5}$ up to the term in x³ is:

$$\frac{1}{32} - \frac{5}{8}x + 5x^2 - 20x^3 + ...$$

This expansion forms the basis for the next part of the problem, where we will use it to find the coefficient of x³ in a more complex expression. The ability to accurately expand binomial expressions is essential for various mathematical applications, including finding approximations and solving equations involving polynomials.

Finding the Coefficient of x^3 in (1 + ax + 3x^2)(1/2 - 2x)^5

Now, let's move on to the second part of the problem. We are given the expression ${(1 + ax + 3x^2)(\frac{1}{2} - 2x)^5}$ and we know the coefficient of x² in its expansion is ${\frac{13}{2}}$. Our goal is to find the coefficient of x³. To do this, we will use the expansion we derived in the previous section:

$$(\frac{1}{2} - 2x)^5 = \frac{1}{32} - \frac{5}{8}x + 5x^2 - 20x^3 + ...$$

We multiply this expansion by ${(1 + ax + 3x^2)}$:

$$(1 + ax + 3x^2)(\frac{1}{32} - \frac{5}{8}x + 5x^2 - 20x^3 + ...)$$

To find the coefficient of x², we look at the terms that multiply to give x²:

• 1 multiplied by 5x² gives 5x²
• ax multiplied by -\frac{5}{8}x gives -\frac{5}{8}ax²
• 3x² multiplied by \frac{1}{32} gives \frac{3}{32}x²

So the x² term is:

$$\left(5 - \frac{5}{8}a + \frac{3}{32}\right)x^2$$

We are given that the coefficient of x² is ${\frac{13}{2}}$, so:

$$5 - \frac{5}{8}a + \frac{3}{32} = \frac{13}{2}$$

Now we solve for a:

$$\frac{160}{32} - \frac{20}{32}a + \frac{3}{32} = \frac{208}{32}$$

$$160 - 20a + 3 = 208$$

$$-20a = 208 - 163$$

$$-20a = 45$$

$$a = -\frac{45}{20} = -\frac{9}{4}$$

Now that we have found a, we can find the coefficient of x³. We look at the terms that multiply to give x³:

• 1 multiplied by -20x³ gives -20x³
• ax multiplied by 5x² gives 5ax³
• 3x² multiplied by -\frac{5}{8}x gives -\frac{15}{8}x³

Substituting ${a = -\frac{9}{4}}$, the x³ term is:

$$\left(-20 + 5\left(-\frac{9}{4}\right) - \frac{15}{8}\right)x^3$$

$$\left(-20 - \frac{45}{4} - \frac{15}{8}\right)x^3$$

$$\left(-\frac{160}{8} - \frac{90}{8} - \frac{15}{8}\right)x^3$$

$$-\frac{265}{8}x^3$$

Thus, the coefficient of x³ is ${-\frac{265}{8}}$.

This process demonstrates how combining binomial expansion with algebraic manipulation allows us to solve for specific coefficients in polynomial expressions. Understanding these techniques is crucial for advanced mathematical problem-solving and applications in various fields.

Conclusion

In summary, we have successfully expanded the binomial expression ${(\frac{1}{2} - 2x)^5}$ up to the x³ term using the binomial theorem. The expansion yielded the expression ${\frac{1}{32} - \frac{5}{8}x + 5x^2 - 20x^3 + ...}$. We then applied this result to a more complex problem, where we needed to find the coefficient of x³ in the expansion of ${(1 + ax + 3x^2)(\frac{1}{2} - 2x)^5}$, given that the coefficient of x² was ${\frac{13}{2}}$. By carefully equating coefficients and solving for the unknown variable a, we found that ${a = -\frac{9}{4}}$. Subsequently, we calculated the coefficient of x³ to be ${-\frac{265}{8}}$.

These exercises highlight the importance of mastering the binomial theorem and coefficient manipulation in polynomial algebra. The ability to expand binomials and solve for coefficients is fundamental in many areas of mathematics, including calculus, combinatorics, and numerical analysis. Moreover, these skills are highly applicable in various scientific and engineering disciplines where polynomial models are frequently used.

By working through these problems, we have reinforced the understanding of binomial expansion, algebraic manipulation, and problem-solving strategies. This comprehensive approach not only enhances mathematical proficiency but also cultivates analytical thinking, a crucial skill in any field. Continued practice and application of these techniques will further solidify these skills, enabling the tackling of increasingly complex mathematical challenges.