Problem 10 - Algebra and Trigonometry

Question

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (x+4)^{3} $$

Step-by-Step Solution

Answer

\( (x+4)^{3} = x^{3}+12x^{2}+48x+64 \)

Step 1: Identify the Parts of the Binomial

The binomial to be expanded is \( (x+4)^{3} \). Here, the binomial is \( (x+4) \), the term \( x \) is 'a', the term \( 4 \) is 'b', and the power of the binomial \( 3 \) is 'n'.

Step 2: Apply the Binomial Theorem

Apply the Binomial Theorem given by \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^{k}\). Collecting terms, the expansion becomes: \(_{3}C_{0}\cdot x^{3} \cdot 4^{0} + _{3}C_{1} \cdot x^{2} \cdot 4^{1} + _{3}C_{2} \cdot x^{1} \cdot 4^{2} + _{3}C_{3}\cdot x^0 \cdot 4^{3}\). The \(_{n}C_{k}\) notation stands for 'n choose k', which is a way of calculating the coefficients in the binomial expansion.

Step 3: Calculate the Terms

Now calculate each term: \(_{3}C_{0} \cdot x^{3} \cdot 4^{0}= 1 \cdot x^{3} \cdot 1 = x^{3}, _{3}C_{1} \cdot x^{2} \cdot 4^{1} = 3 \cdot x^{2} \cdot 4 = 12x^{2}, _{3}C_{2} \cdot x^{1} \cdot 4^{2} = 3\cdot x \cdot 16 = 48x, _{3}C_{3} \cdot x^{0} \cdot 4^{3} = 1 \cdot 1 \cdot 64 = 64. \)

Step 4: Combine the Terms

Finally, combine all the calculated terms to get the expanded form of the ORIGINAL binomial: \((x+4)^{3} = x^{3}+12x^{2}+48x+64.\)