Question

expand the function.
f(x) = (2x + 2)^3
?x^3 + x^2 + x +

Explanation:

Step1: Apply binomial expansion formula

The binomial expansion formula is \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\). Here \(a = 2x\), \(b=2\) and \(n = 3\). So \((2x + 2)^3=\binom{3}{0}(2x)^{3}(2)^{0}+\binom{3}{1}(2x)^{2}(2)^{1}+\binom{3}{2}(2x)^{1}(2)^{2}+\binom{3}{3}(2x)^{0}(2)^{3}\).

Step2: Calculate binomial - coefficients

We know that \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\). So \(\binom{3}{0}=\frac{3!}{0!(3 - 0)!}=1\), \(\binom{3}{1}=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\), \(\binom{3}{2}=\frac{3!}{2!(3 - 2)!}=3\) and \(\binom{3}{3}=\frac{3!}{3!(3 - 3)!}=1\).

Step3: Simplify each term

• For the first - term: \(\binom{3}{0}(2x)^{3}(2)^{0}=1\times(2x)^{3}\times1 = 8x^{3}\).
• For the second - term: \(\binom{3}{1}(2x)^{2}(2)^{1}=3\times4x^{2}\times2=24x^{2}\).
• For the third - term: \(\binom{3}{2}(2x)^{1}(2)^{2}=3\times2x\times4 = 24x\).
• For the fourth - term: \(\binom{3}{3}(2x)^{0}(2)^{3}=1\times1\times8 = 8\).

Answer:

\(8x^{3}+24x^{2}+24x + 8\)