Apply (a+b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ for n=4.

Expand (x + 2)⁴ Using Binomial Theorem

Question

Expand $(x + 2)^4$ using the Binomial Theorem.

Solution — Step by Step

For any $(a + b)^n$ , the expansion is:

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

Here, $a = x$ , $b = 2$ , and $n = 4$ . We’ll get $n + 1 = 5$ terms total.

Each term is $\binom{4}{r} x^{4-r} \cdot 2^r$ for $r = 0, 1, 2, 3, 4$ .

T_{r+1} = \binom{4}{r} x^{4-r} \cdot 2^r

List them out:

$r$	$\binom{4}{r}$	$x^{4-r}$	$2^r$
0	1	$x^4$	1
1	4	$x^3$	2
2	6	$x^2$	4
3	4	$x$	8
4	1	1	16

$r=0$ : $1 \cdot x^4 \cdot 1 = x^4$
$r=1$ : $4 \cdot x^3 \cdot 2 = 8x^3$
$r=2$ : $6 \cdot x^2 \cdot 4 = 24x^2$
$r=3$ : $4 \cdot x \cdot 8 = 32x$
$r=4$ : $1 \cdot 1 \cdot 16 = 16$

(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Final answer: $x^4 + 8x^3 + 24x^2 + 32x + 16$

Why This Works

The Binomial Theorem is essentially a structured way to count how many times each combination of $a$ ‘s and $b$ ‘s appears when you multiply $(a+b)$ by itself $n$ times. The coefficient $\binom{n}{r}$ counts the number of ways to pick $r$ copies of $b$ from $n$ brackets — that’s exactly why Pascal’s triangle emerges from these coefficients.

The $2^r$ factor is what many students forget to compute carefully. Every time $b = 2$ appears in a term, it contributes a factor of 2, and for the $r$ -th term, $b$ appears $r$ times, giving $2^r$ .

This structure — alternating between choosing how many $b$ ‘s and what power of $a$ remains — is what makes the theorem so predictable. Once you know $n$ , you can write any term directly without expanding step by step.

Alternative Method

We can expand by repeated multiplication. Start with $(x+2)^2$ , then square the result.

(x+2)^2 = x^2 + 4x + 4

Now $(x+2)^4 = \left[(x+2)^2\right]^2 = (x^2 + 4x + 4)^2$ .

Expanding $(x^2 + 4x + 4)^2$ :

= (x^2)^2 + 2(x^2)(4x) + 2(x^2)(4) + (4x)^2 + 2(4x)(4) + 4^2

= x^4 + 8x^3 + 8x^2 + 16x^2 + 32x + 16

= x^4 + 8x^3 + 24x^2 + 32x + 16

This “square the square” trick is useful for $n = 4$ and $n = 6$ when you want a quick check. For odd powers like $n = 5$ , you’d need $(x+2)^2 \cdot (x+2)^3$ , which is more work — stick to the formula there.

Common Mistake

The most common error here is forgetting to raise $b = 2$ to the power $r$ . Students write the $\binom{4}{r}$ correctly, but treat each term as just $\binom{4}{r} x^{4-r} \cdot 2$ instead of $\binom{4}{r} x^{4-r} \cdot 2^r$ . This gives wrong coefficients from $r = 2$ onward.

For example, the $r = 2$ term becomes $6 \cdot x^2 \cdot 2 = 12x^2$ instead of the correct $6 \cdot x^2 \cdot 4 = 24x^2$ . Always write $b^r$ explicitly in each term before multiplying.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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