Question

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

Answer 1

((х+2)²)²+((х-2)²)²=(((х²+2х)+1)²+((х²-2х)+1)²)=х⁴+4х³+4х²+2х²+4х+1+

х⁴-4х³+4х²+2х²-4х+1=2х⁴+12х²+2=2*(х⁴+6х²+1)

Применил формулу квадрата суммы и разности

(а±в)²=а²±2ав+в²

Answer 2

To expand the expression $(x + 2)^4 + (x - 2)^4$, you can use the binomial theorem or the properties of exponents. I'll use the binomial theorem to expand it step by step:

The binomial theorem states that for any real numbers $a$ and $b$ and a non-negative integer $n$, the expansion of $(a + b)^n$ is given by:

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

where $\binom{n}{k}$ is the binomial coefficient given by $\frac{n!}{k!(n-k)!}$.

For the given expression, $a = x$ and $b = 2$, and we need to expand it to the 4th power. Let's do the calculations step by step:

$(x + 2)^4 + (x - 2)^4$

Using the binomial theorem for each term:

$(x + 2)^4 = \binom{4}{0} x^{4} 2^{0} + \binom{4}{1} x^{3} 2^{1} + \binom{4}{2} x^{2} 2^{2} + \binom{4}{3} x^{1} 2^{3} + \binom{4}{4} x^{0} 2^{4}$

Now, let's calculate the binomial coefficients:

$\binom{4}{0} = 1$,

$\binom{4}{1} = 4$,

$\binom{4}{2} = 6$,

$\binom{4}{3} = 4$,

$\binom{4}{4} = 1$.

Plugging the coefficients back in:

$= x^{4} + 4x^{3} + 6x^{2} + 4x + 1$.

Now, let's calculate $(x - 2)^4$ using the binomial theorem again:

$(x - 2)^4 = \binom{4}{0} x^{4} (-2)^{0} + \binom{4}{1} x^{3} (-2)^{1} + \binom{4}{2} x^{2} (-2)^{2} + \binom{4}{3} x^{1} (-2)^{3} + \binom{4}{4} x^{0} (-2)^{4}$

Calculate the binomial coefficients:

$\binom{4}{0} = 1$,

$\binom{4}{1} = 4$,

$\binom{4}{2} = 6$,

$\binom{4}{3} = 4$,

$\binom{4}{4} = 1$.

Plugging the coefficients back in:

$= x^{4} - 4x^{3} + 6x^{2} - 4x + 1$ 0 0