Expand Using the Binomial Theorem (2x^2-5)^2

${(2 x^{2} - 5)}^{2}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(2 x^{2})}^{2 - k} \cdot {(- 5)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(2 x^{2})}^{2 - 0} \cdot {(- 5)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(2 x^{2})}^{2 - 1} \cdot {(- 5)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(2 x^{2})}^{2 - 2} \cdot {(- 5)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x^{2})}^{2} \cdot {(- 5)}^{0} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Step 4

Simplify each term.

Multiply ${(2 x^{2})}^{2}$ by $1$ .

${(2 x^{2})}^{2} \cdot {(- 5)}^{0} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Apply the product rule to $2 x^{2}$ .

$2^{2} {(x^{2})}^{2} \cdot {(- 5)}^{0} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Raise $2$ to the power of $2$ .

$4 {(x^{2})}^{2} \cdot {(- 5)}^{0} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply the exponents in ${(x^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 x^{2 \cdot 2} \cdot {(- 5)}^{0} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply $2$ by $2$ .

$4 x^{4} \cdot {(- 5)}^{0} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Anything raised to $0$ is $1$ .

$4 x^{4} \cdot 1 + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply $4$ by $1$ .

$4 x^{4} + 2 \cdot {(2 x^{2})}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Simplify.

$4 x^{4} + 2 \cdot (2 x^{2}) \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply $2$ by $2$ .

$4 x^{4} + 4 x^{2} \cdot {(- 5)}^{1} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Evaluate the exponent.

$4 x^{4} + 4 x^{2} \cdot - 5 + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply $- 5$ by $4$ .

$4 x^{4} - 20 x^{2} + 1 \cdot {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply ${(2 x^{2})}^{0}$ by $1$ .

$4 x^{4} - 20 x^{2} + {(2 x^{2})}^{0} \cdot {(- 5)}^{2}$

Apply the product rule to $2 x^{2}$ .

$4 x^{4} - 20 x^{2} + 2^{0} {(x^{2})}^{0} \cdot {(- 5)}^{2}$

Anything raised to $0$ is $1$ .

$4 x^{4} - 20 x^{2} + 1 {(x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply ${(x^{2})}^{0}$ by $1$ .

$4 x^{4} - 20 x^{2} + {(x^{2})}^{0} \cdot {(- 5)}^{2}$

Multiply the exponents in ${(x^{2})}^{0}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$4 x^{4} - 20 x^{2} + x^{2 \cdot 0} \cdot {(- 5)}^{2}$

Multiply $2$ by $0$ .

$4 x^{4} - 20 x^{2} + x^{0} \cdot {(- 5)}^{2}$

Anything raised to $0$ is $1$ .

$4 x^{4} - 20 x^{2} + 1 \cdot {(- 5)}^{2}$

Multiply ${(- 5)}^{2}$ by $1$ .

$4 x^{4} - 20 x^{2} + {(- 5)}^{2}$

Raise $- 5$ to the power of $2$ .

$4 x^{4} - 20 x^{2} + 25$

Expand Using the Binomial Theorem (2x^2-5)^2

Related Questions

Support LCMGCF.com by disabling your adblocker.