# How do you find the product of (x + 5)^2?

Mar 20, 2018

${x}^{2} + 10 x + 25$

#### Explanation:

We can use Binomial Expansion to find the product.

The Binomial Theorem States that,

For Any Integer $n > 0$,

${\left(x + y\right)}^{n} =$ (n combination 0)x^n + (n combination 1)x^(n-1)y^1 + .............. + (n combination n-1)x^1y^(n-1) + (n combination n)y^n

So, Here, use can use the formula.

${\left(x + 5\right)}^{2} =$ (2 combination 0)x^2 + (2 combination 1)x^(2-1)5^1 + (2 combination 1)x^1 5^(2-1) + (2 combination 2)5^2

= $1 \times {x}^{2} + 1 \times x \cdot 5 + 1 \times x \cdot 5 + 1 \times {5}^{2}$

[You should learn Permutations And Combinations Prior to this step.]

= ${x}^{2} + 5 x + 5 x + 25$

=${x}^{2} + 10 x + 25$

Hope this helps.

Mar 20, 2018

${x}^{2} + 10 x + 25$

#### Explanation:

$\textcolor{w h i t e}{\times} {\left(x + 5\right)}^{2}$

$= \left(x + 5\right) \left(x + 5\right)$ [Break it up]

$= x \left(x + 5\right) + 5 \left(x + 5\right)$ [Multiply]

$= \left({x}^{2} + 5 x\right) + \left(5 x + 25\right)$ [Distributive Property]

$= {x}^{2} + 5 x + 5 x + 25$

$= {x}^{2} + 10 x + 25$ [Add everything up]

Hence Explained.

Mar 20, 2018

FOIL(First, Outer, Inner, Last)
Answer: ${x}^{2} + 10 x + 25$

#### Explanation:

${\left(x + 5\right)}^{2} = \left(x + 5\right) \left(x + 5\right)$
First- Multiply $x \cdot x$ to get ${x}^{2}$
Outer- $x \cdot 5 = 5 x$
Inner- $5 \cdot x = 5 x$
Last- $5 \cdot 5 = 25$
We know have ${x}^{2} + 5 x + 5 x + 25$
$5 x + 5 x = 10 x$
${x}^{2} + 10 x + 25$