# How do you express (x-5)^2 in standard form?

Jan 13, 2017

${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25$

#### Explanation:

The standard form of a polynomial in one variable is a sum of terms in decreasing order of degree.

For brevity, terms with negative coefficients are usually written as subtractions rather than adding the additive inverses. That is, we would write ${x}^{2} - 3 x + 2$ rather than ${x}^{2} + \left(- 3\right) x + 2$

In order to express ${\left(x - 5\right)}^{2}$ in standard form, we need to multiply it out, combining terms of like degree. We can use the FOIL mnemonic to help:

${\left(x - 5\right)}^{2} = \left(x - 5\right) \left(x - 5\right)$

$\textcolor{w h i t e}{{\left(x - 5\right)}^{2}} = {\overbrace{\left(x\right) \left(x\right)}}^{\text{First" + overbrace((x)(-5))^"Outside" + overbrace((-5)(x))^"Inside" + overbrace((-5)(-5))^"Last}}$

$\textcolor{w h i t e}{{\left(x - 5\right)}^{2}} = {x}^{2} - 5 x - 5 x + 25$

$\textcolor{w h i t e}{{\left(x - 5\right)}^{2}} = {x}^{2} - 10 x + 25$

Alternatively, we can recognise a pattern and use it.

For example:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

So we could take the second of these and put $a = x$, $b = 5$ to find:

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

$\textcolor{w h i t e}{{\left(x - 5\right)}^{2}} = {x}^{2} - 10 x + 25$