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

1 Answer
Jan 13, 2017

Answer:

#(x-5)^2 = x^2-10x+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^2-3x+2# rather than #x^2+(-3)x+2#

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

#(x-5)^2 = (x-5)(x-5)#

#color(white)((x-5)^2) = overbrace((x)(x))^"First" + overbrace((x)(-5))^"Outside" + overbrace((-5)(x))^"Inside" + overbrace((-5)(-5))^"Last"#

#color(white)((x-5)^2) = x^2-5x-5x+25#

#color(white)((x-5)^2) = x^2-10x+25#

Alternatively, we can recognise a pattern and use it.

For example:

#(a+b)^2 = a^2+2ab+b^2#

#(a-b)^2 = a^2-2ab+b^2#

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

#(x-5)^2 = x^2-2(x)(5)+5^2#

#color(white)((x-5)^2) = x^2-10x+25#