How do you solve #x^2-4x-32>0# using a sign chart?

1 Answer
Nov 12, 2016

Answer:

The answer is #x in] -oo,-4 [ uu ] 8, oo[#

Explanation:

Let's factorise the expression #f(x)=x^2-4x-32=(x-8)(x+4)#

let's do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##8##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+4##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-8##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore, #f(x)>0# when # x in] -oo,-4 [ uu ] 8, oo[#

graph{x^2-4x-32 [-74, 74.05, -37, 37.14]}