How do you factor #t^2 + 16t + 60#?

1 Answer
Mar 26, 2018

# (t + 10)(t +6)#

Explanation:

This is simple. No need to worry about.

If we have #ax^2 + bx + c# to factor, we should break #b# to #p + q# such that #pq = ac#.

We have,

#color(white)(xxx)t^2 +16t + 60#

#= t^2 + (10 + 6)t + 60# [As #6 xx 10 = 60#]

#= t^2 + 10t + 6t + 60# [Distributive Property]

#= t(t + 10) + 6(t + 10)# [Grouping like terms]

#= (t + 10)(t +6)# [Grouping Again]

And Done. Hope this helps.