How do you factor: #y= m^2-4m-21 #?

Question

1281 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-10T13:41:20

#y= (m+3)(m-7)#

The things to note is that:
#(+3)xx(-7)=-21#
#-7+3=-4#

#color(blue)("Considering the RHS:")#

There are two condition to give #m^2# and they are #(-m)xx(-m) and (+m)xx(+m)#

Lets try the #+m#

So we have: #(m+?)(m+?) " giving " m^2 + "something"#

Considering the numbers we look at first lets try:
#(m+3)(m-7) -> m^2-7m+3m-21 #

But #-7m+3m = -4m#

#color(blue)("Putting it all together:")#

#y= (m+3)(m-7) =m^2-4m-21#