#(4p^2-1)/(p-4)# subtracted from #(4p+4)/(4-p)#?

Question

rational expressions

720 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-14T06:22:53

#(4p^2+4p+3)/(4-p)#

We need to subtract two fractions.

#(4p+4)/(4-p) - (4p^2-1)/(p-4)" "larr# make the same denominators

#=(4p+4)/(4-p) color(blue)(+) (4p^2-1)/color(blue)(4-p)" "larr( xx-1)#

#=(4p+4+4p^2-1)/(4-p)#

#=(4p^2+4p+3)/(4-p)#