How do you simplify #(2t^2+7t-4)/(-2t^2-5t+3)#?

Question

2984 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T05:33:49

#(t+4)/(-1(t+3))#

Factorize first as described below:

#2t^2+7t-4#

Factor by splitting the middle term

Step-1 : Multiply the coefficient of the first term by the constant # 2 xx -4 = -8#

Step-2 : Find two factors of #-8# whose sum equals the coefficient of the middle term, which is #7#.

#-1 + 8 = 7#

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, #-1# and # 8#

#2t^2-1t+8t-4#

#t(2t-1)+4(2t-1)#

#(2t-1)(t+4)# -----> Factors!

Now lets do the same for #-2t^2-5t+3#

Re-write the equation as #color(red)(-1(2t^2+5t-3))#

Factor by splitting the middle term

Step-1 : Multiply the coefficient of the first term by the constant #2 xx -3 = -6#

Step-2 : Find two factors of #-6# whose sum equals the coefficient of the middle term, which is #5# .

#-1 + 6 = 5#

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step #2# above, #-1# and #6#.

#2t^2-1t+6t-3#

#t(2t-1)+3(2t-1)#

#-1(2t-1)(t+3)# ----- Factors!

So now we get:

#(2t^2+7t-4)/(-2t^2-5t+3)# = #((2t-1)(t+4))/(-1(2t-2)(t+3)#

#((2t-1)(t+4))/(-1(2t-1)(t+3)# = #(cancel(2t-1)(t+4))/(-1cancel(2t-1)(t+3)#

#(t+4)/(-1(t+3))#