I'm trying to solve the equation: log base 3(x+2) + log base 3(4x-1)=5 and I have it to 0=4x^2+7x-245 and I can't figure out how to factor the trinomial. Can someone please help me?!?!

Question

anonymous · Answer

oh 3^5 = 4x^2+7x-2

anonymous · Answer

because I ended up with: 3^5=4x^2+7x-2 
so that went to 243=4x^2+7x-2
then I subtracted 243 from both sides and ended up with: 0=4x^2+7x-245

anonymous · Answer

one of the terms is (4x+35) this might be of help: http://www.algebrahelp.com/lessons/factoring/trinomial/ http://www.mesacc.edu/~scotz47781/mat120/notes/factoring/trinomials/a_is_not_1/trinomials_a_is_not_1.html

anonymous · Answer

thank you so much! I'll look at them and see if they help :D

anonymous · Answer

I listed the factors of 980 and followed the directions given in that second link to get the following:

28,35:
4x^2+7x-245
4x^2+35x-28x-245
x(4x+35)-7(4x-35)
(4x+35)(x-7)

anonymous · Answer

ok that's what I was having trouble with :P I couldn't find the right factors

anonymous · Answer

x(4x+35)-7(4x+35)* oopsies

anonymous · Answer

tbh I had no idea how to factor things with leading coefficients until just now. You've learned me

anonymous · Answer

awesome both of us :D so could you help me finish out the equation to find what x is?

anonymous · Answer

$$\huge \log _3(x+2) + \log _3(4x-1)=5$$$$\huge \log _3(x+2)(4x-1)=5$$
$$\huge \log _3(4x^2+7x-2)=5$$$$\huge (4x^2+7x-2)=3^5$$

anonymous · Answer

$$\huge (4x^2+7x-2)=243$$$$\huge 4x^2+7x-245=0$$
$$\huge (4x+35)(x-7)=0$$

anonymous · Answer

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