Posted by Elise on Thursday, November 13, 2008 at 4:21pm.
You have two different terms (3x and -5) that both multiply (4x + 5) . You can add these two terms together into (3x - 5)
You are using the "distributive" rule that
(x + y)z = xz + yz
but in this case you are using it "backwards".
x, y and z can be constants or algebraic terms
Related Questions
bobpursley,math - about this problem , I figured it out and i did get the answer...
math,bobpursly, - 4x-2x^2+4x^3 from 4x^3+x-3x^2 4x^3+x-3x^2 -( 4x-2x^2+4x^3) ...
calc - 6(4x+3)^5(4)(3x-5)^5+ 5(3x-5)^4(3)(4X+3)^6 common factors: (4x+3)^5 (3x-5...
8th Grade Algebra: Answer Check #2 - Add or Subtract (Radicals) { this means ...
BobPursley3...another one.. - Problem 3: 4x/(x+2) - 8/(x-1) What I did: Common ...
algebra - simplify (6x^3-16x^2+11x-5)/(3x-2) (3x-2)(2x-1)^2-3 my muliple choice ...
math - how do you factor the polynomial 3x^4-4x^3+4x^2-4x+1? i got (x-1)(3x-1)(...
Math - Can someone please check my answers below for accurateness? The ...
math - sums and differences of rational algebraic expression please answer the ...
algebra - thank everyone that site helped but now i just have to practice. for ...
For Further Reading