Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

burhan101Best ResponseYou've already chosen the best response.0
\[\huge x^33x^2=4x12\]
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.1
Rearrange as \[x^33x^24x+12 = 0\]Our factors will be of the form \[(x+a)(x+b)(x+c) = 0\] to generate the x^3 term, and because the 12 is positive, we know that either all of the values are positive, or two of them are negative, because those are the only combinations of three numbers that can be multiplied together and yield a positive number. However, because the middle terms of our polynomial are negative, we must have some negative values, so 2 are negative and 1 is positive. What are the prime factors of 12? A bit of experimentation will give you the combination that unlocks the factoring of the polynomial. Once you've factored the polynomial, you'll have (x+a) = 0, (x+b) = 0, (x+c) = 0 and the values of a, b, and c that make those expressions = 0 are the solutions or roots or zeros.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.