A community for students.
Here's the question you clicked on:
 0 viewing
burhan101
 3 years ago
Solve :
burhan101
 3 years ago
Solve :

This Question is Closed

burhan101
 3 years ago
Best ResponseYou've already chosen the best response.0\[\huge x^33x^2=4x12\]

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Rearrange 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
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.