A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Given that 5x^3 + 6x^2 x + 2 = (x+1) (x2) Q(x) + ax + b find the value of a and b.
Question on Identities :)
anonymous
 4 years ago
Given that 5x^3 + 6x^2 x + 2 = (x+1) (x2) Q(x) + ax + b find the value of a and b. Question on Identities :)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Q(x) is an unknown polynomial?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then... a = 1, b = 2?...dont think it should be this simple

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that isnt the answer lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer is a = 20 b = 24

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0o..o maybe you could work out the full solution and see if it matches my answer :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ (x+1)*(x+2)*(5*x9) = 5x^3 + 6x^2 17x  18 + 16x +20=5x^3 + 6x^2 x + 2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oo then how should i carry on

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm a is 16 and b is 20 for me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0check for any careless mistake? :O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you upload the picture of your working?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i didn't do it on my notebook, in head.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x+1)(x+2)(5x−9) +ax + b = 5x^3+6x^2−17x−18+ ax+b\]

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.2i did it also and got same result as ishaan

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need the working :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or rather simple explanation towards the steps

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.2equating coefficients of x gives 1 = 10  27 + a a = 2711 = 16

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use q(x) as px +q, suggested by you earlier if you really need to show working

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can i let Q(x) = px + q will i be able to get the answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh but i cant seem to get the answer somehow :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get decimal numbers :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0show me your working i will try to guide you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well you can get multiple answers, maybe.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(x+1) (x2) *********

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ishaan i edited the question

cwrw238
 4 years ago
Best ResponseYou've already chosen the best response.2i can see how ishaan got p = 5 because of the 5x^3 but the q = 9 is not obvious = 5x^3 + 15x^2 + 10x + qx^2 + 3qx + 2q + ax + b = 5x^3 + (15 + q)x^2 + (10 + 3q + a)x + 2q + b compare coefficients with 5x^3 + 6x^2  x + 2 of x^2: q + 15 = 6 so q = 9 of x: 10 + a + 3q = 10 + a  27 = a  17 and a 17 = 1 so a = 16 of x^0 : 2q + b = 2 so 18 + b = 2 so b = 20 a=16, b = 20
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.