A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
3. What is a polynomial function in standard form with zeroes 1, 2, –3, and –3? (1 point)g(x) = x4 + 3x3 –7x2 – 15x + 18
g(x) = x4 + 3x3 –7x2 + 2x + 18
g(x) = x4 – 3x3 + 7x2 + 15x + 18
g(x) = x4 – 3x3 –7x2 + 15x + 18
i think its b
anonymous
 4 years ago
3. What is a polynomial function in standard form with zeroes 1, 2, –3, and –3? (1 point)g(x) = x4 + 3x3 –7x2 – 15x + 18 g(x) = x4 + 3x3 –7x2 + 2x + 18 g(x) = x4 – 3x3 + 7x2 + 15x + 18 g(x) = x4 – 3x3 –7x2 + 15x + 18 i think its b

This Question is Closed

tamtoan
 4 years ago
Best ResponseYou've already chosen the best response.0why don't you write that out and multiply them and see what you get ? ...if equation have the root of a ...then you can write x  a = 0, use this concept to write out your g(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x1)(x2)(x+3)(x3)=(x1)(x2)(x^29)\] is a start then more multiplication is needed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im tring to see if i am right that why i said its is b so am i right yes or no

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i wrote it out wrong. does it really say 3 twice?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is this one \[x^4+3 x^37 x^215 x+18\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can u show me ach step u did
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.