Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Can someone help me create a problem that has a binomial which IS a factor of the polynomial being divided?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
If 2(3)=6 what is 6/3?
2.
Is 3 a factor of 6?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Yes.
So perhaps we could use that idea and apply it to binomials so we end up with a polynomial that has a binomial factor.
Make up a binomial and post it.
y = 2x + 10
Make up one which is not factorable.
See why you can't get help. I have been trying to help you for more than 2 hours and you have disappeared yet again.
Sorry, I keep forgetting to check back into Openstudy.
A non factorable binomial could be \[v^{2} + 16\]
Make up a second binomial
\[(x - 1)^{2}\]
That is a binomial squared. What is the matter with something simple like x+2 or something?
Okay, x + 2
Multiply your two binomials.
Use the same variable in both. Lose the v and put x
Could you hold on please, I have to make a short phone call?
I'm outta here.

Not the answer you are looking for?

Search for more explanations.

Ask your own question