Solve: x^3-2x^2-5x+6=0

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Solve: x^3-2x^2-5x+6=0

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Please help!
first thing I do in these situations is check if you get lucky and have +/-1 be a solution...
Answer: x= -2 , x= 1 and x = 3

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Did you try synthetic division?
No, I haven't tried synthetic division because I was unsure how to start it? Do I just pick number I think will work?
The factors of the constant term, which is 6, are +-1, +-2, +-3, +-6. Would you agree?
The coefficient of the x^3 term is 1. So the possible rational roots are the factors of the constant term divided by the factors of 1.
Yes that makes sense. Okay, so now I try all of those numbers?
\[\pm1, \pm2,\pm3, \pm 6\]
i usually start with the small ones.
|dw:1327802691317:dw|
Look at that. We stumbled onto one first thing. That means that one factor is (x-1) And the other factor is x^2-x-6 which can then be factored by the usual methods.
So we have: (x-1)(x^2-x-6)= (x-1)(x-3)(x+2)=0 So x=1, 3, or -2
I'm sorry, but could you explain how we got x^2-x-6? I understand how we found the first one but now I am rather lost
I will repost the synthetic drawing and then you will see.
|dw:1327802958637:dw|
It's the same as when you divide anything. Like 12 divided by 4 = 3 because 3 times 4 = 12
Okay thank you. I understand now. You are a life saver!
S= x^3-2x^2-5x+6 divided by x-1 = x^2-x-6 because (x-1)(x^2-x-6)=x^3-2x^2-5x+6
You are very welcome.

Not the answer you are looking for?

Search for more explanations.

Ask your own question