Do not answer, Explain how to do the following.2x^4 + 11x^3 + x^2 – 10x - 4, what are possible rational roots?

Question

I would respect a step be step explanation.
Breaking each step down so a lame person can understand.

'Mathfriend' enlightened me more of how to do the problem.
But I still don't understand. There's no point in anyone giving me the answer to this problem. How exactly do u solve it?

2x^4 + 11x^3 + x^2 – 10x - 4
-what are possible rational roots?
-Actual rational roots?
-Other real roots?
-Divide by (x-r) where r is a root. Repeat with as many roots as you know. Can you find the other roots?

y is x-1 a factor?

Dr. D I want to thank u for explaining. But please note I still had trouble figuring it out. It is now 3am.

I was very specific in my instructions. EXPLAIN IT TO A LAME PERSON!
If x=1, then x-1 is a factor is not an explanation!
Think outside the box when answering a question.
Give detailed, detailed examples!
Are you trying to help me or just getting points?
I figured it out, thank you.

Answer 1

Just look at the equation, and you'll observe that x=1 is a solution. So x-1 is a factor.
Now divide the polynomial by x-1 to get a cubic function.

Then you'll observe that x = -1/2 is a solution, so (2x+1) is also a factor.
Divide again to obtain a quadratic.

Then use the quadratic formula to find the remaining roots.

Answer 2

Write F(x) = 2x^4 + 11x³ + x² - 10x - 4. And be conscious that, via trial, F(a million) = 0 and F(- ½ ) = 0. for this reason it follows, from the rest Theorem, that F(x) has a minimum of two real, rational aspects (x - a million) & (2x + a million). to locate others, divide F(x), via long branch, via the fabricated from (x - a million) & (2x + a million) i.e. via (2x² - x -a million). for that reason we get the the rest factor as (x² + 6x + 4). so as that each and one and all rational aspects of F(x) are (x - a million), (2x + a million) & (x² + 6x + 4). The roots of the quadratic factor x² + 6x + 4 = 0 are - 3 ± ?5 that are no longer rational. subsequently the only obtainable rational roots of F(x) are x = a million and x = - ½. it may now be much less confusing to respond to the the rest areas of the question.

Answer 3

You have written an expression not an equation. Expressions cannot have roots. Did you mean to have the expression equal to zero???????????
If thats so, then:

If you replace x with 1 and work out the value to zero.

By the remainder/ factor theorem this means that x-1 is a factor. Divide the long expression to get
2x^3 + 13x^2 +14x +4 as the other factor.

So we have
(x - 1)(2x^3 + 13x^2 +14x + 4) = 0

Answer 4

f(1) = 2 + 11 + 1 - 10 - 4 = 0
This means that x - 1 is a factor.
To find other factors, use synthetic division:-
1 |2__11__1__-10__-4
""|____2__13__14__4
""|2__13__14___4__0

(x - 1)(2x³ + 13x² + 14x + 4) = 0