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help with simple algebra

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You can use rational root theorem to find 1 root and then you can use it to factor out the polynomial and then solve for other root
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Factoring polynomials of degree 3 or higher is significantly harder than factoring quadratics. Short of applying (the famously complicated) cubic formula, your best bet for cubics is to either:

* Spot a simply way to break it down to allow obvious factoring, OR
* Apply the rational root theorem and *hope* it has rational roots.

# By Inspection

If you're really lucky, you'll spot this early. But I wouldn't rely on it at all. Because of how factors multiply out, if this polynomial has "nice" linear factors (a big if) then one of them will have a leading coefficient of 2 and the others will have coefficients of 1. With that in mind, we can try to 'chunk' our coefficients into sums/differences of a multiple of 2 and another number. So, let's look at our coefficients:

* 2x^(3) \+ 3x^(2) \+ 0x - 1
* (2)x^(3) \+ (4 - 1)x^(2) \+ (2 - 2)x + (-1)
* (2x^(3) \+ 4x^(2) \+ 2x) - (x^(2) \+ 2x + 1)
* 2x(x^(2) \+ 2x + 1) - (x^(2) \+ 2x + 1)
* (2x - 1)(x^(2) \+ 2x + 1)
* (2x-1)(x+1)^(2)

I think that's really hard to spot and think of. And oftentimes the polynomial just won't factor nicely. So you're better off learning the next method, but sometimes you can grab a quick win by just playing with the coefficients a bit.

# Rational Roots

If a polynomial with integer coefficients has rational roots then the numerator of each root will divide the constant term and the denominators will divide the leading coefficient. So, in our case, we really don't have much to check. The rational roots can only have numerators +1 or -1 (as these are the only integers that divide -1) and they can only have denominators +2, +1 (the negatives also divide 2 but seeing as we're talking about fractions here, we've already accounted for those by including negative numerators). That's it. We don't have too much to check. There are only really 4 possible rational roots it could have.

Also note that if a rational root of a polynomial with integer coefficients is written p/q then it corresponds to a linear factor (qx - p). Now, let's see if we can verify a rational root:

1. If (1/1) is a rational root then p(1) = 0. But 2(1^(3)) + 3(1^(2)) - 1 = 4, so that's not a root.
2. If (1/2) is a rational root then p(1/2) = 0. 2(1/8) + 3(1/4) - 1 = 1/4 + 3/4 - 1 = 0. So 1/2 is a root! That corresponds to a factor of (2x-1), so let's try to factor that out:

We'll factor out the (2x-1) by equating coefficients. We get:

>2x^(3) \+ 3x^(2) \+ 0x - 1 = (2x-1)(ax^(2) \+ bx + c)  
>  
>2x^(3) \+ 3x^(2) \+ 0x - 1 = (2ax^(3) \+ (2b-a)x^(2) \+ (2c-b) - c

So, immediately we see:

1. 2 = 2a, so a = 1.
2. 2b - a = 3, so b = 2.
3. \-1 = -c, so c = 1.

Hence:

>2x^(3) \+ 3x^(2) \- 1 = (2x-1)(x^(2) \+ 2x + 1)

We could try to continue to search for roots like we did, and we'd find them, but the remaining factor is a quadratic (which we can factorize more easily than a cubic) and also it's very clearly a perfect square. So the final step is:

>2x^(3) \+ 3x^(2) \- 1 = (2x-1)(x+1)^(2)

QED.
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Learn about Horner's method, it has helped me so much in school and uni for finding the root of a polynomial of a higher degree than 2. Just keep in mind if the free coefficient(not sure what is the term in English, but I'm talking about the last number in the current equation) is too big it might not be a good idea to use it.
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You can use a method called Horner not sure if it's actually called that but that's how I learned it in school.

So first find a number that when put in makes the function equal 0 example: -1

Because: `2(-1)\^3+3(-1)\^2-1 = 0`

Now you can create a little table using the coefficients of the polynomial example:

|   | 2 | 3 | 0 | -1|
|---|---|---|---|---|
|   |   |   |   |   |
| -1|   |   |   |   |
|   |   |   |   |   |

The 0 is there because there is `0 * x` in the polynomial (`2x^3 + 3x^2 + 0x - 1`). Now all we do is drop down the 2 multiply it by -1 and add it to 3 and keep doing the same so we end up with a table that looks like this:

|   | 2 | 3 | 0 | -1|
|---|---|---|---|---|
|   |   |   |   |   |
| -1|   | -2| -1| 1 |
|   | 2 | 1 | -1| 0 |

Now all we do is take these numbers and use them as a coëfficient for our new polynomial like so: `2x^2+x-1`

Then we may not forget the -1 and we multiply the polynomial we just got with `(x - (a))` where a in this case is the `-1` so with `(x + 1)`

That leaves us with the equation: `(x + 1)(2x^2 +x -1)` and now it's easy to find the roots of this function because when saying that `(x + 1)(2x^2 +x -1) = 0` we may say that either `(x + 1) = 0` or `(2x^2 +x -1) = 0`

So for `(x + 1) = 0 <=> x = -1`
And for `(2x^2 +x -1) = 0` we can use the formule `x = -b +- root(D)/2a`
where `D = b² - 4*a*c` so `D = 1 + 8 = 9`
So we can say that `x = -1 +- 3 / 4` or that `x = -1` or `x = 1/2`

Now we take these 3 roots (since -1 is a root of both factors we are left with 2 roots) and conclude that the roots of this polynomial are:
-1 and +1/2

If you don't fully understand the method don't be afraid for a better explanation
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What is meant by roots idk
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LOTS of help here (congrats to all who spent time explaining their process!!!).  One further comment, from the teacher perspective.  I would ask this type of question to see if the student could use the rational root theorem, followed by division and factoring a trinomial.  It is a relatively easy function to find a root since P(-1) = 0 by inspection.  If I couldn’t find the root, I would go back and make sure I copied down correctly, or decide that the teacher made a mistake.  (Teachers do, and we rarely give ones that can’t be factored, in my experience).
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Thank you i did it very simmilar i looked up some video on youtube called synthetic divison i think its pretty much the same method
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What is meant be roots
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