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Pofe, J. (2017, September 18). YouTube. FACTORIZACIÓN POR EVALUACIÓN (USANDO DIVISIÓN SINTÉTICA) - Ejercicio 1 - YouTube. Retrieved October 19, 2017, from http://www.youtube.com/watch?v=j5AfFpeUBEs

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Author: Rubén Darío Fuentes Zavala: A01067142
Francisco Guzmán: A01067139
Andrés Bolaños: A01067143
Miranda Hernández:
A01067151
Andrea Ruíz:
A01067186
Claudio Anguiano:
A01067196
Juan Luis Díaz:
A01067196
Emiliana Rodríguez: A01067257
André Moreno:
A01067

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By factoring the cuadrinomial with the evaluation method we got the terms:
(X - 5), (X - 2) AND (X - 2)

Now, at last, you have to join them with the common term of step 1 that is "X^3", and what you got is:

X^3 (X - 2)^2 (X - 5)

And with that, you have completely factorized the polynomial that you were given.

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Now let's on the final process of factorization by the  "(X^2 + BX + C)" form:

1. Write a parenthesis:
X^2 - 7X + 10 = (       )(        )

2. Spare the "X":
X^2 - 7X + 10 = (X     )(X     )

3. Apply the sings:
X^2 - 7X + 10 = (X -   )(X -   )

4. Look for 2 negative numbers that multiplied give "+10" and added give "-7":
X^2 - 7X + 10 = (X-5)(X-2)

Whit that, the trinomial is factorized.

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Now you are getting close to the end of the process, because you got a trinomial that can be factorized by the form:
"(X^2 + BX + C)"

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Finally you got the residual of 0 that you were looking for, and correctly organizing it, your result would be:

X^2 - 7X + 10

Also, you conclude that "X=2" produce the factor
"(X - 2)" by passing the "2" to the other side of the equal and you are going to use that term as a factor of the polynomial next.

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Imagine that you kept trying with "X=-1", but you discarded it because you got a residual of "-54". Then you are trying with "X=2", and this is what you got:

X=2

X^3    X^2       X             IT
1         -9         24         -20
1          2        -14          20
-------------------------------------------
1         -7         10           0

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Then, you have to keep testing all the divisors that you got at step 2

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Then, you should start testing the numbers one by one in the synthetic division. For example, let's start by "X=1"

X=1

X^3    X^2       X             IT
1         -9         24         -20
1          1         -8           16
-------------------------------------------
1         -8         16          -4

As we saw, the "1" wasn't useful, because we got a residual of "-4"

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Now you start with your process of synthetic division by writing the terms one by one from the major exponent to the independent term, that in the case of the example that we have would be:

X^3    X^2       X             IT
1         -9         24         -20

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Now is where the "tedious" part arrives. What you have to do is try whit the divisors that you obtained, which one can be used in a synthetic division having a residual of 0.

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Having the expression of the first step, now you can start factorizing by synthetic division. What you should do next is list the divisors of 20, that are:

D20 (+-1, +-2, +-4, +-5, +-10, +-20)

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At first what you should do is to use the common factor, that in this case, as we can see, is the "X^3", so the correct expression would be:

X^3 (X^3 - 9X^2 + 24X - 20)

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Imagine that you need to factorize the next polynomial:

X^6 - 9X^5 + 24X^4 - 20X^3

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This factorization is useful after performing a factorization by common factor