<PRE><B>Question 1170466</B>
&lt;br&gt;
Once again tutor @MathLover1 has shown a method for factoring polynomials without showing HOW the factorization is done.  It looks like magic and thus teaches the student nothing.&lt;br&gt;
Perhaps one day she will provide a response to a similar question in which she explains how her method is actually done....&lt;br&gt;
I will solve these three similar problems by two very different methods.  The first is a formal process which is what will be taught in most resources; the second uses some powerful mathematical ideas but also involves some playing with numbers.&lt;br&gt;
Here is the general process for the standard formal method:
(1) Use the rational roots theorem to find the possible rational roots;
(2) use substitution and/or synthetic division to find the actual roots;
(3) when the remaining polynomial has been reduced to second degree, solve by factoring, or by using the quadratic formula.&lt;br&gt;
example 1: {{{x^3-2x^2-5x+6 = 0}}}&lt;br&gt;
The possible rational roots are (plus or minus) 1, 2, 3, and 6.&lt;br&gt;
It&#39;s always easy to check roots 1 and -1 by substituting.  In this case, x=1 satisfies the equation, so 1 is a root and (x-1) is a factor of the polynomial.  Remove that factor using synthetic division.&lt;br&gt;&lt;pre&gt;

   1 | 1  -2  -5   6
     |     1  -1  -6
     ----------------
       1  -1  -6   0&lt;/pre&gt;
So&lt;br&gt;
{{{x^3-2x^2-5x+6 = (x-1)(x^2-x-6)}}}&lt;br&gt;
Then factoring the quadratic gives us the final factored form of the equation:&lt;br&gt;
{{{x^3-2x^2-5x+6 = 0}}}
{{{(x-1)(x-3)(x+2) = 0}}}&lt;br&gt;
example 2: {{{x^3-x^2-10x-8 = 0}}}&lt;br&gt;
The possible rational roots are (plus or minus) 1, 2, 4, and 8.&lt;br&gt;
In this case, direct substitution shows that x=1 does not satisfy the equation, but x=-1 does.  So again, with x=-1 a root, remove the factor (x+1) using synthetic division.&lt;br&gt;&lt;pre&gt;

  -1 | 1  -1  -10  -8
     |    -1    2   8
     ----------------
       1  -2   -8   0&lt;/pre&gt;
So&lt;br&gt;
{{{x^3-x^2-10x-8 = (x+1)(x^2-2x-8)}}}&lt;br&gt;
And factoring the quadratic gives us the final factored form of the equation:&lt;br&gt;
{{{x^3-x^2-10x-8 = 0}}}
{{{(x+1)(x-4)(x+2) = 0}}}&lt;br&gt;
example 3: {{{2x^4+5x^3-5x-2 = 0}}}&lt;br&gt;
we are given that x=1 is a root, so remove the factor (x-1) using synthetic division.&lt;br&gt;&lt;pre&gt;

   1 | 2   5   0  -5  -2
     |     2   7   7   2
     -------------------
       2   7   7   2   0&lt;/pre&gt;
So&lt;br&gt;
{{{2x^4+5x^3-5x-2 = (x-1)(2x^3+7x^2+7x+2)}}}&lt;br&gt;
The possible rational roots are now (plus or minus) 1, 2, 1/2.&lt;br&gt;
Direct substitution shows that x=-1 is a root, so remove the factor of (x+1) using synthetic division.&lt;br&gt;&lt;pre&gt;

  -1 | 2  7  7  2
     |   -2 -5 -2
     ------------
       2  5  2  0&lt;/pre&gt;
So&lt;br&gt;
{{{2x^4+5x^3-5x-2 = (x-1)(x+1)(2x^2+5x+2)}}}&lt;br&gt;
Factoring the quadratic then gives us the final factored form of the equation:&lt;br&gt;
{{{2x^4+5x^3-5x-2 = (x-1)(x+1)(2x+1)(x+2) = 0}}}&lt;br&gt;
There are the solutions of the three examples using the standard process.&lt;br&gt;
Now here are solutions using Vieta&#39;s Theorem, which tells us that, in a cubic polynomial equation&lt;br&gt;
{{{ax^3+bx^2+cx+d=0}}}&lt;br&gt;
The sum of the three roots is -b/a and the product of the three roots is -d/a.&lt;br&gt;
We can use this along with the rational roots theorem to play around with the possible rational roots to find the factored forms of the three example equations.&lt;br&gt;
example 1: {{{x^3-2x^2-5x+6 = 0}}}&lt;br&gt; 
The possible rational roots are (plus or minus) 1, 2, 3, and 6.&lt;br&gt;
The sum of the three roots is 2 and the product is -6.&lt;br&gt;
Playing around with the possible rational roots we can find the roots are 1, -2, and 3: (1)+(-2)+(3) = 2; (1)(-2)(3) = -6.  Then with those roots the factored form of the equation is (as before,of course!)&lt;br&gt;
{{{(x-1)(x+2)(x-3) = 0}}}&lt;br&gt;
example 2: {{{x^3-x^2-10x-8 = 0}}}&lt;br&gt;
The possible rational roots are (plus or minus) 1, 2, 4, and 8.&lt;br&gt;
The sum of the roots is 1 and the product of the three roots is 8.  Again playing around with the possible rational roots, we can find the roots are -1, -2, and 4: (-1)+(-2)+(4) = 1; (-1)(-2)(4) = 8.  Then with those roots the factored form, agreeing with the result above, is&lt;br&gt;
{{{x^3-x^2-10x-8 = (x+1)(x+2)(x-4) = 0}}}&lt;br&gt;
example 3: {{{2x^4+5x^3-5x-2 = 0}}}&lt;br&gt;
For this one, we will not repeat the process of removing the (x-1) factor corresponding to the given root of x=1.  We will start with the cubic polynomial equation&lt;br&gt;
{{{(2x^3+7x^2+7x+2) = 0}}}&lt;br&gt;
The possible rational roots are again (plus or minus) 1, 2, 1/2.&lt;br&gt;
The sum of the three roots is -7/2 and the product of the three roots is -1.  This one is harder because of the fractions; but again playing around with the possible rational roots gives us the roots -1, -2, and -1/2.  And that again gives us a factored form of the equation that agrees with our earlier result:&lt;br&gt;
{{{(x-1)(x+1)(x+2)(2x+1) = 0}}}&lt;br&gt;
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