Question 1140166
<pre>1.   Factor:  {{{x^4-5x^3+5x^2+5x-6}}}

All potential roots or zeros are ± the factors of 6, which are ±1,±2,±3,±6

We try the easiest one first, which is 1 with synthetic division,  We are
dividing by x-1

1 | 1 -5  5  5 -6
  |<u>    1 -4  1  6</u>
    1 -4  1  6  0

Luckily that gave 0 remainder, telling us that 1 is a zero or root and that
x-1 is a factor,  So the first partial factorization using (x-1) and the bottom
numbers is:

(x-1)(x³-4x²+x+6)

Next we factor x³-4x²+x+6

All potential roots or zeros are again ± the factors of 6, which are ±1,±2,±3,±6

We try the easiest one first again, which is 1, with synthetic division,  We are
again dividing by x-1

1 | 1 -4  1  6
  |<u>    1 -3 -2</u>
    1 -3 -2  4

That does not give 0 remainder, so we try the next easiest one, which is -1,
with synthetic division,  We are dividing this time by x+1

-1 | 1 -4  1  6
   |<u>   -1  5 -6</u>
     1 -5  6  0

Luckily that gave 0 remainder, telling us that -1 is a zero or root and that
x+1 is a factor,  So the next partial factorization using (x+1) and the bottom
numbers is:

(x-1)(x+1)(x²-5x+6)

Now we can complete the factorization without synthetic division:

(x-1)(x+1)(x-2)(x-3)

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Factor:  {{{x^9-y^9}}}, we rewrite as {{{(x^3)^3+(y^3)^3}}} and
then we can use the formula: {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}
with A=x³ and B=y³

{{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}
{{{(x^3)^3-(y^3)^3=((x^3)-(y^3)^"")((x^3)^2+(x^3)(y^3)+(y^3)^2)}}}
{{{x^9-y^9=(x^3-y^3)(x^6+x^3y^3+y^6)}}}

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3.solve the equation {{{x^4-5x^3+5x^2+5x-6=0}}}

We have already factored the left side in the first problem, so
we have:

(x-1)(x+1)(x-2)(x-3) = 0

We set each factor equal to 0:

x-1=0;  x+1=0;  x-2=0;  x-3=0
  x=1;    x=-1;   x=2;    x=3

Those are the four solutions.

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4.  {{{7e^x - 3 = 2e^(2x)}}}

Let e<sup>x</sup> = u, then e<sup>2x</sup> = u<sup>2</sup>

    {{{7u-3=2u^2}}}
    {{{0=2u^2-7u+3}}}
    {{{2u^2-7u+3=0}}}
    {{{(2u-1)(u-3)=0}}}
    2u-1=0;   u-3=0
      2u=1      u=3
       u={{{1/2}}}

Substitute e<sup>x</sup> for u

{{{e^x=1/2}}};   {{{e^x=3}}}
{{{x=ln(1/2)}}};  {{{x=ln(3)}}}
{{{x=ln(2^(-1))}}};
{{{x=-ln(2)}}}

Here ln() means natural logarithm.

Edwin</pre>