Question 83397
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Here are the 1st and 3rd.
I haven't figured out the 2nd one yet.  
Is that the way it was given?  No errors?
Is it supposed to be a division?
You were asked to "expand a quotient"?  
That doesn't seem right.

Anyway here are the first and third ones:

Q1 (i) Factor the expression                          

{{{4y^2-12xy-8yz+9x^2+12xz+4z^2}}}                              

Rearrange the terms as

{{{9x^2-12xy+4y^2}}} + {{{12xz - 8yz}}} + {{{4z^2}}}

The first three terms factor as {{{(3x-2y)(3x-2y)}}} or {{{(3-2x)^2}}}

We can take {{{4z}}} out of the 
3rd and 4th terms and we now have

{{{(3x-2y)^2 + 4z(3x-2y) + 4z²}}}

Now let {{{w = (3x-2y)}}}

Then the above becomes

{{{w^2 +4zw + 4z^2}}}

which factors as

{{{(w + 2z)(w+2z)}}} or

{{{(w + 2z)^2}}}

Now replace the w by (3x-2y)

{{{(3x-2y+2z)^2}}}

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(iii) Solve the equation:
      
{{{x^5-2x^4+x^3- 2x^2 - 2x + 4 = 0}}} 

We notice that the coefficients of the
1st, 3rd, and 5th terms are 1,1,-2, and that
the coefficients of the 2nd, 4th, and 6th
terms are -2,-2,4 and that these are
proportional, so we rearrange the terms:

{{{x^5+x^3-2x-2x^2-2x+4=0}}}

Factor x out of the first 3 terms and -2
out of the last 3 terms:

{{{x(x^4+x^2-2)-2(x^4+x^2-2)=0}}} 

Now we can take out the common factor 
{{{(x^4+x^2-2)}}} and we have:

{{{(x^4+x^2-2)(x-2)=0}}}

Now we can factor the first parenthetical 
expression:

{{{(x^2+2)(x^2-1)(x-2)=0}}}

and finally factor the expression in 
the middle parentheses as the difference 
of two squares:

{{{(x^2+2)(x-1)(x+1)(x-2)=0}}}

Now we set each factor = 0:

Setting first factor = 0

{{{x^2+2=0}}}
{{{x^2=-2}}}

Use the principle of square roots  

    x = ±{{{sqrt(-2)}}}
    x = ±{{{i*sqrt(2)}}}

Setting second factor = 0
    x - 1 = 0 gives x = 1

Setting third factor = 0
    x + 1 = 0 gives x = -1

Setting fourth factor = 0
    x - 2 = 0 gives x = 2

So the 5 solutions are ±i{{{sqrt(2)}}},1,-1, and 2  

Edwin</pre>