Question 74388
Find the product of (x^2 - 3x + 5) with the quotient of (18x^6 - 27x^5 - 9x^3) ÷ 9x^3. 
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First we'll work on finding the quotient of:
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{{{(18x^6 - 27x^5 - 9x^3)/ 9x^3}}}
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You do that by dividing the denominator into each of the terms in the numerator, so your
answer has three terms.  And the way you do the division is you divide the 9 into the number
in the term and you also subtract the exponents on their exponents of x.  So to divide
{{{9x^3}}} into {{{18x^6}}} you divide the 9 into the 18 to get 2, and then you subtract
the exponents of the x terms to get as the x part of the answer {{{x^(6-3) = x^3}}}.
Putting this together results in the division of the first term by the denominator 
being {{{2x^3}}}
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We use the same process to divide {{{-27x^5}}} by {{{9x^3}}}.  Divide the 9 into the -27 and
get -3.  Then divide the {{{x^3}}} into the {{{x^5}}} and get {{{x^(5-3) = x^2}}}. 
Put these two divisions together and the answer for the second term is {{{-3x^2}}}.
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And finally, divide the {{{9x^3}}} into the third term {{{-9x^3}}}. The 9 into the -9 results
in -1 and the {{{x^3}}} into the {{{x^3}}} results in {{{x^(3-3) = x^0}}} and by definition
anything raised to the 0 power is 1.  So the answer for this division is -1 times 1 or
just -1.
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Putting all three terms together, the quotient is {{{2x^3 -3x^2 -1}}}. Halfway home to being
done.
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Next the problem requires that this quotient be multiplied by {{{x^2 - 3x + 5}}}
You can do
this by picking one term in the quotient and multiplying it by all the terms in {{{x^2 - 3x + 5}}}.
You will get 3 answers.  Then you pick a second term from the quotient and multiply it
by all three terms in {{{x^2 - 3x + 5}}} to get 3 more answers.  Finally, you take the final
term in the quotient and multiply it by all three terms in {{{x^2 - 3x + 5}}} to get the final 
group of 3 answers.  Then you add all 9 answers together, look for common terms that can be 
combined, combine them and what you end up with is the answer to the problem.
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Before we begin this process, let's establish the multiplication ground rules.  Just the opposite
of the division process.  This time we multiply the numbers, not divide them. And we add
the exponents, not subtract them.
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OK, let's go.  From the quotient let's first take the term {{{2x^3}}} and we'll multiply it
by all the terms in {{{x^2 - 3x + 5}}}.  The first multiplication is of {{{2x^3}}} times 
{{{1x^2}}} and the answer is {{{2*1}}} times {{{x^(3+2) = x^5}}} to give us {{{2x^5}}}.
The next multiplication is {{{2x^3}}} times {{{-3x}}} and its answer is {{{2*(-3)}}} 
and {{{x^(3+1) = x^4}}} which gives us a total result of {{{-6x^4}}}. And finally we 
multiply {{{2x^3}}} times {{{+5}}} which results in {{{2*5}}} and {{{x^(3+0) = x^3}}} for
an answer of {{{10x^3}}}.  We now have the first 3 of the 9 multiplication answers and
they are {{{2x^5 - 6x^4 +10x^3}}}.
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Next we take the second term from the quotient.  It is {{{-3x^2}}} and we multiply it 
by all three terms in {{{x^2 - 3x + 5}}}.  Maybe by now you have enough of the idea to see 
that the three answers are {{{-3x^4 + 9x^3 -15x^2}}}. (Multiply the coefficients and add the 
exponents for each term.)
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Finally multiply the last term of the quotient {{{-1}}} and multiply it by each term in
{{{x^2 - 3x + 5}}} and you get {{{-x^2 + 3x - 5}}} for these 3 answers.
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So in one long string the answers are:
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{{{2x^5 - 6x^4 + 10x^3 - 3x^4 + 9x^3 - 15x^2 - x^2 + 3x -5}}}
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You have 2 terms containing {{{x^4}}}. They are {{{-6x^4}}} and {{{-3x^4}}}. Combine them
into {{{-9x^4}}}.  You also can combine the {{{x^3}}} terms. {{{+10x^3 + 9x^3 = 19x^3}}}. And you 
can combine the {{{x^2}}} terms. {{{-15x^2 -x^2 = -16x^2}}}.  None of the other terms can
be combined so the resulting string of terms is:
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{{{2x^5 -9x^4 + 19x^3 - 16x^2 +3x -5}}} and that should be the answer to this problem.
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It's been a long journey.  I hope that this exercise has given you some insight into multiplying
and dividing algebraic terms and how you combine terms that have like powers of x.