Question 1136943
<pre>a.) {{{(15a^3-9a+a^2-24)/(6a^2)}}} 
b.) {{{(21x^3y^3-7xy^3-4xy^2-35y)/(-14xy^3)}}} 
c.) {{{(6x^6+32x^4+3x^3-14x^2-8)/(x^2+5)}}}

The first two are DIVISION BY A MONOMIAL because the denominator has just one
term.
The third one is DIVISION BY A POLYNOMIAL because the denominator has more than
one term.

They are done differently.

a.) {{{(15a^3-9a+a^2-24)/(6a^2)}}} 

Put each term of the numerator over the denominator separately:

{{{(15a^3)/(6a^2)-(9a^"")/(6a^2)+(a^2)/(6a^2)-(24^"")/(6a^2)}}} 

Then simplify each fraction as though it were a separate problem

{{{(5a^"")/(2^"")-(3^"")/(2a^"")+(1^"")/(6^"")-(4^"")/(a^2)}}} 



b.) {{{(21x^3y^3-7xy^3-4xy^2-35y)/(-14xy^3)}}} 

Put each term of the numerator over the denominator separately:

{{{(21x^3y^3)/(-14xy^3)-(7xy^3)/(-14xy^3)-(4xy^2)/(-14xy^3)-(35y^"")/(-14xy^3)}}}

Remove the negative signs from the denominator using the rule of
multiplication and division of signs: 

{{{-(21x^3y^3)/(14xy^3)+(7xy^3)/(14xy^3)+(4xy^2)/(14xy^3)+(35y^"")/(14xy^3)}}}

Then simplify each fraction as though it were a separate problem:

{{{-(3x^2)/(2^"")+(1^"")/(2^"")+(2^"")/(7y^"")+(5^"")/(2xy^2)}}}



c.) {{{(6x^6+32x^4+3x^3-14x^2-8)/(x^2+5)}}}

Here the denominator contains more than one term, so it is done by long
division. Be sure to insert 0 times any power of x that's missing:

         <u>         6x⁴+0x³+ 2x²+ 3x¹- 24</u>
x²+0x¹+5)6x⁶+0x⁵+32x⁴+3x³-14x²+ 0x¹-  8
         <u>6x⁶+0x⁵+30x⁴</u>
             0x⁵+ 2x⁴+3x³
             <u>0x⁵+ 0x⁴+0x³</u>
                  2x⁴+3x³-14x²
                  <u>2x⁴+0x³+10x²</u>
                      3x³-24x²+ 0x¹
                      <u>3x³+ 0x²+15x¹</u>   
                         -24x²-15x¹-  8
                         <u>-24x²+ 0x¹-120</u>
                              -15x¹+112 


Answer: {{{6x^4+2x^2+3x-24+(-15x+112^"")/(x^2+5)}}}


Edwin</pre>