Question

(5x^3-8x^2+9x+12)/(x-3) please answer with sentences on how to do it step by step. thanks!

Answer 1

Answer:

5x^2 + 7x + 30 + 102/x-3

Step-by-step explanation:

     _________________

x-3| 5x^3 - 8x^2 + 9x + 12      

In case you would like a more visual representative with simple explanations.

First you would want to set it up as either long polynomial division or using synthetic division. Whichever one is easier for you. However, I am going to use long polynomial division since you might not be familiar with synthetic division just yet.

    _5x^2_____________

x-3|  5x^3 - 8x^2 + 9x + 12        Because 5x^3 divided by x gives you 5x^2

     - (5x^3 - 15x^2)______       and -3 times 5x^2 gives you -15x^2

              0 + 7x^2 +9x                But since the negative is outside the   parentheses then you distribute it turning he -15x^2 it into +15x then adding it to the -8x^2 above it thus giving you 7x^2. Afterwards bring down the 9x.

A negative times a negative is a positive (just as a reminder)

Next,

    _5x^2+7x_________

x-3| 5x^3 - 8x^2 + 9x + 12

   - (5x^3 - 15x^2)______                    7x^2 divided by x gives you 7x

            0 + 7x^2 + 9x

               - (7x^2 - 21x)___                  and 7x times -3 gives you -21x

                       0 + 30x +12                  you distribute the negative in the parentheses again.  

 

Then,

     _5x^2   +7x  +30____

x-3| 5x^3 - 8x^2 + 9x + 12

   - (5x^3 - 15x^2)______

            0 + 7x^2 + 9x

              - (7x^2 - 21x)___

                       0 + 30x + 12                30x divided by x gives you 30.

                          - (30x - 90)                Distribute the negative again.

                                0  + 102                102 is the remainder.  

When writing remainders in long polynomial equations, it is expressed by writing the remainder over the divisor. Which is 102/x-3.

So the answer is 5x^2 + 7x + 30 + 102/x-3.      

Answer 2

Answer:

$5x^2+7x+30+\frac{102}{x-3}$

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}$