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aleksley [76]
2 years ago
14

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

Mathematics
1 answer:
umka21 [38]2 years ago
5 0

Answer:

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

Step-by-step explanation:

     _________________

x-3| 5x^3 - 8x^2 + 9x + 12      

   

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.

    _<u>5x^2</u>_____________

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

     <u>- (5x^3 - 15x^2)</u>______       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,

    _<u>5x^2+7x</u>_________

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

   <u>- (5x^3 - 15x^2)</u>______                    7x^2 divided by x gives you 7x

            0 + 7x^2 + 9x

               <u>- (7x^2 - 21x)</u>___                  and 7x times -3 gives you -21x

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

 

Then,

     _<u>5x^2   +7x  +30</u>____

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

   <u>- (5x^3 - 15x^2</u>)______

            0 + 7x^2 + 9x

              <u> - (7x^2 - 21x)</u>___

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

                          <u>- (30x - 90)</u>                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.      

                   

               

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Step-by-step explanation:

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Firlakuza [10]

Answer:

  • Domain: All the real values except x = 2 and x = 4: R - {2, 4}
  • Holes: x = 2
  • VA, vertical asymptores: x = 4
  • HA: horizontal asymptotes: there are not horizontal asymptotes
  • OA: oblique asymptotes: x + 6 [note that OH does not stand for any known feature, and so it is understood that it was intended to write OA]
  • Roots: x = 2
  • Y-intercept: -1

Step-by-step explanation:

1. <u>Given</u>:

f(x)=\frac{x^3-8}{x^2-6x+8}

  • Note that the number 8 in the numerator is not part of the power.
  • Type of function: rational function

2. <u>Domain</u>: is the set of x-values for which the function is defined.

The given function is defined for all x except those for which the denominator equals 0.

  • Denominator:  x² -6x + 8 = 0
  • Solve for x:

        Factor. (x - 4 )(x - 2) = 0

        Zero product property: (x - 4) = 0 or (x - 2) = 0

        x - 4 = 0 ⇒ x = 4

        x - 2 = 0 ⇒ x = 2

  • Domain:

        All the real values except x = 2 and x = 4: x ∈ R / x ≠ 2 and x ≠ 4.

3. <u>Holes</u>:

The holes on the graph of a rational function are at those x-values for which both the numerator and denominator are zero.

  • Find the values for which the numerator is zero:

        Numerator: x³ - 8 = 0

        Factor using difference of cubes property:

                   a³ - b³ = (a - b)(a² + ab + b²)

                   x³ - 8 = (x - 2)(x² + 2x + 4) = 0

        Zero product property:  (x - 2)(x² + 2x + 4) = 0

                    x - 2 = 0 ⇒ x = 2                    

                    x² + 2x + 4 = 0 (this has not real solution)

  • The values for which the denominator is zero were determined above: x = 2 and x = 4.

  • Conclusion: for x = 2 both numerator and denominator equal 0, so this is a hole.

4. <u>VA: Vertical asymptotes</u>.

The vertical asymptotes on the graph of a rational function are the vertical lines for which only the denominator (and not the numerator) equals zero.

  • In the previous part it was determined that happens when x = 4.

5. <u>HA: Horizontal asymptotes</u>.

In rational functions, if the numerator is a higher degree polynomial than the denominator, there is no horizontal asymptote.

6. <u>OA: oblique asymptotes</u>

  • Find the quotient and the remainder.

                       x + 6

                  _______________

x² - 6x + 8 )   x³ + 0x² + 0x - 8

                  - x³ + 6x² - 8x

                   ___________

                          6 x² -   8x -  8

                        - 6x² + 36x - 48

                        _____________

                                    28x  - 56

Result: (x + 6) + (28x - 56) / (x² - 6x + 8)

  • Find limit x → ∞

\lim_{x \to \infty}(x + 6) + \frac{28x-56}{x^2-6x+8}=x+6

<u>7. Roots</u>:

Roots are the values for which f(x) = 0.

That happens when the numerator equals 0, and the denominator is not 0.

As determined earlier: x³ - 8 = 0 ⇒ x = 2.

8. <u>Y-Intercept</u>

The y-intercepts of any function are the y-values when x = 0

  • Substitute x = 0 into the function:

         f(x)=\frac{x^3-8}{x^2-6x+8}=\frac{0^3-8}{0^2-6(0)+8}}=\frac{-8}{8} =-1

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