(x^2 + 7x - 5) / (4x - 3): how to divide

[ryana]

How do you divide when the coefficient of x in the numerator is greator than it is in the denominator?

 

[stapel]

How do you devide[sic] when the coefficient of x in the numerator is greator than it is in the denominator?

The division works in the same way as always; you'd just lead off with a fraction instead of a whole number. :wink:

To learn how to do long polynomial division, try some of the many great lessons available online:

. . . . .Google results for "polynomial long division"

In general, you look at the leading terms of the divisor and the dividend. Then you put, up top, whatever you'd have to multiply the divisor's leading term by, to get the leading term of the dividend (or whatever is the leading term you're working with, underneath the division symbol in the long division). In your case, you'd start off with:

         x
         -
         4
       ---------------
4x - 3 ) x^2 +  7x - 5
               3
         x^2 - - x
               4
       ---------------
              31
              -- x - 5
               4

Hope that helps!

Eliz.

 

[soroban]

Re: (x^2+7x-5)/(4x-3)

Hello, ryana!

\(\displaystyle \frac{x^2+7x-5}{4x-3}\)

How do you devide when the coefficient of x in the denominator is greator than it is in the numerator?

I know of two ways . . . neither is pleasant.



\(\displaystyle \text{[1] Divide as usual, introducing fractions.}\)


\(\displaystyle \begin{array}{ccccc} & & +\frac{1}{4}x & + \frac{31}{16} \\ & -- & --- & --- \\ 4x-3\ & x^2 & +7x & -5 \\ & x^2 & -\frac{3}{4}x \\ & -- & --- \\ & & \frac{31}{4}x & -5 \\ & & \frac{31}{4}x & -\frac{93}{16} \\ & & --- & --- \\ & & & \frac{13}{16} \end{array}\)


\(\displaystyle \text{Answer: }\;\;\frac{1}{4}x + \frac{31}{16} + \frac{13}{16(4x-3)}\)




\(\displaystyle \text{[2] Factor first: }\;\frac{x^2+7x-5}{4\left(x-\frac{3}{4}\right)} \;=\;\frac{1}{4}\cdot\frac{x^2+7x-5}{x - \frac{3}{4}}\;\;\cdots\;\;\text{then divide.}\)


\(\displaystyle \begin{array}{ccccc} & & }x & + \frac{31}{4} \\ & -- & --- & --- \\ x-\frac{3}{4}\ & x^2 & +7x & -5 \\ & x^2 & -\frac{3}{4}x \\ & -- & --- \\ & & \frac{31}{4}x & -5 \\ & & \frac{31}{4}x & -\frac{93}{16} \\ & & --- & --- \\ & & & \frac{13}{16} \end{array}\)



\(\displaystyle \text{Answer: }\;\;\frac{1}{4}\left[x + \frac{31}{4} + \frac{\frac{13}{16}}{x-\frac{3}{4}}\right] \;\;=\;\;\frac{1}{4}x + \frac{31}{16} + \frac{13}{16(4x-3)}\)

 

[ryana]

The purpose of the exercise was to build a new function.
The problem: Given the functions h(x) = x^2 + 7x - 5 and k(x) = 4x -3, evaluate (h/k)(-1)

To find the correct answer (without building a function) I evaluate each function:
h(-1) = -11
k(-1) = -7

The result (h/k)(-1) must be -11/-7 = 11/7

Building the function (h/k) to evaluate (-1) requires that ugly long division process with fractions then a bunch of work to solve.

1    31   13
-X + -- + --------
4    16   16(4x-3)

-1   31   13
-- + -- + --------
 4   16   16(-4-3)

-4   31   13
-- + -- + ------
16   16   16(-7)

27   13
-- - ---
16   112

378   26
--- - ---
224   224

352
---
224

11
--
 7

Thanks for help. I know what method I won't use on tests. That's for sure.

 

[Deleted member 4993]

How do you divide when the coefficient of x in the numerator is greator than it is in the denominator?

Just to remind you the question you had posted....

Did you mention - you needed to find (h/k)(-1)????

Your sloppiness made us waste a lot of time that could have helped others....

 

[stapel]

The purpose of the exercise was to build a new function.
The problem: Given the functions h(x) = x^2 + 7x - 5 and k(x) = 4x -3, evaluate (h/k)(-1)

Ah. This is why the "Read Before Posting" thread asked you to post the entire exercise. Had you done so, we could have pointed out that all you needed to do was find h(-1) and k(-1), and then divide.

Eliz.

 

[ryana]

Please don't misunderstand.

I didn't have trouble evaluating (h/k)(-1). I wanted to learn how to build a new function from (h/k).
Dividing (x^2 + 7x - 5) / (4x - 3) was throwing me off because of the large coefficient in the numerator and I needed to know how to deal with that.

This question was inline and your time and efforts were not wasted.