Algebra question

Question

x^2+x-4 /    1-x
---------      ------
x^2-16      3x^2-11x-4

Answer 1

Hello Des,

It looks like you have typed that the first fraction is being divided by the second:

(x² + x - 4)            (1 - x)

------------    /  -----------------

   (x² - 16)         (3x² - 11x - 4)

 

Let's first change it to multiplication by flipping the second fraction (multiply by the reciprocal):

(x² + x - 4)      (3x² - 11x - 4)

------------   *   ----------------
 (x² - 16)                (1 - x)

 

The top left quadratic is not factorable, so we will leave it as it.

 

The lower left is a difference of squares, generally speaking a² - b² = (a - b)(a + b)

→ x² - 16 = (x - 4)(x + 4)

 

The upper right I am going to use what is typically called the 'ac method',

from the quadratic form: ax² + bx + c,

where you multiply a*c to figure out how to break up the middle term,

and then factor by grouping.

So from  3x² - 11x - 4, we have

a * c = (3)(4) = 12.

Since 1 * 12 = 12, and 1 - 12 = -11, we will break up the middle term as follows:

3x² + 1x - 12x - 4

= x(3x +1) - 4(3x + 1)

= (3x + 1)(x - 4)

 

For the bottom right, factor out a (-1)

→ 1 - x = (-1)(x - 1)

 

So the problem now looks like the below, with cancelling:

  (x² + x - 4)           (3x + 1)(x - 4)
---------------      *   ----------------
(x - 4)(x + 4)            (-1)(x - 1)

 

which simplifies to:

          (x² + x - 4) (3x + 1)
(-1) * -----------------------

             (x + 4)(x - 1)

 

I hope the above helps.  Should you have any questions about any of the steps, please let me know.

 

Thank you for posting the question.

Michael Ehlers