The only factoring you need to do is already done for you:
<em>x</em>² + <em>x</em> - 12 = (<em>x</em> + 4) (<em>x</em> - 3)
What you're asked to do is decompose
(3<em>x</em> - 4) / (<em>x</em>² + <em>x</em> - 12)
into partial fractions, i.e. find <em>a</em> and <em>b</em> such that
(3<em>x</em> - 4) / (<em>x</em>² + <em>x</em> - 12) = <em>a</em> / (<em>x</em> + 4) + <em>b</em> / (<em>x</em> - 3)
Multiply both sides by <em>x</em>² + <em>x</em> - 12 :
3<em>x</em> - 4 = <em>a</em> (<em>x</em> - 3) + <em>b</em> (<em>x</em> + 4)
3<em>x</em> - 4 = (<em>a</em> + <em>b</em>) <em>x</em> + (-3<em>a</em> + 4<em>b</em>)
So we have
<em>a</em> + <em>b</em> = 3
-3<em>a</em> + 4<em>b</em> = -4
and solving this system gives
<em>a</em> = 16/7 and <em>b</em> = 5/7
so you should submit the numbers in bold:
(3<em>x</em> - 4) / (<em>x</em>² + <em>x</em> - 12) = 16 / (7 (<em>x</em> + 4)) + 5 / (7 (<em>x</em> - 3))
Step-by-step explanation:
Chord AB and Chord CD are equidistant from the center of the circle.

The detailed explanation of the question is provided in the attached image.
Hope this helps :)
Answer:
Step-by-step explanation:
(x+2)(3x-3)
= 3x^2 -3x + 6x - 6
= 3x^2 + 3x - 6
2 candies will be left over after you but ten in each bag.