SOLUTION: Please help me begin to solve: 1-2/x-8/x^2=0

Question 77539: Please help me begin to solve:
1-2/x-8/x^2=0
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Multiply both sides by , this eliminates the denominator

Distribute on both sides

Now factor the quadratic:

Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)

In order to factor , first we need to ask ourselves: What two numbers multiply to -8 and add to -2? Lets find out by listing all of the possible factors of -8

Factors:

1,2,4,8,

-1,-2,-4,-8,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -8.

(-1)*(8)=-8

(-2)*(4)=-8

Now which of these pairs add to -2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -2

||||

First Number	\|	Second Number	\|	Sum
1	\|	-8	\|	1+(-8)=-7
2	\|	-4	\|	2+(-4)=-2
-1	\|	8	\|	(-1)+8=7
-2	\|	4	\|	(-2)+4=2

We can see from the table that 2 and -4 add to -2.So the two numbers that multiply to -8 and add to -2 are: 2 and -4 Now we substitute these numbers into a and b of the general equation of a product of linear factors which is: substitute a=2 and b=-4 So the equation becomes: (x+2)(x-4) Notice that if we foil (x+2)(x-4) we get the quadratic again

Now set each factor equal to zero

So our solution is:
or
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