SOLUTION: x^2+10x-128. I've had this problem a few times now (I'm a tutor myself) and it asks to solve for x. I just cannot factor this. Please help me refresh my algebra skills and help me

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: x^2+10x-128. I've had this problem a few times now (I'm a tutor myself) and it asks to solve for x. I just cannot factor this. Please help me refresh my algebra skills and help me       Log On


   

Question 274329: x^2+10x-128. I've had this problem a few times now (I'm a tutor myself) and it asks to solve for x. I just cannot factor this. Please help me refresh my algebra skills and help me find a way to solve for x on this one. Thank you.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
For more factoring help, check out this quadratic formula solver.

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression x%5E2%2B10x-128, we can see that the first coefficient is 1, the second coefficient is 10, and the last term is -128.



Now multiply the first coefficient 1 by the last term -128 to get %281%29%28-128%29=-128.



Now the question is: what two whole numbers multiply to -128 (the previous product) and add to the second coefficient 10?



To find these two numbers, we need to list all of the factors of -128 (the previous product).



Factors of -128:

1,2,4,8,16,32,64,128

-1,-2,-4,-8,-16,-32,-64,-128



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to -128.

1*(-128) = -128
2*(-64) = -128
4*(-32) = -128
8*(-16) = -128
(-1)*(128) = -128
(-2)*(64) = -128
(-4)*(32) = -128
(-8)*(16) = -128


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 10:



First NumberSecond NumberSum
1-1281+(-128)=-127
2-642+(-64)=-62
4-324+(-32)=-28
8-168+(-16)=-8
-1128-1+128=127
-264-2+64=62
-432-4+32=28
-816-8+16=8




From the table, we can see that there are no pairs of numbers which add to 10. So x%5E2%2B10x-128 cannot be factored.



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Answer:



So x%5E2%2B10%2Ax-128 doesn't factor at all (over the rational numbers).



So x%5E2%2B10%2Ax-128 is prime.





So you must use the quadratic formula


So let's use the quadratic formula to solve x%5E2%2B10x-128=0


For more help with the quadratic formula, check out this quadratic formula solver.

Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2%2B10%2Ax-128=0 ( notice a=1, b=10, and c=-128)





x+=+%28-10+%2B-+sqrt%28+%2810%29%5E2-4%2A1%2A-128+%29%29%2F%282%2A1%29 Plug in a=1, b=10, and c=-128




x+=+%28-10+%2B-+sqrt%28+100-4%2A1%2A-128+%29%29%2F%282%2A1%29 Square 10 to get 100




x+=+%28-10+%2B-+sqrt%28+100%2B512+%29%29%2F%282%2A1%29 Multiply -4%2A-128%2A1 to get 512




x+=+%28-10+%2B-+sqrt%28+612+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-10+%2B-+6%2Asqrt%2817%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-10+%2B-+6%2Asqrt%2817%29%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%28-10+%2B+6%2Asqrt%2817%29%29%2F2 or x+=+%28-10+-+6%2Asqrt%2817%29%29%2F2



Now break up the fraction



x=-10%2F2%2B6%2Asqrt%2817%29%2F2 or x=-10%2F2-6%2Asqrt%2817%29%2F2



Simplify



x=-5%2B3%2Asqrt%2817%29 or x=-5-3%2Asqrt%2817%29



So the solutions are:

x=-5%2B3%2Asqrt%2817%29 or x=-5-3%2Asqrt%2817%29