SOLUTION: I'm doing Factoring by Trial and Error/Grouping TRINOMIALS:
THE PROBLEM IS:
A^2-9A-18
Algebra.Com
Question 819008: I'm doing Factoring by Trial and Error/Grouping TRINOMIALS:
THE PROBLEM IS:
A^2-9A-18
Found 2 solutions by Alan3354, josgarithmetic:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A^2-9A-18
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Can't be factored.
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
3 and 6.
-3 and 6 multplies to -18, adds to +3.
3 and -6 multiplies to -18, adds to -3.
Not work.
What about 2 and 9?
-2 and 9 multiplies to -18, adds to 7.
2 and -9 multiplies to -18, adds to -7.
A^2-9A-18 is not factorable.
You would generally want INTEGERS x and y so that xy=-18 and x+y=-9.
y=-x-9;
x(-x-9)=-18
-1*x(x+9)=-18
x(x+9)=18
x^2+9x=18
x^2+9x-18=0
Resorting to general solution of quadratic formula:
x=(-9+- sqrt(81-4(-18))/(2)
x=(-9+- sqrt(81+72))/2
x=(-9+- sqrt(153))/2
x=(-9+- 3*sqrt(17))/2
..... You really do not want this.
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