SOLUTION: fractoring trinomials by grouping 60p^4+81p^3+27p^2

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Question 57330: fractoring trinomials by grouping
60p^4+81p^3+27p^2
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
60p^4 + 81p^3 + 27p^2
3p^2(20p^2 + 27p + 9)
3p^2(20p^2 + 27p + 9)
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ap%5E2%2Bbp%2Bc=0 (in our case 20p%5E2%2B27p%2B9+=+0) has the following solutons:

p%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2827%29%5E2-4%2A20%2A9=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-27%2B-sqrt%28+9+%29%29%2F2%5Ca.

p%5B1%5D+=+%28-%2827%29%2Bsqrt%28+9+%29%29%2F2%5C20+=+-0.6
p%5B2%5D+=+%28-%2827%29-sqrt%28+9+%29%29%2F2%5C20+=+-0.75

Quadratic expression 20p%5E2%2B27p%2B9 can be factored:
20p%5E2%2B27p%2B9+=+20%28p--0.6%29%2A%28p--0.75%29
Again, the answer is: -0.6, -0.75. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+20%2Ax%5E2%2B27%2Ax%2B9+%29