SOLUTION: factoring second polynomials 7pē+18p=9

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Question 912345: factoring second polynomials
7pē+18p=9
Found 2 solutions by mananth, josgarithmetic:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
7p%5E2%2B18p-9=0

7p%5E2%2B21p-3p-9=0


7p%28p%2B3%29-3%28p%2B3%29=0

%28p%2B3%29%287p-3%29=0
either (p+3)=0
p=-3
OR
7p-3=0
7p=3
p=3/7


Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
You just need to try all the combinations (until or unless you know another way).

7p%5E2%2B18p-9=0

Try these:

(7p+u)(p+v) and you must have 7pv+up=(7v+u)p=18p and u*v=9.
Pick your possible combinations u and v for the product 9.
You have 1 and 9
OR
you have 3 and 3.
You must expect one of the numbers positive and one of them negative.

%287p%2B-+1%29%28p%2B-+9%29, check this.
OR
%287p%2B-+3%29%28p%2B-+3%29, check this.
OR
%287p%2B-+9%29%28p%2B-+1%29, check this.

Which combination (including signs) works?

(Note that the rendering is working but is difficult to read).



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ANSWER: (7p-3)(p+3)