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3p^2-2p-5 factor completely

Sagot :

LoganTutor99
[tex]a x^{2} +bx+c[/tex] Here is the basis equation. [tex]3p ^{2} -2p-5[/tex] Let us use the AC method. Multiple the A and C values, or 3 and -5. This is -15. So now ask yourself... what multiplies to -15 but adds to -2. that would be, -5 and 3. So now reconstruct your equation ...[tex](3p^2+3p)+(-5p-5)[/tex] The larger number will go first (3>-5) Now you can factor each group. [tex]3p(p+1)-5(p+1)[/tex] Because what is now in the parentheses are qualm and they are being added, you can now make the coeffecients a group [tex](3p-5) (p+1)[/tex] If you are solving for p, set equal to zero and then the answers are -1 and (5/3). If you are just factoring, then that is your answer.
3p^2-2p-5
(3p-5) (p+1)

analyzing
-5 tells me the signs are different, therefore use subtraction
5 and 1 are the only possible combinations for 5
the same for 3
trial and error
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