Question

3p^2-2p-5 factor completely

Answer 1

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

Answer 2

$a x^{2} +bx+c$ Here is the basis equation. $3p ^{2} -2p-5$ 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 ...$(3p^2+3p)+(-5p-5)$ The larger number will go first (3>-5) Now you can factor each group. $3p(p+1)-5(p+1)$ Because what is now in the parentheses are qualm and they are being added, you can now make the coeffecients a group $(3p-5) (p+1)$ 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.