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Mathematics 13 Online
OpenStudy (anonymous):

solve the equation by factoring (2p-3)(p-4)=0

jimthompson5910 (jim_thompson5910):

(2p-3)(p-4) is already factored

jimthompson5910 (jim_thompson5910):

Now you use the zero product property to go from (2p-3)(p-4) = 0 to 2p-3 = 0 or p-4 = 0

OpenStudy (anonymous):

my teacher said it wasn't ?

jimthompson5910 (jim_thompson5910):

The goal now is to solve 2p-3 = 0 and p-4 = 0 for p

jimthompson5910 (jim_thompson5910):

well (2p-3)(p-4) is factored already

OpenStudy (anonymous):

okay thanks for the help anyways ? how about (k+2)(k-1)=0

jimthompson5910 (jim_thompson5910):

same idea, (k+2)(k-1)=0 turns into k+2 = 0 or k-1=0

jimthompson5910 (jim_thompson5910):

when you use the zero product property

jimthompson5910 (jim_thompson5910):

afterwards you solve k+2 = 0 and k-1=0 for k

OpenStudy (anonymous):

yeah when u use the 0 property its always zero right?

jimthompson5910 (jim_thompson5910):

it's the idea that if two things multiply to zero, then one of those things must be zero so in algebraic form, if A*B = 0 then either A = 0 or B = 0

OpenStudy (anonymous):

okay I got it now thank you !

jimthompson5910 (jim_thompson5910):

you're welcome

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