How do you factor #12p^2+16p+5#?

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Answer 1 · 2015-06-02T14:46:51

#(6p+5)(2p+1)#

Use the quadratic formula for solving #ax^2+bx+c=0#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Taking #b^2-4ac# this is #16^2-4.12.5 = 256-240=16#, #sqrt(16)=4#

This gives #p=(-16+-4)/(2.12)# thus #p =-12/24 or -20/24#

Simplifying gives #p=-1/2 or p=-5/6#

These can be written as #2p+1=0 or 6p+5=0#

Hence #12p^2+16p+5=(6p+5)(2p+1)#