Question 19892
p^4, p^3, p^2 are variable terms present in the expression.  we can see p^2 is present in all the terms.

and 18, 63, 27 are the constants we see in the expression
the common factor for all of them is 9 (Reason!!!...we see them in the 9 table)

so, we can take out 9p^2 from the expression and see 
9p^2(2p^2+7p+3)

the remaining quadratic factor can be factorized using the usual method of factorisation for quadratics
this is how it goes

9p^2(2p^2+7p+3)=9p^2(2p^2+p+6p+3)=9p^2[p(2p+1)+3(2p+1)]=9p^2(p+3)(2p+1)

hence the factorisation over.