Factor completely using the difference of squares along with factoring by grouping. $16 p ^ { 3 } - 48 p ^ { 2 } - 25 p + 75$
A) $( p - 3 ) \left( 16 p ^ { 2 } - 25 \right)$
B) $( p - 3 ) ( 4 p - 5 ) ^ { 2 }$
C) $( p - 3 ) ( 4 p + 5 ) ( 4 p - 5 )$
D) Prime

Question

Quizplus · Accepted Answer

First, factor by grouping: $16p^3 - 48p^2 - 25p + 75 = 16p^2(p - 3) - 25(p - 3) = (16p^2 - 25)(p - 3)$. Then, notice $16p^2 - 25$ is a difference of squares: $(4p)^2 - 5^2 = (4p + 5)(4p - 5)$. Thus, the complete factorization is $(p - 3)(4p + 5)(4p - 5)$.

Factor Completely Using the Difference of Squares Along with Factoring 16p3−48p2−25p+7516 p ^ { 3 } - 48 p ^ { 2 } - 25 p + 7516p3−48p2−25p+75

Factor Completely Using the Difference of Squares Along with Factoring $16 p ^ { 3 } - 48 p ^ { 2 } - 25 p + 75$