64 z^2 - 1
That is the difference of two squares
remember
a^2-b^2 = (a-b)(a+b)
so
(8z-1)(8z+1)
How do I completely factor this expression?
256z^2 -4 -192z^2 +3
I appreciate your help.
2 answers
256z^2 -4 -192z^2 +3
= (256z^2-4) - 3(64z^2-1)
= (16z-2)(16z+2)-3(8z-1)(8z+1)
= 4(8z-1)(8z+1)-3(8z-1)(8z+1)
= (8z-1)(8z+1)
cool, eh?
= (256z^2-4) - 3(64z^2-1)
= (16z-2)(16z+2)-3(8z-1)(8z+1)
= 4(8z-1)(8z+1)-3(8z-1)(8z+1)
= (8z-1)(8z+1)
cool, eh?