Yes
factor out the 2z^2 in each term
(2z^2)(z^2-5z+4)
factor some more
z^2-5z+4
find what 2 numbers multiply to get 4 and add to get -5
the numbers are -1 and -4
(z-1)(z-4)
the factored form is
(2z^2)(z-1)(z-4)
5.
the equation looks like:
((2x+1)+4)/((x)+4)= 5/3
then you do cross multiplication and get:
6x+3+12=5x+20
subtract 5x on each side. subrtract the 3 and 12 from the 20.
and you get your answer 5=x
(4x-3y)^2(4x-3y)
=(16x^2-24xy+9y^2)(4x-3y)
=64x^3-96x^2y+36xy^2-48x^2y+72xy^2-27y^3
=64x^2-144x^2y+108xy^2-27y^3
Answer:
(x+2) (5x)
Step-by-step explanation:
the way I factor is the products of a*c added together equals b. so the products if (5*2) 10 equals 11. so 10 and 1 are the 2 products that add into 11. Now we put that into the equation. 5x^2+10x+1x+2 now take the two haves until you can't factor them any more 5x(x+2) (x+2). now take the repeated factor and outside factors to get (5x) and (x+2)