How do you divide $(16z x ^ { 3} - 20z ^ { 4} x ^ { 7} ) \div ( 4z ^ { 2} x ^ { 5} )$ ?

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Answer 1 · 2017-08-30T19:59:50

$(4)/(zx^2) - 5z^2x^2$

$(16zx^3 - 20z^4x^7)/(4z^2x^5)$

Since the numerator has a term being subtracted from another, we cannot cancel yet. First, we can factor the numerator. The GCF is $4zx^3$ .

$(4zx^3(4 - 5z^3x^4))/(4z^2x^5)$

We can now cancel $4$ , $z$ , and $x^3$ .

$(cancel4cancelzcancel(x^3)(4 - 5z^3x^4))/(cancel4 cancel(z^2)^z cancel(x^5)^(x^2))$

We are left with

$(4 - 5z^3x^4)/(zx^2)$

We can now split this into

$(4)/(zx^2) - (5z^3x^4)/(zx^2)$

Canceling, we get

$(4)/(zx^2) - (5cancel(z^3)^(z^2) cancel(x^4)^(x^2))/(cancelz cancel(x^2))$

$= (4)/(zx^2) - 5z^2x^2$

How do you divide (16z x ^ { 3} - 20z ^ { 4} x ^ { 7} ) \div ( 4z ^ { 2} x ^ { 5} )(16z x ^ { 3} - 20z ^ { 4} x ^ { 7} ) \div ( 4z ^ { 2} x ^ { 5} )?