How do you divide #(18x ^ { 4} - 14x ^ { 3} + 8x ^ { 2} + 7) \div 2x ^ { 2}#?

1 Answer
May 27, 2017

See a solution process below:

Explanation:

First, rewrite this expression as:

#(18x^4 - 14x^3 + 8x^2 + 7)/(2x^2)#

We can then rewrite this again as:

#(18x^4)/(2x^2) - (14x^3)/(2x^2) + (8x^2)/(2x^2) + 7/(2x^2)#

We can once again rewrite this as:

#(9x^4)/x^2 - (7x^3)/x^2 + (4x^2)/x^2 + 7/(2x^2)#

We can now use these rule for exponents to divide the #x# terms in the first three fractions:

#x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))# and #a^color(red)(1) = a# and #a^color(red)(0) = 1#

#(9x^color(red)(4))/x^color(blue)(2) - (7x^color(red)(3))/x^color(blue)(2) + (4x^color(red)(2))/x^color(blue)(2) + 7/(2x^2) =>#

#9x^(color(red)(4)-color(blue)(2)) - 7x^(color(red)(3)-color(blue)(2)) + 4x^(color(red)(2)-color(blue)(2)) + 7/(2x^2) =>#

#9x^2 - 7x^1 + 4x^0 + 7/(2x^2) =>#

#9x^2 - 7x + 4 + 7/(2x^2)#

Or, we can use this rule of exponents to eliminate the exponent in the denominator of the last term:

#1/x^color(red)(a) = x^color(red)(-a)#

#9x^2 - 7x + 4 + 7/(2x^color(red)(2)) =>#

#9x^2 - 7x + 4 + 7/2x^color(red)(-2)#