How do you divide #2x ^ { 3} + 4x ^ { 2} + 5# by #x-3#?

1 Answer
Jim G.
Jul 4, 2017

#2x^2+10x+30+95/(x-3)#

Explanation:

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(2x^2)(x-3)color(magenta)(+6x^2)+4x^2+5#

#=color(red)(2x^2)(x-3)color(red)(+10x)(x-3)color(magenta)(+30x)+5#

#=color(red)(2x^2)(x-3)color(red)(+10x)(x-3)color(red)(+30)(x-3)color(magenta)(+90)+5#

#=color(red)(2x^2)(x-3)color(red)(+10x)(x-3)color(red)(+30)(x-3)+95#

#"quotient "=color(red)(2x^2+10x+30)," remainder "=95#

#rArr(2x^3+4x^2+5)/(x-3)#

#=2x^2+10x+30+95/(x-3)#

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