How do you divide #(2x^3-5x^2+4x+12)/(x-7) #?
2 Answers
Explanation:
#"one way is to use the divisor as a factor in the numerator"#
#"consider the numerator"#
#color(red)(2x^2)(x-7)color(magenta)(+14x^2)-5x^2+4x+12#
#=color(red)(2x^2)(x-7)color(red)(+9x)(x-7)color(magenta)(+63x)+4x+12#
#=color(red)(2x^2)(x-7)color(red)(+9x)(x-7)color(red)(+67)(x-7)color(magenta)(+469)+12#
#=color(red)(2x^2)(x-7)color(red)(+9x)(x-7)color(red)(+67)(x-7)+481#
#"quotient "=color(red)(2x^2+9x+67)", remainder "=481#
#rArr(2x^3-5x^2+4x+12)/(x-7)#
#=2x^2+9x+67+481/(x-7)#
Explanation: