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

2 Answers
May 17, 2017

Answer:

#3x^2-12x+32-93/(x+3)#

Explanation:

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

#"consider the numerator"#

#color(red)(3x^2)(x+3)color(magenta)(-9x^2)-3x^2-4x+3#

#=color(red)(3x^2)(x+3)color(red)(-12x)(x+3)color(magenta)(+36x)-4x+3#

#=color(red)(3x^2)(x+3)color(red)(-12x)(x+3)color(red)(+32)(x+3)color(magenta)(-96)+3#

#=color(red)(3x^2)(x+3)color(red)(-12x)(x+3)color(red)(+32)(x+3)-93#

#"quotient "=color(red)(3x^2-12x+32)," remainder "=-93#

#rArr(3x^3-3x^2-4x+3)/(x+3)=3x^2-12x+32-93/(x+3)#

May 17, 2017

Answer:

The quotient is #=3x^2-12x+32# and the remainder is #=-93#

Explanation:

We perform a long division

#color(white)(aaaa)##x+3##|##color(white)(aaaa)##3x^3-3x^2-4x+3##color(white)(aaaa)##|##3x^2-12x+32#

#color(white)(aaaaaaaaaaaaaaa)##3x^3+9x^2#

#color(white)(aaaaaaaaaaaaaaaaa)##0-12x^2-4x#

#color(white)(aaaaaaaaaaaaaaaaaaa)##-12x^2-36x#

#color(white)(aaaaaaaaaaaaaaaaaaaaaa)##+0+32x+3#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaa)##+32x+96#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)##-0-93#

Therefore,

#(3x^3-3x^2-4x+3)/(x+3)=3x^2-12x+32-93/(x+3)#