How do you find the quotient of #(x^3+27) / (x+3)#?

1 Answer
Nov 9, 2015

Answer:

#x^2-3x+9#

Explanation:

You can use long division of polynomials. To start set up the problem in long division format, with the denominator out in front and the numerator under the division sign.

#color(white)(x/color(black)(x+3)) ()/(")" x^3 +0x^2 + 0x + 27)#

To start the division, look at the first term of each term. What do you need to multiply #x# by to get #x^3#? The answer is #x^2#, so that goes on the top.

#color(white)(x/color(black)(x+3)) (x^2color(white)(-3x -9 +27x))/(")" x^3 +0x^2 + 0x + 27)#

Now multiply the divisor, #x+3#, by #x^2# and subtract from the dividend. Remember, subtraction happens in columns.

#color(white)(x/color(black)(x+3)) (x^2 color(white)(-3x -9 +27x))/(")" x^3 +0x^2 + 0x + 27)#
#color(white)(XXXx)(x^3 + 3x^2)/(color(white)(x^3)-3x^2)#

Now we need to know what to multiply #x# by to get #-3x^2#. Its #-3x#. Write #-3x# on top, multiply by #x+3# and subtract.

#color(white)(x/color(black)(x+3)) (x^2-3x color(white)(+9 +27x))/(")" x^3 +0x^2 + 0x + 27)#
#color(white)(XXXx)(x^3 + 3x^2)/(color(white)(x^3)-3x^2)#
#color(white)(XXXXX|)(-3x^2-9x)/(color(white)(-3x^2-)9x#

Last one, we multiply #x# by #9# to get #9x#.

#color(white)(x/color(black)(x+3)) (x^2-3x +9 color(white)(+27x))/(")" x^3 +0x^2 + 0x + 27)#
#color(white)(XXXx)(x^3 + 3x^2)/(color(white)(x^3)-3x^2)#
#color(white)(XXXXX|)(-3x^2-9x)/(color(white)(-3x^2-)9x#
#color(white)(XXXXXXXXXx)(9x+27)/(color(white)(9x+)0)#

There is no remainder, so the quotient is;

#x^2-3x+9#