How do you divide: #(x^3 - 6x^2 + 12x - 8) ÷ (x - 2)#?

2 Answers
May 3, 2016

Answer:

#(x^3-6x^2+12x-8)/(x-2)=x^2-4x+4#

Explanation:

One way to solve this problem is to perform polynomial long division. We write the polynomial and its divisor down in long division form and work through the normal steps of long division. Then we guess our first term in the quotient, which should subtract from the first term in the dividend shown in #color(red)"red"# below:

#{:("",{:("",color(red)(x^2),color(blue)(-4x),color(green)(+4)):}),(x-2,bar(")"{:(x^3,-6x^2,+12x,-8):})),(,{:(color(red)(-x^3 +2x^2),):}),(,{:(bar(" "-4x^2),+12x,):}),(,{:(" ",color(blue)(4x^2),color(blue)(-8x),):}),(,{:(" ",bar(" "4x),-8):}),(,{:(" "," ",color(green)(-4x),color(green)(+8)):}),(,{:(" ",bar(" "0)):}):}#

We then repeat this step for the next powers of #x# shown in #color(blue)"blue"# and #color(green)"green"#. We end up with a zero remainder.

#(x^3-6x^2+12x-8)/(x-2)=x^2-4x+4#

May 3, 2016

Answer:

#(x^3-6x^2+12x-8)/(x-2)=(x-2)^3/(x-2) = (x-2)^2 = x^2-4x+4#

Explanation:

Another way to solve this problem is to factor the dividend polynomial to see if there are any factors which are the divisor!

One clue to try this is that the first term in the dividend, #x^3#, is the cube of the first term in the divisor, #x# and the last term, #-8# is the cube of the last term of the divisor #(-2)^3#. It seems that perhaps, if we are lucky, our polynomial is the cube of the divisor:

#(x-2)^3 = color(red)((x^2-4x+4))(x-2)=x^3-6x^2+12x-8#

In fact it is - and we can use the intermediate result in red above to find the answer to our question:

#(x^3-6x^2+12x-8)/(x-2)=x^2-4x+4#