How do you solve #(x ^ { 5} - 15x ^ { 4} + 90x ^ { 3} - 270x^ { 2} + 405x - 243) -: ( x - 3)#?

2 Answers
Dec 28, 2016

#(x^5-15x^4+90x^3-270x^2+405x-243)/(x-3) = x^4-12x^3+54x^2-108x+81#

Explanation:

What follows is not the quickest way, but it does exercise the binomial theorem a bit...

The quintic looks a suspiciously like #(x-3)^5#. Let's check...

By the binomial theorem:

#(a+b)^5 = sum_(k=0)^5 ((5),(k)) a^(5-k) b^k#

where #((5),(k)) = (5!)/((5-k)! k!)#

These binomial coefficients can be read from the sixth row of Pascal's triangle...

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So we find:

#(a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5#

Putting #a=x# and #b=-3# we get:

#(x-3)^5 = x^5+5(-3)x^4+10(-3)^2x^3+10(-3)^3x^2+5(-3)^4x+(-3)^5#

#color(white)((x-3)^5) = x^5-15x^4+90x^3-270x^2+405x-243#

Similarly, using the fifth row of Pascal's triangle, we have:

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

#color(white)((x-3)^4) = x^4-12x^3+54x^2-108x+81#

Hence:

#(x^5-15x^4+90x^3-270x^2+405x-243)/(x-3)#

#= (x-3)^5/(x-3)#

#= (x-3)^4#

#= x^4-12x^3+54x^2-108x+81#

Dec 28, 2016

#(x^5-15x^4+90x^3-270x^2+405x-243)/(x-3)#

#= x^4-12x^3+54x^2-108x+81#

Explanation:

Here's how I would do it myself.

Write the beginning of the factorisation:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(#

Then examine each power of #x# in turn in descending order to work out what the next term needs to be.

The first term we need to add is #x^4# in order to get #x^5# when multiplied:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(color(blue)(x^4)#

Examining what we have so far, the #(-3)*x^4# will result in #-3x^4#, when what we want for the next product is #-15x^4#. So we want an extra #-12x^4#. We can get this by writing #-12x^3# as our next term:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4color(blue)(-12x^3)#

Then #(-3)(-12x^3) = 36x^3#, but we want #90x^3#, so we need an extra #54x^3#. We can get this by making our next term #54x^2#:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+color(blue)(54x^2)#

Then #(-3)(54x^2) = -162x^2#, but we want #-270x^2#, so we need an extra #-108x^2#. We can get this by making our next term #-108x#:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+54x^2color(blue)(-108x)#

Then #(-3)(-108x) = 324x#, but we want #405x#, so we need an extra #81x#. We can get this by making our next term #81#:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+54x^2-108x+color(blue)(81))#

Finally, note that #(-3)(81) = -243#, just as we want. So the factorisation is exact.

I have used quite a few words to describe it, but most of the calculations can be done in your head without having to write down more than:

#(x^5-15x^4+90x^3-270x^2+405x-243) = (x-3)(x^4-12x^3+54x^2-108x+81)#

Hence:

#(x^5-15x^4+90x^3-270x^2+405x-243)/(x-3)#

#= x^4-12x^3+54x^2-108x+81#