How do you factor #(x^3 - 7x^2 + 14x - 8)#?

1 Answer
Jim H
Apr 8, 2015

There doesn't seem to be a grouping that will work.

The rational zero theorem tells us that tlhe possible rational zeros are #+-1, +-2, +-4, +-8#.

We can see that #(1)^3-7(1)^2+14(1)-8=1-7+14-8=0#, so 1 is a zero.

The factor theorem then tells us that #x-1# is a factor.
Do the division:

#(x^3-7x^2+14x-8)/(x-1) = x^2-6x+8#

So
#(x^3-7x^2+14x-8) =(x-1) (x^2-6x+8)=(x-1)(x-2)(x-4)#

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