How do you factor completely #P(x)=x^3-6x^2+11x-6#?

1 Answer
George C.
May 3, 2017

#x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)#

Explanation:

Given:

#P(x) = x^3-6x^2+11x-6#

Note that linear factors correspond to zeros. That is, #x=a# is a zero if and only if #(x-a)# is a factor.

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Rational Roots Theorem

By the Rational Roots Theorem, any rational zeros of #P(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-6# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros of #P(x)# are:

#+-1, +-2, +-3, +-6#

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Descartes' Rule of Signs

Note that the pattern of the signs of the coefficients of #P(x)# is #+ - + -#. With #3# changes, this means that #P(x)# has #3# or #1# positive real zeros.

The pattern of the signs of the coefficients of #P(-x)# is #- - - -#. With no changes, this means that #P(x)# has no negative real zeros.

So, in combination with the Rational Roots Theorem, we can deduce that the only possible rational zeros of #P(x)# are:

#1, 2, 3, 6#

We could try each of these values in turn and find that:

#P(1) = P(2) = P(3) = 0#

Hence #P(x) = (x-1)(x-2)(x-3)#

but here's another way...

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Sum of coefficients shortcut

Note that the sum of the coefficients of #P(x)# is zero. That is:

#1-6+11-6 = 0#

Hence we can deduce that #x=1# is a zero of #P(x)# and #(x-1)# a factor:

#x^3-6x^2+11x-6 = (x-1)(x^2-5x+6)#

Then to factor the remaining quadratic note that #5 = 2+3# and #6 = 2*3#, and hence:

#x^2-5x+6 = (x-2)(x-3)#

Either way we arrive at our result:

#x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)#

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