How do you factor #x^3-8x^2+17x-10#?

1 Answer
Alan P.
Apr 10, 2015

Polynomials of degree #3# and above can be very difficult to factor.

One trick is to hope for an integer solution.

If an integer solution exists the constant term in the factors must be one of the factors of the constant term of the expression.

In this case the factors of #10# are #+-1, +-2, +-5#

Substituting #1# into the expression #x^3-8x+17x-10#
gives #1-8+17-10 =0#
so #x=1# is a root and #(x-1)# is a factor of this expression.

Using synthetic division to divide #(x^3-8x^2+17x-10)/(x-1)#
we get #x^2-7x+10#
(sorry I can't see any way to neatly demonstrate synthetic division here)
with obvious factors #(x-2)(x-5)#

So
#(x^3-8x^2+17x-10) = (x-1)(x-2)(x-5)#

Late addition: Here is an image to help explain the synthetic division
enter image source here

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