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

Question

Savannah Anderson · Accepted Answer

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

Abigail Allen · Answer

undefined To factor the polynomial $x^3 - 8x^2 + 17x - 10$, you can use techniques like grouping or synthetic division. However, since this polynomial doesn't seem to have any rational roots, we can use a different approach.

One method is to look for factors of the constant term (-10) that sum up to the coefficient of the linear term (17). The factors of -10 are ±1, ±2, ±5, and ±10. By trial and error, we find that (-2) and 5 are the factors that sum up to 17.

Then, we rewrite the polynomial with four terms:

$x^3 - 8x^2 + 17x - 10 = (x^3 - 2x^2) + (5x - 10)$

Now, we factor by grouping:

$x^2(x - 2) + 5(x - 2)$

Now, we factor out the common factor $x - 2$:

$(x^2 + 5)(x - 2)$

So, the factored form of $x^3 - 8x^2 + 17x - 10$ is $(x^2 + 5)(x - 2)$.