How do you factor #x^3-8x^2+19x-12#?

Question

Savannah Anderson · Accepted Answer

We have the polynomial #color(red)(p(x) = x^3-8x^2+19x-12#
As the degree of the polynomial is greater than 2, we need to use Synthetic Division to factorize it.
In Synthetic Division, we first find a value of #x# that makes this polynomial equal to zero. That is, we need to find the ZERO of the polynomial. We do that by Trial and Error.
For #x=color(blue)1#, this Polynomial will equal zero.

#p(1) = 1^3-8*1^2+19*1-12 = 1 - 8 + 19 -12 = 0#

As #x=1# makes this polynomial zero, we know for sure that #color(green)((x-1))# is one of the factors of the polynomial.

To find the other two factors, we use the process of Synthetic Division. To explain the entire process of Synthetic Division using text is a bit difficult, but this image should help:

The 1, -7 and 12 give us another factor of the polynomial:
#x^2 - 7x + 12#
We have reduced p(x) to

#p(x) = (x-1)(x^2-7x+12)#
But #x^2-7x+12# can be factorised further
#=x^2-3x-4x+12#
#=x(x-3)-4(x-3)#
#=(x-3)(x-4)#
So finally, we get

#p(x) = (x-1)(x-3)(x-4)#
#x^3-8x^2+19x-12 = color(green)( (x-1)(x-3)(x-4)#

Jace Anderson · Answer

undefined To factor $x^3 - 8x^2 + 19x - 12$, you can use various methods like synthetic division, grouping, or trial and error. One approach is to look for factors of the constant term (-12) that might add up or subtract to give the coefficient of the $x^2$ term (-8). Then, using those factors, you can split the middle term and factor by grouping. This process may involve trying different factor combinations until finding the correct ones.