Question

How do you factor `x^3 - 7x^2 = -5x - 35`?

Answer 1

`(x+1.74)(x+4.37-1.04i)(x-4.37+1.04i)approx0`

Explanation:

First, graph the left and right sides of the equation to determine approximately where they intersect:

graph{(y-x^3+7x^2)(y+5x+35)=0 [-5, 8, -55, 20]}

They intersect around `x=-1.74`, which creates our first factor `x+1.74 = 0`. A graphing calculator can give you even greater accuracy (e.g., my TI-83 says `xapprox-1.7358`)

Next, use long division to determine the remaining quadratic. The result of long division (ignoring the very small remainder) gives,

`(x^3-7x^2+5x+35)/(x+1.74)approxx^2-8.74x+20.16`

Plugging this into the quadratic equation gives you the remaining imaginary factors:

`xapprox-4.37+1.04i` or `xapprox4.37-1.04i`

All three factors multiplied together gives

`(x+1.74)(x+4.37-1.04i)(x-4.37+1.04i)approx0`