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

1 Answer
Mar 14, 2016

Answer:

#(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#