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

Mar 14, 2016

#### Answer:

$\left(x + 1.74\right) \left(x + 4.37 - 1.04 i\right) \left(x - 4.37 + 1.04 i\right) \approx 0$

#### 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 $x \approx - 1.7358$)

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

$\frac{{x}^{3} - 7 {x}^{2} + 5 x + 35}{x + 1.74} \approx {x}^{2} - 8.74 x + 20.16$

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

$x \approx - 4.37 + 1.04 i$ or $x \approx 4.37 - 1.04 i$

All three factors multiplied together gives

$\left(x + 1.74\right) \left(x + 4.37 - 1.04 i\right) \left(x - 4.37 + 1.04 i\right) \approx 0$