solve the equation x^3+13x^2+32x+20
To solve the equation, we will use the rational root test by looking at the constant term "20" and test for 1,-1,2,-2,...
By testing -2 , we found it is a root, Then (x+2) is a factor.
Now we will divide the equations by (x+2) to find other factors:
then the solution is :
To solve the equation x^3+13x^2+32x+32x+20 = 0 (editted).
Let f(x) = x^3+13x^2+32x+20
We see that f(-2) = (-2)^3+13(-2)^2+32(-2)+20 = 72-72 = 0.
So, x=-2 is a root of the given expression.
Also f(x) = (x+2)Q(x), where Q(x) is an expression like ax^2+bx+c of 2nd degree. But if you equate the coefficient of x^3 in the product and on the left, we get a = 1. and similarly c = 20/2 = 10. Therefore,
x^3+13x^2+32x+20 = (x+2)(x+bx+10). Put x =1, we get:
1+13+32+20 = (1+2)(1+b+10). Or
66 = 3(b+11). Or b = 66/3-11 = 22-11 = 11.
So the x^3+13x^2+32x+20 = (x+2)(x^2+11x+10). Or the other two roots of x^3+13x^2+32x+20 are the roots of x^2+11x+10 = 0. Or
x^2+1ox+x+10 = 0
x(x+10)+1(x+10) = 0.
(x+10)(x+1) = 0.
x = -10 and x= -1 are the other two roots in addtion to x = -2.