solve  the equation x^3+13x^2+32x+20

Expert Answers
hala718 eNotes educator| Certified Educator

x^3+13x^2+32x+20 =0

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:

(x+2)(x^2+11x+10)=0

(x+2)(x+10)(x+1) =0

then the solution is :

x1=-2

x2=-10

x3=-1

neela | Student

To solve the equation x^3+13x^2+32x+32x+20 = 0 (editted).

Solution:

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.

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