If f(x)=x^3 +11x^2 +20x−32 and f(−4)=0, then find all of the zeros of f(x) algebraically.

Answer 1

If f(-4) = 0, then (x-(-4)) = (x+4) is a factor of f(x).

Use either polynomial long division or synthetic division to factor this out of f(x). This results in f(x)=g(x)*(x+4) where g(x)= x²+7x-8.

You can either use basic quadratic factoring or the quadratic formula to arrive at: g(x)=(x+8)*(x-1). Setting each of these two factor equal to zero and solving for x completes the solution process.

So, the zeros of f(x) are -8, -4, and 1.

Answer 2

Answer 3

f(x) = x^3 +11x^2+20x -32

f(0) =-32

f(1) =1+11+20-32 = 0

(x-1) is a factor as x=1 is a zero

by Descartes rule of signs there is 1 positive real zero and either 2 negative real zeros or 2 imaginary zeros

f(-1) = -1+11-20-32 = -42

f(-2) = -8+44-40 -32=-36

f(-3) = -27 +99 -60-32=-34

f(-4)=-64+172-80-32=0

f(-5)= -125+275-100-32= 18

f(-8)=-512+704-160-32=0

zeros are 1,-4,-8

factors are (x-1), (x+4), (x+8)

since f(-4)=0, -4 is a zero. divide -4 into -32 to get 8. look for other integer factors of 8, where one is positive and the other negative, such as 1 and -8.

another approach: use a graphing calculator and see where the graph intersects the x axis

Answer 4

When f(-4) = 0, it means x+4 is a factor of f(x)

So, one of the zeros is x = - 4

Since the highest degree is three there will be two more solutions. To find the other two zeros we use synthetic division and divide f(x) = x^3+11x^2+20x-32 by x+4.

We get x^2 +7x - 8

Factoring the above equation gives (x-1)(x+8)

So the other two zeros are -8 and 1.