SOLUTION: What are the zeroes of f(x) = x^3 + 5x^2 - 7x + 1?

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Question 885552: What are the zeroes of f(x) = x^3 + 5x^2 - 7x + 1?
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = x^3 + 5x^2 - 7x + 1
possible zeros are -1 and 1.
use synthetic division to check
1 | 1 5 -7 1
1 6 -1
-----------------
1 6 -1 0
since the remainder is zero it is a factor:
(x-1)(x^2+6x-1)
normally, you would factor the quadratic but since it can't be factored we use the "quadratic formula" to get
zeroes are at:
x = 1; x = 0.1623; and x = -6.1623
.
details of quadratic equation follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B6x%2B-1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%286%29%5E2-4%2A1%2A-1=40.

Discriminant d=40 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+40+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%286%29%2Bsqrt%28+40+%29%29%2F2%5C1+=+0.16227766016838
x%5B2%5D+=+%28-%286%29-sqrt%28+40+%29%29%2F2%5C1+=+-6.16227766016838

Quadratic expression 1x%5E2%2B6x%2B-1 can be factored:
1x%5E2%2B6x%2B-1+=+1%28x-0.16227766016838%29%2A%28x--6.16227766016838%29
Again, the answer is: 0.16227766016838, -6.16227766016838. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B6%2Ax%2B-1+%29