F(x)= x^4 - 11x^3 + 40x^2 - 55x + 25 has zeros at x=1 and x=5 as well as two other values. What are these two other values and where is f(x)>_0

Question

Factor theorems and synthetic division

Answer 1

We use synthetic division to divide by the given roots:

1 | 1 -11..40 -55..25

.....1....1 -10..30 -25

5 | 1 -10..30 -25 | 0

.....1....5 -25..25

.....1...-5....5 | 0

(The dots have been added as spacers for legibility.)

We're left with x^2-5x+5 which has irrational zeroes given by the quadratic formula:

(5-sqrt(5))/2=1.382 and (5+sqrt(5))/2=3.618. These are the missing zeroes.

A sketch of the graph shows f(0)=25, i.e., f(x)>0 when x<1; it dips below the x axis between x=1 and (5-sqrt(5))/2 and for (5-sqrt(5))/2<x<(5+sqrt(5))/2 f(x)>0; f(x)>0 when x>5. This makes 3 zones where f(x)>0.

 

Answer 2

Given F(x)= x^4 - 11x^3 + 40x^2 - 55x + 25
has zeros at x=1 and x=5
1 |1 -11 40 -55 25
...1 1 -10 30 -25
.........................
5 | 1 -10..30 -25 | 0

.....1....5 -25..25

.....1...-5....5 | 0

so x^2-5x+5=0
x=1/2(5 ± root5)

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Comments

Where is f(x)=>0? is the second part of the question!