Finding all the zeros of a function help?

Question

Find all the zeros of the function

f(x) = x^4 -10x^3 +34x^2 -42x +9


O.k. I used synthetic division and have X=3 as one solution.
and the left over equation is: X^3 -7x^2 +13x -3

What is the other zero?

O.K. I'm now at

X^2 -4x +1=0

How do I get this into the quadratic formula?

Answer 1

By inspection, it looks like you have another x = 3...
(3)^3 - 7(3²) + 13(3) - 3 = 0
27 - 63 + 39 - 3 = 0

so divide again by (x - 3):
................... x² - 4x + 1
x - 3 ) x^3 - 7x² + 13x - 3
......... x^3 - 3x²
................- 4x² + 13x
................- 4x² + 12x
............................... x - 3
............................... x - 3
.................................... 0

That gives you: x² - 4x + 1

And solving with the quadratic formula gives you two other roots.

x = [ -b ± sqrt( b² - 4ac) ] / 2a

a = 1
b = -4
c = 1

x = [ -(-4) ± sqrt( (-4)² - 4(1)(1)) ] / 2(1)
x = [ 4 ± sqrt( 16 - 4 ) ] / 2
x = [ 4 ± sqrt( 12 ) ] / 2
x = [ 4 ± sqrt( 4 * 3 ) ] / 2
x = [ 4 ± sqrt(4) * sqrt(3) ) ] / 2
x = [ 4 ± 2 sqrt(3) ] / 2
x = 4/2 ± 2 sqrt(3)/2
x = 2 ± sqrt(3)

So the four roots are:
x = 3
x = 3
x = 2 + sqrt(3)
x = 2 - sqrt(3)

Answer 2

Since it's a quartic equation, you'll have four zeros.

Remember that a function can have more than one zero at the same X value. Try dividing out another factor of (x-3).

As for the others, once you've found a second zero, you'll have reduced it to a quadratic equation, and you can use the quadratic formula for the last two zeros.

Answer 3

There is no other integer root. There is at least 1 more real root somewhere between 1/4 & 1/3.