Question 756643
<pre>
F(x)=x^3-8x^2+29x-52

The possible rational zeros are ± the factors of 52,
which are ±1,±2,±4,±13,±26, and ±52

You can keep trying those and the only one that gives
a zero remainder and is therefore a rational zero is 4.

4|1 -8  29 -52
 |<u>   4 -16  52</u>
  1 -4  13   0

So we have factored F(x) this way:

F(x) = (x-4)(x²-4x+13)

To find the other zeros we set the factor
x² - 4x + 13 = 0 and use the quadratic formula:


x = {{{(-b +- sqrt( b^2-4ac ))/(2a) }}}

x = {{{(-(-4) +- sqrt( (-4)^2-4(1)(13) ))/(2(1)) }}} 

x = {{{(4 +- sqrt(16-52 ))/2 }}}

x = {{{(4 +- sqrt(-36 ))/2 }}}

x = {{{(4 +- 6i)/2 }}}

x = {{{(2(2 +- 3i))/2 }}}

x = {{{(cross(2)(2 +- 3i))/cross(2) }}}

x = 2 ± 3i

So the other zeros are 2 + 3i and 2 - 3i

So the factored form of F(x) is

F(x) = (x - 4)[x - (2 + 3i)][x - (2 - 3i)]

or 

F(x) = (x - 4)(x - 2 - 3i)(x - 2 + 3i)

Edwin</pre>