Question 960773
<pre>
{{{"f(x)"= 1x^3-1x^2-10x+6}}}   -3 is a zero.

-3 | 1  -1  -10   6
   |<u>    -3   12  -6</u>
     1  -4    2   0

So we have factored f(x) as

{{{"f(x)"= (x+3)(x^2-4x+2)}}}

Now we must factor {{{x^2-4x+2}}} into two linear factors.

We find its zeros by setting it equal to 0

{{{x^2-4x+2=0}}}

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

{{{x = (4 +- sqrt(16-8))/2 }}}

{{{x = (4 +- sqrt(8))/2 }}}

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

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

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

{{{x = (cross(2)(2 +- sqrt(2)))/cross(2) }}}

{{{x = 2 +- sqrt(2) }}}

so the other two zeros are {{{2 + sqrt(2) }}} and {{{2 - sqrt(2) }}}

So the remaining linear factors are {{{(x^"" -( 2 + sqrt(2))) }}} and {{{(x^"" -( 2 - sqrt(2))) }}}

{{{"f(x)"= (x+3)(x^"" -( 2 + sqrt(2)))(x^"" -( 2 - sqrt(2))) }}}

{{{"f(x)"= (x+3)(x - 2 - sqrt(2))(x - 2 + sqrt(2))) }}}

Edwin</pre>