Question 977232
p(x)=x^4-2x^3-8x+16
factor the polynomial and use the factored form to find the zeros
x=?
<pre>

p(x) = x&#8308;-2x³-8x+16

Out of the first two terms on the right, factor out x³

p(x) = x³(x-2)-8x+16

Out of the last two terms on the right, factor out -8,
being careful to remember that when you factor a negative
out of a positive, -8 out of +16, you get a negative inside 
the parentheses:

p(x) = x³(x-2)-8(x-2)

Now there is a common factor of (x-2) in both terms so we
factor (x-2) out of both terms:

p(x) = (x-2)(x³-8)

Now since 8 = 2³, we have

p(x) = (x-2)(x³-2³)

And now we use the rule for factoring the difference of two cubes:

p(x) = (x-2)(x-2)(x²+2x+2²)

p(x) = (x-2)²(x²+2x+4)

The only real zero is found by setting factor x-2 equal 0

x-2 = 0
  x = 2

It has multiplicity 2 because p(x) has two factors (x-2)(x-2) which 
when set = 0, provide the zero 2.

There are also two imaginary zeros which we find by setting the other
factor x²+2x+4 equal to 0 and using the quadratic formula since it
does not factor:

x²+2x+4 = 0
   
{{{x}}}{{{""=""}}}{{{(-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

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

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

{{{x}}}{{{""=""}}}{{{(-2 +- sqrt(-12))/2 }}}

{{{x}}}{{{""=""}}}{{{(-2 +- sqrt((-1)(12)))/2 }}}

{{{x}}}{{{""=""}}}{{{(-2 +- i*sqrt(12))/2 }}}

{{{x}}}{{{""=""}}}{{{(-2 +- i*sqrt(4*3))/2 }}}

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

{{{x}}}{{{""=""}}}{{{(2(-1 +- i*sqrt(3)))/2 }}}

{{{x}}}{{{""=""}}}{{{(cross(2)(-1 +- i*sqrt(3)))/cross(2) }}}

{{{x}}}{{{""=""}}}{{{-1 +- i*sqrt(3) }}}

So there are three zeros, 1 real solution of 2 which has multiplicity 2,
and two conjugate imaginary or complex zeros {{{-1 + i*sqrt(3) }}} and {{{-1 - i*sqrt(3) }}}.

Edwin</pre>