Question 962342

For the polynomial below, {{{-1}}} is a zero.

{{{F(x)= x^3+7x^2+24x+18 }}}

Express {{{f(x)}}} as a product of linear factors
since given the polynomial of degree {{{3}}}, there will be three zeros and we can write it as:

{{{x^3+7x^2+24x+18 =(x-x[1])(x-x[2])(x-x[3])}}}

since given {{{-1}}} is a zero, we have {{{(x-x[1])=(x-(-1))=(x+1)}}} as one linear factor

{{{x^3+7x^2+24x+18 =(x+1)(x-x[2])(x-x[3])}}} ....to find other two linear factors, use long division and divide {{{(x^3+7x^2+24x+18)}}} by {{{(x+1)}}}

{{{(x^3+7x^2+24x+18)/(x+1)=(x-x[2])(x-x[3])}}}

-------({{{x^2+6x+18}}}
{{{(x+1)}}}|({{{x^3+7x^2+24x+18}}})
..........{{{x^3+x^2}}}.....................subtract
.................{{{6x^2+24x}}}
.................{{{6x^2+6x}}}.....................subtract
...........................{{{18x+18}}}
...........................{{{18x+18}}}.....................subtract
.....................................{{{0}}}

so, we have

{{{x^2+6x+18=(x-x[2])(x-x[3])}}}......now factor completely left side

first complete square {{{x^2+6x+18=(x+3)^2+9}}}

so, solve for {{{x}}}:

{{{(x+3)^2+9=0}}} if  {{{(x+3)^2=-9}}}  => {{{x+3=sqrt(-9)}}} => {{{x+3}}}=±{{{3i}}}

=>solutions {{{x[2]=-3+3i}}} and {{{x[3]=-3-3i}}}

then, your polynomial is:

{{{x^3+7x^2+24x+18 =(x+1)(x-(-3+3i))(x-(-3-3i))}}}

{{{x^3+7x^2+24x+18 =(x+1)(x+3-3i)(x+3+3i)}}}

or

{{{f(x) =(x+1)(x+3-3i)(x+3+3i)}}}