Question 1001720

4 polynomial with integer coefficients has zeros:
 {{{-1}}},and {{{1}}}, with 1 a zero of multiplicity 2,=>{{{(x+1)^2}}}
if you know that polynomial is degree {{{4}}} polynomial, you cannot have {{{(x-1)^2}}}; that will make your polynomial a polynomial of degree {{{5}}} because
if {{{-3i}}} is a zero, don't forget that complex zeros always come in pairs; so, you have {{{3i}}} too



{{{f(x)=(x+1)^2(x-3i)(x+3i)}}}

{{{f(x)=(x+1)^2(x^2-(3i)^2)}}}

{{{f(x)=(x^2+2x+1)(x^2-9(i)^2)}}}

{{{f(x)=(x^2+2x+1)(x^2-9(-1))}}}

{{{f(x)=(x^2+2x+1)(x^2+9)}}}

{{{f(x)=x^4+2x^3+1x^2+9x^2+18x+9}}}

{{{f(x)=x^4+2x^3+10x^2+18x+9}}}


{{{ graph( 600, 600, -10, 10, -10, 10, x^4+2x^3+10x^2+18x+9) }}}