Question

if  $p(x) = x^{2} +x +1$  and [tex]q(x) = x^{2} -x +1
[/tex]  then  [tex]p(x) q(x) =
[/tex]_______

Answer 1

$p(x)=x^2+x+1;\ q(x)=x^2-x+1\\\\p(x)q(x)=(x^2+x+1)(x^2-x+1)\\\\=x^4-x^3+x^2+x^3-x^2+x+x^2-x+1\\\\=x^4+x^2+1$

Answer 2

p(x) = x² + x + 1  &  q(x) = x² - x + 1

Since we are to find p(x).q(x),
p(x).q(x) = (x² + x + 1)(x² - x + 1) 
p(x).q(x) = x⁴ - x³ + x² + x³ - x² + x + x² - x + 1
Cancelling the terms,
We get,
p(x)q(x) = x⁴ + x² + 1