$x^2- x- (a^2+ 3a + 2)$ এবং $a^6-64$ দুইটি বীজগণিতীয় রাশি।

a = 1 হলে, প্রদত্ত রাশি

$x^2-x-\left\{\right(1{)}^2+3\times 1+2\}=x^2-x-(1+3+2)$

$=x^2-x-6=x^2-3x+2x-6=x(x-3)+2(x-3)$

= (x - 3)(x + 2)

প্রদত্ত রাশি$=x^2-x-(a^2+3a+2)$

                   $=x^2-x-(a^2+2a+a+2)$

                      $=x^2-x-\left\{a\right(a+2)+1(a+2\left)\right\}$

                         $=x^2-x-(a+2)(a+1)$

                                $=x^2+\left\{\right(a+1)-(a+2\left)\right\}x-(a+1)(a+2)$

                                      $=x^2+\left[\right(a+1)+\{-(a+2)\left\}\right]*x+(a+1)\{-(a+2\left)\right\}$

ইহাই প্রদত্ত রাশিটির  $x^2+(p+q)x+pq$ আকারের প্রকাশ যেখানে, p = a + 1 এবং q = - (a+2)

$x^2+\left[\right(a+1)+\{-(a+2)\left\}\right]x+(a+1)\{-(a+2\left)\right\}$

$=x^2+\left\{\right(a\text{'}+1)-(a+2\left)\right\}x-(a+1)(a+2)$

$=x^2+(a+1)x-(a+2)x-(a+1)(a+2)$

$= x(x + a + 1) - (a + 2)(x + a + 1)$

$= (x + a + 1)(x - a - 2)$

$a^6-64=(a^2{)}^3-(4{)}^3$

                  $=(a^2-4)(a^4+4a^2+16)$

                    $=\left\{\right(a)²-{\left(2\right)}^{২}\left\}\right\{(a²)²+\left(4\right)²+4a^{২}\}$

                      $=(a+2)(a-2) (a^2+4{)}^2-2.a^{2}4+4a^2$

                        $=(a+2)(a-2)\left\{\right(a^2+4{)}^2-8a^2+4a^2\}$

                            $=(a+2)(a-2)\left\{\right(a^2+4{)}^2-\left(2a{)}^2\right\}$

                                $=(a+2)(a-2)(a^2+4+2a)(a^2+4-2a)$

                                        $=(a+2)(a-2)(a^2.+2a+4)(a^2-2a+4)$