প্রমাণ কর যে, $(x - a), f\left(x\right)$ এর উৎপাদক হবে, যদি এবং কেবল যদি $f\left(\mathfrak{a}\right)=0$ হয়।

দেওয়া আছে, $f\left(x\right)=x^2-7x+12$

প্রশ্নমতে, $f(x )=0$

বা , $x^2-7x+12=0$

বা , $x^2-4x-3x+12=0$

বা, $x(x - 4) - 3(x - 4) = 0$

বা, $(x - 3)(x - 4) = 0$                                                                    অথবা , $x - 4 = 0$

হয়, $x - 3 = 0$                                                                                                      $\therefore x = 4$

  $\therefore x = 3$

$\therefore x = 3 অথবা, x = 4$ এর জন্য f(x) = 0 হবে।

দেওয়া আছে, ।$P\left(a\right)=a^4-4a^3+5a^2+11ak+6k$

$\therefore P(-1=(-1{)}^4-4(-1{)}^3+5(-1{)}^2+11(-1)k+6k$

                     $= 1 + 4 + 5 - 11k + 6k = 10 - 5k$

   প্রশ্নমতে, $P(- 1) = 0$

বা, $10-5k=0$ বা , $5k = 10$ বা , $k = \frac{10}{5}= 2$

$\therefore k = 2$ এর জন্য $P(- 1) = 0$ হবে।

দেওয়া আছে, $Q\left(x\right)=\frac{1-3x^2+x^3}{x(1-x)}$

$Q\left(\frac{1}{m}\right)=\frac{1-3{\left({\displaystyle \frac{1}{m}}\right)}^2+{\left({\displaystyle \frac{1}{m}}\right)}^1}{{\displaystyle \frac{1}{m}}\left(1-{\displaystyle \frac{1}{m}}\right)}=\frac{1-3{\displaystyle \frac{1}{m^2}}+{\displaystyle \frac{1}{m^3}}}{{\displaystyle \frac{1}{m}}-{\displaystyle \frac{1}{m^2}}}=\frac{{\displaystyle \frac{m^3-3m+1}{m^3}}}{{\displaystyle \frac{m-1}{m^2}}}$

               $=\frac{m^3-3m+1}{m^3}\times \frac{m^2}{m-1}=\frac{m^3-3m+1}{m(m-1}$

আবার, $Q(1-m)=\frac{1-3(1-m{)}^2+(1-m{)}^3}{(1-m)\{1-(1-m\left)\right\}}$

                               $=\frac{1-3(1-2m+m^2)+(1-3m+3m^2-m^3)}{(1-m)(1-1+m)}$

                                 $=\frac{1-3+6m-3m^2+1-3m+3m^2-m^3}{m(1-m)}$

                                    $=\frac{-1+3m-m^3}{m(1-m)}$

                                    $=\frac{-(m^3-3m+1)}{-m(m-1)}=\frac{m^3-3m+1}{m(m-1)}$

                                      $\therefore Q = Q(1-m).$ (প্রমাণিত)

দেওয়া আছে, $P\left(a\right)=a^3+a^2+10a-8$

ভাগশেষ উপপাদ্য অনুসারে $p\left(a\right)$ কে $(a - 2)$ দ্বারা ভাগ করলে ভাগশেষ হবে $P\left(2\right)$

$\therefore P\left(2\right)=2^3+2^2+10\times 2-8=8+4+20-8=24$

  নির্ণেয় মান 24

দেওয়া আছে, $f\left(x\right)=54x^4+27x^{3}a-16x-8a$

                                   $=27x^3(2x+a)-8(2x+a)$

                                     $=(2x+a)(27x^3-8)=(2x+a)\left\{\right(3x{)}^3-\left(2{)}^3\right\}$

                                    $=(2x+a)(3x-2)\left\{\right(3x{)}^2+3x.2+\left(2{)}^2\right\}$

                                      $=(2x+a)(3x-2)(9x^2+6x+4)$

এখন, $(2x + 1)$ বা, $2\left(x + \frac{1}{2}\right),g\left(x\right) এর$ একটি উৎপাদক হবে যদি $s\left(- \frac{1}{2} \right)=0$ হয়।

$\therefore g\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^4+12{\left(-\frac{1}{2}\right)}^3+7{\left(-\frac{1}{2}\right)}^2-3\left(-\frac{1}{2}\right)-2.$

                          $= 4 \times \frac{1}{16}- 12 \times \frac{1}{8} + 7 \times \frac{1}{4}+ \frac{3}{2}- 2$

                          $= \frac{1}{4} - \frac{3}{2} + \frac{7}{4} + \frac{3}{2}- 2$

                           $= \frac{1}{4}+ \frac{7}{4}- 2 = \frac{1 + 7 - 8}{4}= \frac{8 - 8}{4} = \frac{0}{4} = 0$

$\therefore (2x + 1), g\left(x\right)$ এর একটি উৎপাদক।

আবার, $(2x - 1)$

বা, $2\left(x - \frac{1}{2}\right), g\left(x\right)$ এর একটি উৎপাদক হবে যদি $g\left(\frac{1}{2}\right)=0$ হয়।

         $\therefore g\left(\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^4+12{\left(\frac{1}{2}\right)}^3+7{\left(\frac{1}{2}\right)}^2-3\left(\frac{1}{2}\right)-2$

            $=4\times \frac{1}{16}+12\times \frac{1}{8}+\frac{7}{4}-\frac{3}{2}-2=\frac{1}{4}+\frac{3}{2}+\frac{7}{4}-\frac{3}{2}-2$

            $=\frac{1}{4}+\frac{7}{4}-2=\frac{1+7-8}{4}=\frac{8-8}{4}=\frac{0}{4}=0$

$\therefore (2x-1), g\left(x\right)$ এর একটি উৎপাদক।

অতএব , (2x + 1) ও (2x-1) উভয়ই g(x) এর উৎপাদক। (দেখানো হলো)