সংক্ষিপ্ত প্রশ্ন

কোনো বীজগণিতীয় রাশি দুই বা ততোধিক রাশির গুণফল হলে শেষোক্ত রাশিগুলোর প্রত্যেকটিকে প্রথম রাশির উৎপাদক বা গুণনীয়ক বলে। যেমন, $a^2-b^2=(a+b)(a-b)$

এখানে, (a+b) ও (a - b) বীজগণিতীয় রাশিদ্বয় হলো $a^2-b^2$ এর দুটি উৎপাদক বা গুণনীয়ক।

4x + 16y = $4 \times x + 4 \times 4y$

= 4(x + 4y) [বষ্টন বিধি]

3x - 12y = $3 \times x - 3 \times 4y$

= 3(x - 4y) [বণ্টন বিধি]

3ax + 4by + 2ax + by

= 3ax + 2ax + 4by + by

= 5ax + 5by = 5(ax + by)

$7x+14x^2=7x\times 1+0.7x\times 2x$

= 7x(1 + 2x)

$3a-9a^2=3a\times 1-3a\times 3a$

= 3a(1 - 3a)

6m + 24n = $6\times m + 6\times 4n$

= 6(m + 4n)

$35m^2-5m=5m\times 7m-5m\times 1$

= 5m(7m - 1)

12x - 15y = $3 \times 4x - 3 \times 5y$

= 3(4x - 5y)

9a + 12b = $3 \times 3a + 3 \times 4b$ = 3(3a + 4b)

$5x^4{y}^2-7x^2{y}^4=x^2{y}^2\times 5x^2-x^2{y}^2\times 7y^2$

$=x^2{y}^2 (5x^2-7y^2)$

$7m{n}^3-4m^{3}n=mn\times 7n^2-mn\times 4m^2$

$=mn(7n^2-4m^2)$

$16x^2-9=(4x{)}^2-(3{)}^2$

= (4x + 3)(4x - 3)

$36-25y^2=(6{)}^2-(5y{)}^2$

= (6 + 5y)(6 - 5y)

$49a^2-64b^2=(7a{)}^2-(8b{)}^2$

= (7a + 8b)(7a - 8b)

$81p^2-100q^2=(9p{)}^2-(10q{)}^2$

= (9p + 10q)(9p - 10q)

$9m^2-36n^2=9(m^2-4n^2)$

$=9\left\{\right(m{)}^2-\left(2n{)}^2\right\}$

= 9(m + 2n)(m - 2n)

$4z^2-1=(2z{)}^2-(1{)}^2$

= (2z + 1)(2z - 1)

$x^2+7x+12=x^2+(3+4)x+3\times 4$

= (x + 3)(x + 4)

$4x^2+16xy+15y^2=(2x{)}^2+(3y+5y)\times 2x+3y\times 5y$

= (2x + 3y)(2x + 5y)

$a^2+13a+42=a^2+(6+7)a+6\times 7$

= (a + 6)(a + 7)

$a^2+5ab+6b^2=a^2+(2b+3b)\times a+2b\times 3b$

= (a + 2b)(a + 3b)

$p^2+13pq+42q^2=p^2+(6q+7q)\times p+6q\times 7q$

= (p + 6q)(p + 7q)

$p^2+14p+48=p^2+(6+8) p+6\times 8$

= (p + 6)(p + 8)

$m^2+10m+21=m^2+(3+7) m+3\times 7$

= (m + 3)(m + 7)