Koshe dekhi 14 Class 8

2. নীচের বীজগাণিতিক সংখ্যামালাগুলির ল.সা.গু. নির্ণয় করি—

(i) \( 2x^{2}y^{3}, 10x^{3}y \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=2 x^{2} y^{3} \\
=2 \times x \times x \times y \times y \times y
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=10 x^{3} y \\
=2 \times 5 \times x \times x \times x \times y
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=2 \times 5 \times x \times x \times x \times y \times y \times y \\
=10 x^{3} y^{3}
\end{array}
\]

(ii) \( 7p^{2}q^{3}, 35p^{3}q, 42pq^{4} \)

সমাধানঃ

প্রথম সংখ্যামালা
\[
\begin{array}{l}
=7 p^{2} q^{3} \\
=7 \times p \times p \times q \times q \times q
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=35 p^{3} q \\
=7 \times 5 \times p \times q \times p \times q
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=42 p q^{4} \\
=7 \times 2 \times 3 \times p \times q \times q \times q \times q
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=7 \times 5 \times 2 \times 3 \times p \times p \times p \times q \times q \times q \times q \\
=210 p^{3} q^{4}
\end{array}
\]

(iii) \( 5a^{5}b, 15ab^{2}c, 25a^{2}b^{2}c^{2} \)

সমাধানঃ

প্রথম সংখ্যামালা
\[
\begin{array}{l}
=5 a^{5} b \\
=5 \times a \times a \times a \times a \times a \times b
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=15 a^{2} b c \\
=5 \times 3 \times a \times a \times b \times c
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=25 a^{2} b^{2} c^{2} \\
=5 \times 5 \times a \times a \times b \times b \times c \times c
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=5 \times 3 \times 5 \times a \times a \times a \times a \times a \times b \times b \times c \times c \\
=75 a^{5} b^{2} c^{2} \\
\end{array}
\]

(iv) \( 11a^{2}bc^{2}, 33a^{2}b^{2}c, 55a^{2}bc^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=11 a^{2} b c^{2} \\
=11 \times a \times a \times b \times c \times c
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=33 a^{2} b^{2} c \\
=11 \times 3 \times a \times a \times b \times b \times c
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=55 a^{2} b c^{2} \\
=11 \times 5 \times a \times a \times b \times c \times c
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=11 \times 3 \times 5 \times a \times a \times b \times b \times c \times c \\
=165 a^{2} b^{2} c^{2}
\end{array}
\]

আরও দেখুনঃ

3 . নীচের বীজগাণিতিক সংখ্যামালাগুলির গ.সা.গু. নির্ণয় করি—

(i) \( 5x (x + y), x^{3}− xy^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
=5 x(x+y)
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{3}-x y^{2} \\
=x\left(x^{2}-y^{2}\right) \\
=x(x+y)(x-y)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =x(x+y) \)

(ii) \( x^{3}− 3x^{2}y, x^{2}− 9y^{2} \)

সমাধানঃ

প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{3}-3 x^{2} y \\
=x^{2}(x-3 y)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{2}-9 y^{2} \\
=x^{2}-(3 y)^{2} \\
=(x+3 y)(x-3 y)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =(x-3 y) \)

(iii) \( 2ax (a − x)^{2}, 4a^{2}x (a − x)^{3} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=2 a x(a-x)^{2} \\
=2 a x(a-x)(a-x)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=4 a^{2} x(a-x)^{3} \\
=2^{2} \cdot a^{2} \cdot x(a-x)(a-x)(a-x)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু.
\[
\begin{array}{l}
=2 a x(a-x)(a-x) \\
=2 a x(a-x)^{2}
\end{array}
\]

(iv) \( x^{2}− 1, x^{2}− 2x + 1, x^{3}+ x^{2}− 2x \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{2}-1 \\
=x^{2}-1^{2} \\
=(x+1)(x-1)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{2}-2 x+1 \\
=x^{2}-2 . x .1+1^{2} \\
=(x-1)^{2} \\
=(x-1)(x-1)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{3}+x^{2}-2 x \\
=x\left(x^{2}+x-2\right) \\
=x\left(x^{2}+2 x-x-2\right) \\
=x\{x(x+2)-1(x+2)\} \\
=x(x+2)(x-1)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =(x-1) \)

(v) \( a^{2}− 1, a^{3}− 1, a^{2}+ a − 2 \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=a^{2}-1 \\
=a^{2}-1^{2} \\
=(a+1)(a-1)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=a^{3}-1 \\
=a^{3}-1^{3} \\
=(a-1)\left(a^{2}+a+1\right)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=a^{2}+a-2 \\ =a^{2}+(2-1)a-2 \\
=a^{2}+2 a-a-2 \\
=a(a+2)-1(a+2) \\
=(a+2)(a-1)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =(a-1) \)

আরও দেখুনঃ

(vi) \( x^{2}+ 3x + 2, x^{2}+ 4x + 3, x^{2}+ 5x + 6 \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{2}+3 x+2 \\ =x^{2}+(2+1) x+2 \\
=x^{2}+2 x+x+2 \\
=x(x+2)+1(x+2) \\
=(x+2)(x+1)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{2}+4 x+3 \\ =x^{2}+(3+1) x+3 \\
=x^{2}+3 x+x+3 \\
=x(x+3)+1(x+3) \\
=(x+3)(x+1)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{2}+5 x+6 \\= x^{2}+(3+2) x+6 \\
=x^{2}+2 x+3 x+6 \\
=x(x+2)+3(x+2) \\
=(x+2)(x+3)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =1 \)

(vii) \( x^{2}+ xy, xz + yz, x^{2}+ 2xy + y^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{2}+x y \\
=x(x+y)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=x z+y z \\
=z(x+y)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{2}+2 x y+y^{2} \\
=(x+y)^{2} \\
=(x+y)(x+y)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =(x+y) \)

(viii) \( 8 (x^{2} − 4), 12 (x^{2}+ 8), 36 (x^{2}− 3x − 10) \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=8\left(x^{2}-4\right) \\
=2 \times 2 \times 2 \times\left\{(x)^{2}-(2)^{2}\right\} \\
=2 \times 2 \times 2 \times(x+2)(x-2)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=12\left(x^{2}+8\right) \\
=12\left\{x^{2}+(2)^{3}\right\} \\
=12(x+2)\left(x^{2}-x .2+2^{2}\right) \\
=2 \times 2 \times 3(x+2)\left(x^{2}-2 x+4\right)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=36\left(x^{2}-3 x-10\right) \\ =36\left\{ x^{2}-(5-2) x-10 \right\} \\
=3 \times 3 \times 2 \times 2\left(x^{2}-5 x+2 x-10\right) \\
=3 \times 3 \times 2 \times 2\{x(x-5)+2(x-5)\} \\
=3 \times 3 \times 2 \times 2(x+5)(x+2)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু.
\[
\begin{array}{l}
=2 \times 2 \times(x+2) \\
=4(x+2)
\end{array}
\]

(ix) \( a^{2}− b^{2}− c^{2}+ 2bc, b^{2}− c^{2}− a^{2}+ 2ac, c^{2}− a^{2}− b^{2}+ 2ab \)

সমাধানঃ

প্রথম সংখ্যামালা
\[
\begin{array}{l}
=a^{2}-b^{2}-c^{2}+2 b c \\
=a^{2}-\left(b^{2}-2 b c+c^{2}\right) \\
=a^{2}-\left(b^{2}-2 b c+c^{2}\right) \\
=(a)^{2}-(b-c)^{2} \\ =\left\{ a+(b-c) \right\}\left\{ a-(b-c) \right\} \\
=(a+b-c) \times(a-b+c)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=b^{2}-c^{2}-a^{2}+2 a c \\
=b^{2}-\left(c^{2}-2 a c+a^{2}\right) \\
=(b)^{2}-(c-a)^{2} \\ =\left\{ b+(c-a) \right\}\left\{ b-(c-a) \right\} \\
=(b+c-a)(b-c+a)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=c^{2}-a^{2}-b^{2}+2 a b \\
=c^{2}-\left(a^{2}+b^{2}-2 a b\right) \\
=(c)^{2}-(a-b)^{2} \\ =\left\{ c+(a-b) \right\}\left\{ c-(a-b) \right\} \\
=(c+a-b) \times(c-a+b)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. = 1

আরও দেখুনঃ

(x) \( x^{3}− 16x, 2x^{3}+ 9x^{2}+ 4x, 2x^{3}+ x^{2}− 28x \)

সমাধানঃ

প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{3}-16 x \\
=x\left(x^{2}-16\right) \\
=x\left(x^{2}-4^{2}\right) \\
=x(x+4)(x-4)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=2 x^{3}+9 x^{2}+4 x \\
=x\left(2 x^{2}+9x+4\right) \\ =x\left\{2 x^{2}+(8+1)x+4\right\} \\
=x\left(2 x^{2}+8 x+x+4\right) \\
=x \left\{2 x(x+4)+1(x+4) \right\} \\
=x(x+4)(2 x+1)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
= 2 x^{3}+x^{2}-28 x \\
= x\left(2 x^{2}+x-28\right) \\ = x\left\{2 x^{2}+(8-7)x-28\right\} \\
= x\left(2 x^{2}+8 x-7 x-28\right) \\
= x \left\{2 x(x+4)-7(x+4)\right\} \\
= x(x+4)(2 x-7)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =x(x+4) \)

(xi) \( 4x^{2}− 1, 8x^{3}− 1, 4x^{2}− 4x + 1 \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=4 x^{2}-1 \\
=(2 x)^{2}-(1)^{2} \\
=(2 x+1)(2 x-1)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=8 x^{3}-1 \\
=(2 x)^{3}-(1)^{3} \\
=(2 x-1)\left(2 x^{2}+2 x .1+1^{2}\right) \\
=(2 x-1)\left(4 x^{2}+2 x+1\right)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=4 x^{2}-4 x+1 \\
=(2 x)^{2}-2.2 x \cdot 1+1^{2} \\
=(2 x-1)^{2} \\
=(2 x-1)(2 x-1)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =(2 x-1) \)

(xii) \( x^{3}− 3x^{2}− 10x, x^{3}+ 6x^{2}+ 8x, x^{2}− 5x^{3}− 14x^{2} \)

সমাধানঃ

প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{3}-3 x^{2}-10 x \\
=x\left(x^{2}-3 x-10\right) \\ =x\left\{x^{2}-(5-2) x-10\right\} \\
=x\left(x^{2}-5 x+2 x-10\right) \\
=x \left\{ x(x-5)+2(x-5)\right\} \\
=x(x-5)(x+2)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=x^{3}+6 x^{2}+8 x \\
=x\left(x^{2}+6 x+8\right) \\ =x\left\{x^{2}+(4+2) x+8\right\} \\
=x\left(x^{2}+4 x+2 x+8\right) \\
=x \left\{x(x+4)+2(x+4) \right\} \\
=x(x+2)(x+4)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{4}-5 x^{3}-14 x^{2} \\
=x^{2}\left(x^{2}-5 x-14\right) \\ =x^{2}\left\{x^{2}-(7-2) x-14\right\} \\
=x^{2}\left(x^{2}-7 x+2 x-14\right) \\
=x^{2} \left\{x^{2}(x-7)+2(x-7) \right\}\\
=x^{2}(x-7)(x+2)
\end{array}
\]

\( \therefore \quad \) নির্ণেয় গ.সা.গু. \( =x(x+2) \)

(xiii) \( 6x^{2}− 13xa + 6a^{2}, 6x^{2}+ 11xa − 10a^{2}, 6x^{2}+ 2xa − 4a^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=6 x^{2}-13 x a+6 a^{2} \\ =6 x^{2}-(9+4)x a+6 a^{2} \\
=6 x^{2}-9 x a-4 x a+6 a^{2} \\
=3 x(2 x-3 a)-2 a(2 x-3 a) \\
=(2 x-3 a)(3 x-2 a)
\end{array}
\]

দ্বিতীয় সংখখামালা
\[
\begin{array}{l}
=6 a^{2}+11 x a-10 a^{2} \\ =6 a^{2}+(15-4) x a-10 a^{2} \\
=6 x^{2}+15 x a-4 x a-10 a^{2} \\
=3 x(2 x+5 a)-2 a(2 x+5 a) \\
=(2 a+5 a)(3 x-2 a)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=6 x^{2}+2 x a-4 a^{2} \\
=2\left(3 x^{2}+x a-2 a^{2}\right) \\ =2\left\{3 x^{2}+(3-2)x a-2 a^{2}\right\} \\
=2\left(3 x^{2}+3 x a-2 x a-2 a^{2}\right) \\
=2\left\{3 x(x+a)-2 a(x+a)\right\} \\
=2(x+a)(3 x-2 a)
\end{array}
\]

\( \therefore \quad \) নির্ণেয় গ.সা.গু. \( =(3 x-2a) \)

আরও দেখুনঃ

4. নীচের বীজগাণিতিক সংখ্যামালাগুলির ল.সা.গু. নির্ণয় করি—

(i) \( p^{2}− q^{2}, (p + q)^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
p^{2}-q^{2} \\
=(p+q)(p-q)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
({p}+{q})^{2} \\
=(p+q)(p+q)
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=(p+q)(p-q)(p+q) \\
=(p+q)^{2}(p-q)
\end{array}
\]

(ii) \( (x^{2}y^{2}− x^{2}), (xy^{2}− 2xy + x) \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
\left(x^{2} y^{2}-x^{2}\right) \\
=x^{2}\left(y^{2}-1\right) \\
=x^{2}\left(y^{2}-1^{2}\right) \\
=x^{2}(y+1)(y-1)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=x y^{2}-2 x y+x \\
=x\left(y^{2}-2 y+1\right) \\
=x(y-1)^{2} \\
=x(y-1)(y-1)
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\(
\begin{array}{l}
=x^{2}(y+1)(y-1)(y-1) \\
=x^{2}(y+1)(y-1)^{2}
\end{array}
\)

(iii) \( (p + q) (p + r), (q + r) (r + p), (r + p) (p + q) \)

সমাধানঃ
প্রথম সংখ্যামালা = \((p+q)(q+r)\)
দ্বিতীয় সংখ্যামালা = \( (q+r)(r+p) \)
তৃতীয় সংখ্যামালা \( =(r+p)(p+q) \)

\( \therefore \quad \) নির্ণেয় ল.সা.গু. \( =(p+q)(p+r)(q+r) \)

(iv) \( ab^{4}− 8ab, a^{2}b^{4}+ 8a^{2}b, ab^{4}− 4ab^{2} \)

সমাধানঃ
প্রথম সशখ্যামালা
\[
\begin{array}{l}
=a b^{4}-8 a b \\
=a b\left(b^{3}-8\right) \\
=a b\left\{(b)^{3}-(2)^{3}\right\} \\
=a b(b-2)\left(b^{2}+b .2+2^{2}\right) \\
=a b(b-2)\left(b^{2}+2 b+4\right)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=a^{2} b+8 a^{2} b \\
=a^{2} b\left(b^{3}+8\right) \\
=a^{2} b\left\{(b)^{3}+(2)^{3}\right\} \\
=a^{2} b(b+2)\left(b^{2}-b .2+2^{2}\right) \\
=a^{2} b(b+2)\left(b^{2}-2 b+4\right)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=a b^{4}-4 a b^{2} \\
=a b^{2}\left(b^{2}-4\right) \\
=a b^{2}\left\{(b)^{2}-(2)^{2}\right\} \\
=a b^{2}(b+2)(b-2)
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}
= a^{2} b^{2}(b-2)\left(b^{2}+2 b+4\right) \times(b+2)\left(b^{2}-2 b+4\right) \\
= a^{2} b^{2}(b-2)(b+2) \times\left(b^{2}-2 b+4\right)\left(b^{2}+2 b+4\right)
\end{array}
\]

(v) \( x^{4}+ x^{2}y^{2}+ y^{4}, x^{3}y + y^{4}, (x^{2}− xy)^{3} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{4}+x^{2} y^{2}+y^{4} \\
=x^{4}+2 x^{2} y^{2}+y^{4}-x^{2} y^{2} \\
=\left(x^{2}+y^{2}\right)^{2}-(x y)^{2} \\
=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=x^{3} y+y^{4} \\
=y\left(x^{3}+y^{3}\right) \\
=y(x+y)\left(x^{2}-x y+y^{2}\right)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
= \left(x^{2}-x y\right)^{3} \\
= \{x(x-y)\}^{3} \\
= x^{3}(x-y)^{3}
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=x^{3} y(x+y)(x-y)^{3} \times\left(x^{2}-x y+y\right)\left(x^{2}+x y+y^{2}\right)
\end{array}
\]

আরও দেখুনঃ

(vi) \( p^{2}+ 2p, 2p^{4}+ 3p^{3}− 2p^{2}, 2p^{3}− 3p^{2} − 14p \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=p^{2}+2 p \\
=p(p+2)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=2 p^{4}+3 p^{3}-p^{2} \\ =p^{2}\{2 p^{2}+(4-1)p-2\} \\
=p^{2}\left(2 p^{2}+3 p-2\right) \\
=p^{2}\left(2 p^{2}+4 p-p-2\right) \\
=p^{2}\{2 p(p+2)-1(p+2)\} \\
=p^{2}(p+2)(2 p-1)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
= 2 p^{3}-3 p^{2}-14 p \\
= p\left(2 p^{2}-3 p-14\right) \\ = p\{2 p^{2}-(7-4)p-14\} \\
= p\left(2 p^{2}-7 p+4 p-14\right) \\
= \{p(2 p-7)+2 .(2 p-7)\} \\
= p(2 p-7)(p+2)
\end{array}
\]

( \therefore \) নির্ণেয় ল.সা.গু. \( =p^{2}(p+2)(2 p-1)(2 p-7) \)

(vii) \( x^{2}− y^{2}+ z^{2} − 2xz, x^{2}− y^{2}− z^{2}+ 2yz, xy + zx + y^{2}− z^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{2}-y^{2}+z^{2}-2 x z \\
=x^{2}-2 x z+z^{2}-y^{2} \\
=(x-z)^{2}-y^{2} \\
=(x-z+y)(x-z-y)
\end{array}
\]

দ্বিতীয় সংখামালা
\[
\begin{array}{l}
=x^{2}-y^{2}-z^{2}+2 y z \\
=x^{2}-\left(y^{2}-2 y z+z^{2}\right) \\
=x^{2}-(y-z)^{2} \\=\{x+(y-z)\}\{x-(y-z)\} \\
=(x+y-z)(x-y+z)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
x y+z x+y^{2}-z^{2} \\
=x(y+z)+(y-z)(y+z) \\
=(y+z)(x+y-z)
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু. \( =(y+z)(x+y-z)(x-y-z)(x-y+z) \)

(viii) \( x^{2}− xy − 2y^{2}, 2x^{2}− 5xy + 2y^{2}, 2x^{2}+ xy − y^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{2}-x y-2 y^{2} \\ =x^{2}-(2-1)x y-2 y^{2} \\
=x^{2}-2 x y+x y-2 y^{2} \\
=x(x-2 y)+y(x-2 y) \\
=(x-2 y)(x+y)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=2 x^{2}-5 x y+2 y^{2} \\ =2 x^{2}-(4+1) x y+2 y^{2} \\
=2 x^{2}-4 x y-x y+2 y^{2} \\
=2 x(x-2 y)-y(x-2 y) \\
=(x-2 y)(2 x-y)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=2 x^{2}+x y-y^{2} \\ =2 x^{2}+(2-1)x y-y^{2} \\
=2 x^{2}+2 x y-x y-y^{2} \\
=2 x(x+y)-y(x+y) \\
=(x+y)(2 x-y)
\end{array}
\]

\( \therefore \quad \) নির্ণেয় ল.সা.গু. \( =(x+y)(x-2 y)(2 x-y) \)

আরও দেখুনঃ

(ix) \( 3x^{2}− 15x + 18, 2x^{2}+ 2x − 24, 4x^{2}+ 36x + 80 \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=3 x^{2}-15 x+18 \\
=3\left(x^{2}-5 x+6\right) \\ =3\{x^{2}-(3-2)x+6\} \\
=3\left(x^{2}-3 x-2 x+6\right) \\
=3\{x(x-3)-2(x-3)\} \\
=3(x-3)(x-2)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=2 x^{2}+2 x-24 \\
=2\left(x^{2}+x-12\right) \\ =2\{x^{2}+(4-3)x-12\} \\
=2\left(x^{2}+4 x-3 x-12\right) \\
=2\{x(x+4)-3(x+4)\} \\
=2(x+4)(x-3)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=4 x^{2}+36 x+80 \\
=4\left(x^{2}+9 x+20\right) \\ =4\{x^{2}+(5+4)x+20\} \\
=2\times 2\left(x^{2}+4 x+5 x+20\right) \\
=2 \times 2\{x(x+4)+5(x+4)\} \\
=2\times 2(x+4)(x+5)
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
\begin{array}{l}
=3\times 2 \times 2 \times (x-2)(x-3)(x+4)(x+5) \\
=12(x-2)(x-3)(x+4)(x+5)
\end{array}
\]

(x) \( (a^{2}+ 2a)2, 2a^{3}+ 3a^{2}− 2a, 2a^{4}− 3a^{3}− 14a^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=\left(a^{2}+2 a\right)^{2} \\
=\left\{(a(a+2)\}^{2}\right. \\
=a^{2}(a+2)^{2}
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=2 a^{3}+3 a^{2}-2 a \\
=a\left(2 a^{2}+3 a-2\right) \\ =a\{2 a^{2}+(4-1) a-2\} \\
=a\left(2 a^{2}+4 a-a-2\right) \\
=a\{2 a(a+2)-1(a+2)\} \\
=a(a+2)(2 a-1)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=a^{4}-3 a^{3}-14 a^{2} \\
=a^{2}\left(2 a^{2}-3 a-14\right) \\ =a^{2}\{2 a^{2}-(7-4) a-14\} \\
=a^{2}\left(2 a^{2}-7 a+4 a-14\right) \\
=a^{2}\{a(2 a-7)+2(2 a-7)\} \\
=a^{2}(2 a-7)(a+2)
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
=a^{2}(a+2)^{2}(2 a-1)(2 a-7)
\]

(xi) \( 3a^{2}− 5ab − 12b^{2}, a^{5}− 27a^{2}b^{3}, 9a^{2}+ 24ab + 16b^{2} \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=3 a^{2}-5 a b-12 b^{2} \\ =3 a^{2}-(9-4)a b-12 b^{2} \\
=3 a^{2}-9 a b+4 a b-12 b^{2} \\
=3 a(a-3 b)+4 b(a-3 b) \\
=(a-3 b)(3 a+4 b)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=a^{5}-27 a^{2} b^{3} \\
=a^{2}\left(a^{3}-27 b^{3}\right) \\
=a^{2}\left\{(a)^{3}-(3 b)^{3}\right\} \\
==a^{2}(a-3 b)(a^{2}+3 a b+9 b^{2})
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=9 a^{2}+24 a b+16 b^{2} \\
=(3 a)^{2}+2.3 a .4 b+(4 b)^{2} \\
=(3 a+4 b)^{2}
\end{array}
\]

\( \therefore \) নির্ণেয় ল.সা.গু.
\[
=a^{2}(a-3 b)(3 a+4 b)^{2}\left(a^{2}+3 a b+9 b^{2}\right)
\]

5. নীচের বীজগাণিতিক সংখ্যামালাগুলির গ.সা.গু. ও ল.সা.গু. নির্ণয় করি—

(i) \( x^{3} − 8, x^{2}+ 3x − 10, x^{3}+ 2x^{2} − 8x \)

সমাধানঃ
প্রথম সংখ্যামালা
\[
\begin{array}{l}
=x^{3}-8 \\
=x^{3}-2^{3} \\
=(x-2)\left(x^{2}+2 x+4\right)
\end{array}
\]

দ্বিতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{2}+3 x-10 \\ =x^{2}+(5-2)x-10 \\
=x^{2}+5 x-2 x-10 \\
=x(x+5)-2(x+5) \\
=(x+5)(x-2)
\end{array}
\]

তৃতীয় সংখ্যামালা
\[
\begin{array}{l}
=x^{3}+2 x^{2}-8 x \\
=x\left(x^{2}+2 x-8\right) \\ =x\{x^{2}+(4-2) x-8\} \\
=x\left(x^{2}+4 x-2 x-8\right) \\
=x\{x(x+4)-2(x+4)\} \\
=x(x+4)(x-2)
\end{array}
\]

\( \therefore \) নির্ণেয় গ.সা.গু. \( =(x-2) \)
এবং ল.সা.গু. \( =x(x-2)(x+4)(x+5)\left(x^{2}+2 x+4\right) \)

আরও দেখুনঃ