1. The problem asks to find the LCM (ল.সা.গু.) and GCD (গ.সা.গু.) of the second and third polynomials: $3x^2 + 7x + 2$ and $6x^2 - x - 1$. 2. First, factorize each polynomial. 3. Factorize $3x^2 + 7x + 2$: - Find two numbers that multiply to $3 \times 2 = 6$ and add to $7$. - These numbers are $6$ and $1$. - Rewrite middle term: $3x^2 + 6x + x + 2$. - Group terms: $(3x^2 + 6x) + (x + 2)$. - Factor each group: $3x(x + 2) + 1(x + 2)$. - Factor out common binomial: $(3x + 1)(x + 2)$. 4. Factorize $6x^2 - x - 1$: - Find two numbers that multiply to $6 \times (-1) = -6$ and add to $-1$. - These numbers are $-3$ and $2$. - Rewrite middle term: $6x^2 - 3x + 2x - 1$. - Group terms: $(6x^2 - 3x) + (2x - 1)$. - Factor each group: $3x(2x - 1) + 1(2x - 1)$. - Factor out common binomial: $(3x + 1)(2x - 1)$. 5. Now, the factorizations are: - $3x^2 + 7x + 2 = (3x + 1)(x + 2)$ - $6x^2 - x - 1 = (3x + 1)(2x - 1)$ 6. The GCD is the product of common factors: - Common factor: $(3x + 1)$ - So, $\text{GCD} = 3x + 1$. 7. The LCM is the product of all factors, taking the highest power of each: - Factors: $(3x + 1)$, $(x + 2)$, $(2x - 1)$ - So, $\text{LCM} = (3x + 1)(x + 2)(2x - 1)$. 8. Final answers: - $\boxed{\text{GCD} = 3x + 1}$ - $\boxed{\text{LCM} = (3x + 1)(x + 2)(2x - 1)}$