1. Find HCF by factorization method
(i) \( 21x^2y, 35xy^2 \)
\[21x^2y = 3 \cdot 7 \cdot x^2 \cdot y, \quad 35xy^2 = 5 \cdot 7 \cdot x \cdot y^2\]
\[\text{HCF} = 7xy\]
(ii) \( 4x^2 – 9y^2, 2x^2 – 3xy \)
\[4x^2 – 9y^2 = (2x + 3y)(2x – 3y), \quad 2x^2 – 3xy = x(2x – 3y)\]
\[\text{HCF} = 2x – 3y\]
(iii) \( x^3 – 1, x^2 + x + 1 \)
\[x^3 – 1 = (x – 1)(x^2 + x + 1)\]
\[\text{HCF} = x^2 + x + 1\]
(iv) \( a^3 + 2a^2 – 3a, 2a^3 + 5a^2 – 3a \)
\[a^3 + 2a^2 – 3a = a(a^2 + 2a – 3), \quad 2a^3 + 5a^2 – 3a = a(2a^2 + 5a – 3)\]
\[\text{HCF} = a\]
(v) \( l^2 + 3l – 4, l^2 + 5l + 4, l^2 – 1 \)
\[l^2 + 3l – 4 = (l + 4)(l – 1), \quad l^2 + 5l + 4 = (l + 4)(l + 1), \quad l^2 – 1 = (l – 1)(l + 1)\]
\[\text{HCF} = (l + 1)(l – 1)\]
(vi) \( x^2 + 15x + 56, x^2 + 5x – 24, x^2 + 8x \)
\[x^2 + 15x + 56 = (x + 7)(x + 8), \quad x^2 + 5x – 24 = (x – 3)(x + 8), \quad x^2 + 8x = x(x + 8)\]
\[\text{HCF} = x + 8\]
2. Find HCF by division method
(i) \( 27x^3 + 9x^2 – 3x – 9, 3x – 2 \)
\[\text{HCF} = 3x – 2\]
(ii) \( x^3 – 9x^2 + 21x – 15, x^2 – 4x + 3 \)
\[x^2 – 4x + 3 = (x – 3)(x – 1)\]
Divide \( x^3 – 9x^2 + 21x – 15 \) by \( x^2 – 4x + 3 \):
\[x^3 – 9x^2 + 21x – 15 = (x – 3)(x^2 – 4x + 3)\]
\[\text{HCF} = x – 3\]
(iii) \( 2x^3 + 2x^2 + 2x + 2, 6x^3 + 12x^2 + 6x + 12 \)
\[2x^3 + 2x^2 + 2x + 2 = 2(x^3 + x^2 + x + 1)\]
\[6x^3 + 12x^2 + 6x + 12 = 6(x^3 + 2x^2 + x + 2)\]
\[\text{HCF} = 2\]
(iv) \( 2x^3 – 4x^2 + 6x, x^3 – 2x, 3x^2 – 6x \)
\[\text{Factorize: } 2x^3 – 4x^2 + 6x = 2x(x^2 – 2x + 3), \quad x^3 – 2x = x(x^2 – 2)\]
\[3x^2 – 6x = 3x(x – 2)\]
\[\text{HCF} = x\]
3. Find LCM by prime factorization method
(i) \( 2a^2b, 4ab^2, 6ab \)
\[2a^2b = 2 \cdot a^2 \cdot b, \quad 4ab^2 = 2^2 \cdot a \cdot b^2, \quad 6ab = 2 \cdot 3 \cdot a \cdot b\]
\[\text{LCM} = 12a^2b^2\]
(ii) \( x^2 + x, x^3 + x^2 \)
\[x^2 + x = x(x + 1), \quad x^3 + x^2 = x^2(x + 1)\]
\[\text{LCM} = x^2(x + 1)\]
(iii) \( a^2 – 4a + 4, a^2 – 2a \)
\[a^2 – 4a + 4 = (a – 2)^2, \quad a^2 – 2a = a(a – 2)\]
\[\text{LCM} = a(a – 2)^2\]
(iv) \( x^4 – 16, x^3 – 4x \)
\[x^4 – 16 = (x^2 – 4)(x^2 + 4) = (x – 2)(x + 2)(x^2 + 4)\]
\[x^3 – 4x = x(x – 2)(x + 2)\]
\[\text{LCM} = x(x – 2)(x + 2)(x^2 + 4)\]
(v) \( 16 – 4x^2, x^2 + x – 6, 4 – x^2 \)
\[16 – 4x^2 = 4(4 – x^2) = 4(x – 2)(x + 2)\]
\[x^2 + x – 6 = (x + 3)(x – 2), \quad 4 – x^2 = (2 – x)(2 + x) = -(x – 2)(x + 2)\]
\[\text{LCM} = 4(x – 2)(x + 2)(x + 3)\]
4. The HCF of two polynomials is \( y – 7 \) and their LCM is \( y^3 – 10y^2 + 11y + 70 \). If one of the polynomials is \( y^2 – 5y – 14 \), find the other.
\[\text{HCF} \cdot \text{LCM} = \text{Product of the two polynomials}\]
Let the other polynomial be \( P(y) \):
\[(y – 7) \cdot (y^3 – 10y^2 + 11y + 70) = (y^2 – 5y – 14) \cdot P(y)\]
Factorize \( y^2 – 5y – 14 = (y – 7)(y + 2) \):
\[P(y) = \frac{(y – 7)(y^3 – 10y^2 + 11y + 70)}{(y – 7)(y + 2)}\]
Cancel \( y – 7 \):
\[P(y) = \frac{y^3 – 10y^2 + 11y + 70}{y + 2}\]
\[P(y) = y^2 – 12y + 35\]
5. The LCM and HCF of two polynomials \( p(x) \) and \( q(x) \) are \( 36x^3(x + a)(x^3 – a^3) \) and \( x^2(x – a) \), respectively. If \( p(x) = 4x^2(x^2 – a^2) \), find \( q(x) \).
\[\text{HCF} \cdot \text{LCM} = p(x) \cdot q(x)\]
\[x^2(x – a) \cdot 36x^3(x + a)(x^3 – a^3) = 4x^2(x^2 – a^2) \cdot q(x)\]
\[x^2(x – a) \cdot 36x^3(x + a)(x – a)(x^2 + ax + a^2) = 4x^2(x – a)(x + a) \cdot q(x)\]
Cancel \( x^2(x – a) \):
\[36x^3(x + a)(x – a)(x^2 + ax + a^2) = 4(x + a) \cdot q(x)\]
Cancel \( x + a \):
\[36x^3(x – a)(x^2 + ax + a^2) = 4q(x)\]
Divide by 4:
\[q(x) = 9x^3(x – a)(x^2 + ax + a^2)\]
6. The HCF and LCM of two polynomials is \( (x + a) \) and \( 12x^2(x + a)(x^2 – a^2) \), respectively. Find the product of the two polynomials.
\[\text{HCF} \cdot \text{LCM} = \text{Product of the two polynomials}\]
\[(x + a) \cdot 12x^2(x + a)(x^2 – a^2) = \text{Product of the two polynomials}\]
\[(x + a) \cdot 12x^2(x + a)(x – a)(x + a) = \text{Product of the two polynomials}\]
\[12x^2(x + a)^3(x – a)\]
