Question: Find the HCF of the following by division method.
3. $a^2 + a - 2$, $a^3 + 2a^2 + a + 2$
4. $x^3 + 2x^2 - 4x - 8$, $2x^3 + 7x^2 + 4x - 4$
5. $2x^3 + x^2 - x - 2$, $3x^3 - x^2 + x - 3$
6. $3 + 2p^4 + 5p^2$, $5p + 5p^3 - 3 + 3p^2$
7. $24x^4 - 2x^3 + 60x^2 - 32x$, $18x^4 - 6x^3 - 39x^2 - 18x$
8. $2x^3 + 6x^2 + x + 3$, $3x^3 + 9x^2 - 2x - 6$, $3x^3 + 3x^2 + 2x + 6$
Find the LCM of the following expressions.
9. $9a^2b - b$, $6a^2 + 2a$
10. $p^3q - pq^3$, $5p^2q^2 - p^2q^5$
11. $4x^2y - y$, $2x^2 + x$
12. $x^3 - x - 6$, $x^2 + x - 2$, $x^2 - 4x + 3$
13. $m^6 - 1$, $m^4 - 1$, $m^3 - 1$
14. $x^3 + 2x^2 - x - 2$, $x^2 - x - 2$, $x^2 - 4$
15. $x^2 + x - 20$, $x^2 - 10x + 24$, $x^2 - x - 30$
1. **Problem:** Find the HCF of $a^2 + a - 2$ and $a^3 + 2a^2 + a + 2$ by division method.
2. **Step 1: Write down the polynomials:**
$$f(a) = a^3 + 2a^2 + a + 2$$
$$g(a) = a^2 + a - 2$$
3. **Step 2: Divide $f(a)$ by $g(a)$:**
Divide $a^3 + 2a^2 + a + 2$ by $a^2 + a - 2$.
Leading term division: $\frac{a^3}{a^2} = a$
Multiply divisor by $a$: $a(a^2 + a - 2) = a^3 + a^2 - 2a$
Subtract:
$$(a^3 + 2a^2 + a + 2) - (a^3 + a^2 - 2a) = (2a^2 - a^2) + (a + 2a) + 2 = a^2 + 3a + 2$$
4. **Step 3: Divide the remainder $a^2 + 3a + 2$ by $g(a)$:**
Leading term division: $\frac{a^2}{a^2} = 1$
Multiply divisor by $1$: $a^2 + a - 2$
Subtract:
$$(a^2 + 3a + 2) - (a^2 + a - 2) = (3a - a) + (2 + 2) = 2a + 4$$
5. **Step 4: Now divide $g(a) = a^2 + a - 2$ by the remainder $2a + 4$:**
Leading term division: $\frac{a^2}{2a} = \frac{a}{2}$
Multiply $2a + 4$ by $\frac{a}{2}$:
$$\frac{a}{2} (2a + 4) = a^2 + 2a$$
Subtract:
$$(a^2 + a - 2) - (a^2 + 2a) = (a - 2a) - 2 = -a - 2$$
6. **Step 5: Divide $2a + 4$ by $-a - 2$:**
Leading term division: $\frac{2a}{-a} = -2$
Multiply $-a - 2$ by $-2$:
$$-2(-a - 2) = 2a + 4$$
Subtract:
$$(2a + 4) - (2a + 4) = 0$$
7. **Step 6: Since remainder is zero, the last non-zero remainder is the HCF:**
HCF is $-a - 2$, which can be written as $a + 2$ (ignoring sign).
**Final answer:**
$$\boxed{a + 2}$$
This completes the first problem as per the guest rule to solve only the first question.