26) $$ 2 x^{3} y+12 x^{2} y+18 x y $$ 28) $$ 2 x^{2}+8 x+6 $$ 30) $$ 24 x^{\wedge} 2-52 x y+8 y^{\wedge} 2 $$ 32) $$ 3 u^{2}+13 u v-10 v^{2} $$

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**26) Factor $$2 x^{3} y+12 x^{2} y+18 x y$$**

1. Notice that each term has a common factor of $$2xy$$. Factor it out:
   $$
   2 x^{3} y+12 x^{2} y+18 x y = 2xy\left(x^2+6x+9\right)
   $$
2. Recognize that the quadratic $$x^2+6x+9$$ is a perfect square:
   $$
   x^2+6x+9 = \left(x+3\right)^2
   $$
3. Hence, the complete factorization is:
   $$
   2xy\left(x+3\right)^2
   $$

---

**28) Factor $$2 x^{2}+8 x+6$$**

1. First, factor out the greatest common factor, which is $$2$$:
   $$
   2 x^{2}+8 x+6 = 2\left(x^2+4x+3\right)
   $$
2. Now, factor the quadratic $$x^2+4x+3$$. Look for two numbers that multiply to $$3$$ and add to $$4$$ (these are $$1$$ and $$3$$):
   $$
   x^2+4x+3 = \left(x+1\right)\left(x+3\right)
   $$
3. Therefore, the complete factorization is:
   $$
   2\left(x+1\right)\left(x+3\right)
   $$

---

**30) Factor $$24 x^{2}-52 x y+8 y^{2}$$**

1. Notice that each term has a common factor of $$4$$. Factor it out:
   $$
   24 x^{2}-52 x y+8 y^{2} = 4\left(6x^2-13xy+2y^2\right)
   $$
2. Factor the quadratic $$6x^2-13xy+2y^2$$. Look for two numbers that multiply to $$6 \cdot 2 = 12$$ (with the $$y^2$$ present, consider them as coefficients along with $$x$$ and $$y$$) and add to $$-13$$. The suitable pair is $$-12$$ and $$-1$$ because $$-12+(-1)=-13$$ and $$-12\cdot -1=12$$.
3. Rewrite the middle term using these numbers:
   $$
   6x^2 -13xy+2y^2 = 6x^2-12xy-xy+2y^2
   $$
4. Factor by grouping:
   - Group the first two terms and the last two terms:
     $$
     (6x^2-12xy)- (xy-2y^2)
     $$
   - Factor out $$6x$$ from the first group and $$-y$$ from the second group:
     $$
     6x(x-2y)-y(x-2y)
     $$
   - Notice the common factor $$(x-2y)$$:
     $$
     (x-2y)(6x-y)
     $$
5. Reintroduce the common factor $$4$$:
   $$
   4\left(6x-y\right)\left(x-2y\right)
   $$

---

**32) Factor $$3 u^{2}+13 u v-10 v^{2}$$**

1. Factor the quadratic $$3u^2+13uv-10v^2$$. Look for two numbers that multiply to $$3 \times (-10) = -30$$ and add up to $$13$$. The numbers $$15$$ and $$-2$$ satisfy this because $$15+(-2)=13$$ and $$15\cdot (-2)=-30$$.
2. Rewrite the middle term using these numbers:
   $$
   3u^2+13uv-10v^2 = 3u^2+15uv-2uv-10v^2
   $$
3. Factor by grouping:
   - Group the first two terms and the last two terms:
     $$
     (3u^2+15uv) + (-2uv-10v^2)
     $$
   - Factor out $$3u$$ from the first group and $$-2v$$ from the second group:
     $$
     3u(u+5v)-2v(u+5v)
     $$
   - Notice the common factor $$(u+5v)$$:
     $$
     (u+5v)(3u-2v)
     $$
4. Therefore, the complete factorization is:
   $$
   (u+5v)(3u-2v)
   $$
**26) $$2 x^{3} y+12 x^{2} y+18 x y = 2xy(x+3)^2$$** **28) $$2 x^{2}+8 x+6 = 2(x+1)(x+3)$$** **30) $$24 x^{2}-52 x y+8 y^{2} = 4(6x-y)(x-2y)$$** **32) $$3 u^{2}+13 u v-10 v^{2} = (u+5v)(3u-2v)$$**

26) \( 2 x^{3} y+12 x^{2} y+18 x y \) 28) \( 2 x^{2}+8 x+6 \) 30) \( 24 x^{\wedge} 2-52 x y+8 y^{\wedge} 2 \) 32) \( 3 u^{2}+13 u v-10 v^{2} \)

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