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**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) \(3 u^{2}+13 u v-10 v^{2} = (u+5v)(3u-2v)\)**
Solución
**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:
\[
(u+5v)(3u-2v)
\]
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