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) \( 3 u^{2}+13 u v-10 v^{2} \)

**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: \[ (u+5v)(3u-2v) \]