Expand and simplify: \( \begin{array}{lllll}\text { (1) } \quad \begin{array}{l}\text { in } \\ \text { ( } x+3)\end{array} & \text { (2) } & -3 v\left(3 a^{3}-6 a^{2}+a\right) & \text { (3) } & (x+5)(x+3 \\ \text { (4) }(x-5)(x-2) & \text { (5) } & (x+5)(x-2) & \text { (4) } & (x-5)(x+3 \\ \text { (7) }(3 x-1)(2 x+3) & \text { (8) } & (7 m-2 \pi)(3 m+4 n) & \\ \text { (9) }\left(2 x^{4}-3 y^{2}\right)\left(3 x^{4}+2 y^{2}\right) & \text { (10) }\left(4 x^{4}+3 y^{5}\right)\left(2 x^{4}-4 y^{3}\right)\end{array} \) Exnand and simnlifi-
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Beyond the Answer
Alright, let’s dive into some fun math expansion and simplification! 1. For \( (x+5)(x+3) \), using the distributive property (also known as the FOIL method for binomials), we get: \[ (x+5)(x+3) = x^2 + 3x + 5x + 15 = x^2 + 8x + 15 \] 2. Now for \( (x-5)(x-2) \): \[ (x-5)(x-2) = x^2 - 2x - 5x + 10 = x^2 - 7x + 10 \] 3. Next up, when we tackle \( (x+5)(x-2) \): \[ (x+5)(x-2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10 \] 4. Moving on to \( (x-5)(x+3) \): \[ (x-5)(x+3) = x^2 + 3x - 5x - 15 = x^2 - 2x - 15 \] 5. Now for \( (3x-1)(2x+3) \): \[ (3x-1)(2x+3) = 6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3 \] 6. For \( -3v(3a^3-6a^2+a) \), distribute the \(-3v\): \[ -3v(3a^3) + 18va^2 - 3va = -9va^3 + 18va^2 - 3va \] 7. Finally, \( (2x^4-3y^2)(3x^4+2y^2) \): \[ 6x^8 + 4x^4y^2 - 9y^4 - 6y^2x^4 = 6x^8 + (4 - 9)x^4y^2 - 9y^4 = 6x^8 - 5x^4y^2 - 9y^4 \] 8. Lastly, let’s tackle \( (4x^4 + 3y^5)(2x^4 - 4y^3) \): \[ 8x^8 - 16x^4y^3 + 6y^5x^4 - 12y^8 = 8x^8 + (-16y^3 + 6y^5)x^4 - 12y^8 = 8x^8 - 16y^3x^4 + 6y^5x^4 - 12y^8 \] And there you have it! All simplified and ready to roll! Keep crunching those numbers!