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8. $(2x - 5)(4x^{2} + 6x - 3)$ $8x^{3}-16x^{2}-30$

Question

  1. $(2x - 5)(4x^{2} + 6x - 3)$

$8x^{3}-16x^{2}-30$

Explanation:

Step1: Distribute $2x$ to each term

$2x \cdot 4x^2 + 2x \cdot 6x + 2x \cdot (-3) = 8x^3 + 12x^2 - 6x$

Step2: Distribute $-5$ to each term

$-5 \cdot 4x^2 + (-5) \cdot 6x + (-5) \cdot (-3) = -20x^2 - 30x + 15$

Step3: Combine all resulting terms

$8x^3 + 12x^2 - 6x - 20x^2 - 30x + 15$

Step4: Combine like terms

$8x^3 + (12x^2 - 20x^2) + (-6x - 30x) + 15 = 8x^3 - 8x^2 - 36x + 15$

Answer:

$8x^3 - 8x^2 - 36x + 15$