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\( \left. \begin{array} { | l | l | } \hline \text { Multiply each of the following. } \\ \hline ( x - 1 ) ( 7 x ^ { 2 } + 2 x - 5 ) & { ( x + 4 ) ( 5 x - 7 ) } \\ \hline \end{array} \right. \)

Ask by Gross Romero. in the United States
Mar 18,2025

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The first expression \( (x - 1)(7x^{2} + 2x - 5) \) simplifies to \( 7x^{3} - 5x^{2} - 7x + 5 \), and the second expression \( (x + 4)(5x - 7) \) simplifies to \( 5x^{2} + 13x - 28 \).

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The Deep Dive

To tackle the expression \((x - 1)(7x^2 + 2x - 5)\), use the distributive property (also known as the FOIL method for binomials). 1. Distributing \(x\): \[ x(7x^2) + x(2x) + x(-5) = 7x^3 + 2x^2 - 5x \] 2. Now distributing \(-1\): \[ -1(7x^2) - 1(2x) - 1(-5) = -7x^2 - 2x + 5 \] Combining these results: \[ 7x^3 + (2x^2 - 7x^2) + (-5x - 2x) + 5 = 7x^3 - 5x^2 - 7x + 5 \] Next, for the expression \((x + 4)(5x - 7)\): 1. Distributing \(x\): \[ x(5x) + x(-7) = 5x^2 - 7x \] 2. Now distributing \(4\): \[ 4(5x) + 4(-7) = 20x - 28 \] Combining these results: \[ 5x^2 + (-7x + 20x) - 28 = 5x^2 + 13x - 28 \] So the final expressions are: \[ (x - 1)(7x^2 + 2x - 5) = 7x^3 - 5x^2 - 7x + 5 \] \[ (x + 4)(5x - 7) = 5x^2 + 13x - 28 \]

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