1Cademy - Multiplying $$(x + 4)(2x^2 - 3x + 5)$$ Using the Vertical Method

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Multiplying $(b + 3)(2b^2 - 5b + 8)$ Using the Vertical Method

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Multiplying $(x + 4)(2x^2 - 3x + 5)$ Using the Vertical Method

Multiply $(x + 4)(2x^2 - 3x + 5)$ using the Vertical Method.

Step 1 — Set up vertically: Place the trinomial on top and the binomial on the bottom to minimize partial products:

$2x^2 - 3x + 5$ $imes \quad x + 4$

Step 2 — Multiply $(2x^2 - 3x + 5)$ by $4$ : Compute $4 \cdot 2x^2 = 8x^2$ , $4 \cdot (-3x) = -12x$ , and $4 \cdot 5 = 20$ . Write the first partial product:

$8x^2 - 12x + 20$

Step 3 — Multiply $(2x^2 - 3x + 5)$ by $x$ : Compute $x \cdot 2x^2 = 2x^3$ , $x \cdot (-3x) = -3x^2$ , and $x \cdot 5 = 5x$ . Write the second partial product, aligning like terms beneath the first:

$2x^3 - 3x^2 + 5x$

Step 4 — Add the partial products: Combine like terms column by column. The $x^3$ column: $2x^3$ . The $x^2$ column: $-3x^2 + 8x^2 = 5x^2$ . The $x$ column: $5x - 12x = -7x$ . The constant column: $20$ :

$2x^3 + 5x^2 - 7x + 20$

The result is $2x^3 + 5x^2 - 7x + 20$ .

Updated 2026-04-29

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OpenStax Intermediate Algebra

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