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

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

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

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

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

$y^2 - 5y + 2$ $imes \quad y - 3$

Step 2 — Multiply $(y^2 - 5y + 2)$ by $-3$ : Compute $-3 \cdot y^2 = -3y^2$ , $-3 \cdot (-5y) = 15y$ , and $-3 \cdot 2 = -6$ . Write the first partial product:

$-3y^2 + 15y - 6$

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

$y^3 - 5y^2 + 2y$

Step 4 — Add the partial products: Combine like terms column by column. The $y^3$ column: $y^3$ . The $y^2$ column: $-5y^2 - 3y^2 = -8y^2$ . The $y$ column: $2y + 15y = 17y$ . The constant column: $-6$ :

$y^3 - 8y^2 + 17y - 6$

The result is $y^3 - 8y^2 + 17y - 6$ .

Updated 2026-04-29

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OpenStax Intermediate Algebra

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