Problem

Rewrite \(x^{2}+8x+3\) in vertex form by completing the square. Show the half-coefficient and square calculation, compensate for the added value, and identify the vertex.

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What this problem is really about

Completing the square uses the square of half the signed linear coefficient, with an equal compensation to preserve the expression. We’ll build the perfect-square trinomial, combine the outside constants, read the vertex from x minus h form, and expand to verify exact equivalence.

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Half of 8 is 4, so add and subtract 16: x² + 8x + 16 − 16 + 3 = (x + 4)² − 13. The vertex is (−4, −13), and expansion gives x² + 8x + 3.

(x + 4)² − 13 expands correctly, but the vertex is (4, −13). The vertex x-coordinate is −4.

Add 16 to form (x + 4)² and keep the constant 3, giving (x + 4)² + 3. Adding 16 requires subtracting 16 as compensation.

Half of 8 is 4, so add and subtract 4: x² + 8x + 4 − 4 + 3 = (x + 4)² − 1. The square should use 4² = 16, not 4.

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Rewrite \(x^{2}+8x+3\) in vertex form by completing the square. Show the half-coefficient and square calculation, compensate for the added value, and identify the vertex.

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