1Cademy - Finding the Quotient $$(x^4 - x^2 + 5x

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Inserting Placeholder Terms for Missing Degrees in Polynomial Division

Example

Finding the Quotient $(x^4 - x^2 + 5x - 2) \div (x + 2)$

Apply polynomial long division to divide a degree-4 polynomial that has a missing $x^3$ term by a binomial, yielding a zero remainder: $(x^4 - x^2 + 5x - 2) \div (x + 2)$ .

Step 1 — Insert a placeholder. The dividend $x^4 - x^2 + 5x - 2$ has no $x^3$ term (it skips from degree 4 to degree 2). Add $0x^3$ so the dividend becomes $x^4 + 0x^3 - x^2 + 5x - 2$ . Write this under the long division bracket with $x + 2$ as the divisor.

Step 2 — Divide $x^4$ by $x$ . The result is $x^3$ . Write $x^3$ in the quotient above the $x^3$ column. Multiply $x^3(x + 2) = x^4 + 2x^3$ and write it below. Subtract by changing signs and adding: $(x^4 + 0x^3) - (x^4 + 2x^3) = -2x^3$ . Bring down $-x^2$ to get $-2x^3 - x^2$ .

Step 3 — Divide $-2x^3$ by $x$ . The result is $-2x^2$ . Write $-2x^2$ in the quotient. Multiply $-2x^2(x + 2) = -2x^3 - 4x^2$ and write it below. Subtract: $(-2x^3 - x^2) - (-2x^3 - 4x^2) = 3x^2$ . Bring down $5x$ to get $3x^2 + 5x$ .

Step 4 — Divide $3x^2$ by $x$ . The result is $3x$ . Write $3x$ in the quotient. Multiply $3x(x + 2) = 3x^2 + 6x$ and write it below. Subtract: $(3x^2 + 5x) - (3x^2 + 6x) = -x$ . Bring down $-2$ to get $-x - 2$ .

Step 5 — Divide $-x$ by $x$ . The result is $-1$ . Write $-1$ in the quotient. Multiply $-1(x + 2) = -x - 2$ and write it below. Subtract: $(-x - 2) - (-x - 2) = 0$ .

The quotient is $x^3 - 2x^2 + 3x - 1$ .

Check: Multiply $(x + 2)(x^3 - 2x^2 + 3x - 1)$ ; the result should be $x^4 - x^2 + 5x - 2$ . This example demonstrates how inserting a $0x^3$ placeholder keeps like terms aligned throughout the four division cycles when the dividend has a missing degree term.

Updated 2026-04-21

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OpenStax Elementary Algebra

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