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 $(x^4 - x^2 + 5x - 6) \div (x + 2)$

Apply polynomial long division to divide a degree-four polynomial by a binomial, utilizing a placeholder for a missing term: $(x^4 - x^2 + 5x - 6) \div (x + 2)$ .

Step 1 — Insert placeholders. The dividend is missing an $x^3$ term. Rewrite it in standard form with a placeholder: $x^4 + 0x^3 - x^2 + 5x - 6$ . Set this up under the long division bracket with $x + 2$ outside.

Step 2 — Divide $x^4$ by $x$ . The result is $x^3$ . Write $x^3$ in the quotient. Multiply $x^3(x + 2) = x^4 + 2x^3$ and align it beneath the dividend. Subtract $(x^4 + 0x^3) - (x^4 + 2x^3) = -2x^3$ . 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 $-6$ to get $-x - 6$ .

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 $(-x - 6) - (-x - 2) = -4$ . This is the remainder.

Step 6 — Express the remainder as a fraction. Write the remainder over the divisor: $-\frac{4}{x + 2}$ .

The quotient is $x^3 - 2x^2 + 3x - 1 - \frac{4}{x + 2}$ .

To check, multiply $(x + 2)\left(x^3 - 2x^2 + 3x - 1 - \frac{4}{x + 2}\right)$ ; the result will be $x^4 - x^2 + 5x - 6$ . This example demonstrates the critical importance of adding a $0x^3$ placeholder to maintain proper column alignment when subtracting like terms.

Updated 2026-04-29

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OpenStax Intermediate Algebra

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