1Cademy - Finding the Quotient $$ (8a^3 + 27) \div (2a + 3) $$

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Inserting Placeholder Terms for Missing Degrees in Polynomial Division

Example

Finding the Quotient $(8a^3 + 27) \div (2a + 3)$

Apply polynomial long division to divide a two-term polynomial by a binomial whose leading coefficient is not $1$ , requiring both coefficient division and multiple placeholder terms: $(8a^3 + 27) \div (2a + 3)$ . Step 1 - Insert placeholders. The dividend $8a^3 + 27$ jumps from degree 3 directly to degree 0, so it is missing both the $a^2$ and $a$ terms. Rewrite it as $8a^3 + 0a^2 + 0a + 27$ . Place this under the long division bracket with $2a + 3$ outside. Step 2 - Divide $8a^3$ by $2a$ . The result is $4a^2$ (dividing both the coefficient and the variable: $8 \div 2 = 4$ and $a^3 \div a = a^2$ ). Write $4a^2$ in the quotient. Multiply $4a^2(2a + 3) = 8a^3 + 12a^2$ and write it below. Subtract: $(8a^3 + 0a^2) - (8a^3 + 12a^2) = -12a^2$ . Bring down $0a$ to get $-12a^2 + 0a$ . Step 3 - Divide $-12a^2$ by $2a$ . The result is $-6a$ . Write $-6a$ in the quotient. Multiply $-6a(2a + 3) = -12a^2 - 18a$ and write it below. Subtract: $(-12a^2 + 0a) - (-12a^2 - 18a) = 18a$ . Bring down $27$ to get $18a + 27$ . Step 4 - Divide $18a$ by $2a$ . The result is $9$ . Write $9$ in the quotient. Multiply $9(2a + 3) = 18a + 27$ and write it below. Subtract: $(18a + 27) - (18a + 27) = 0$ . The quotient is $4a^2 - 6a + 9$ . Check: Multiply $(2a + 3)(4a^2 - 6a + 9)$ ; the result should be $8a^3 + 27$ . This example illustrates two new challenges: when the divisor's leading term has a coefficient other than $1$ (here $2a$ ), each "divide" step requires dividing both the coefficient and the variable part of the current leading term. Additionally, the dividend requires two placeholder terms instead of just one.

Updated 2026-06-26

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