P2 2.1 - Dividing polynomials

Quotient and remainder

The quotient is the polynomial produced by division. The remainder is what is left after no further polynomial term can be taken out. With a nonzero linear divisor $ax+b$, the remainder is constant, and $a\ne0$.

Worked example 1: a monic linear divisor

Question. Divide $2x^3+3x^2-5x-6$ by $x+1$.

Plan. Divide leading terms, multiply the whole divisor, subtract, and repeat.

First, $2x^3\div x=2x^2$. Subtract its product:

Next, $x^2\div x=x$:

Finally, $-6x\div x=-6$:

Answer. Quotient $2x^2+x-6$, remainder $0$.

Check. Expanding $(x+1)(2x^2+x-6)$ gives $2x^3+3x^2-5x-6$, the original dividend.

Worked example 2: a non-monic linear divisor

Question. Divide $2x^3+3x^2-5x-6$ by $2x+2$.

Plan. Divide each leading term by $2x$, retaining fractional coefficients where needed.

Answer. Quotient $x^2+\frac12x-3$, remainder $0$.

Check. Since $2x+2=2(x+1)$, multiplying the first quotient by $1/2$ gives exactly this quotient; the reconstructed product is unchanged.

Watch out

Subtract the entire product. For example, $(-6x-6)-(-6x-6)=0$, because both signs change inside the second bracket. Subtracting just its first term gives a false remainder.

Division identity

For a polynomial $f(x)$ divided by $ax+b$, with $a\ne0$,

where $q(x)$ is the quotient and $R$ is constant. As a fraction,

The polynomial identity itself holds for every $x$; only the fractional form excludes a zero denominator.

Not in the formula booklet: learn it.

Worked example 3: a missing power and a remainder

Question. Find the quotient and remainder when $x^3-4x+5$ is divided by $x-2$, and write the division identity.

Plan. Include $0x^2$, then remove the leading term at each stage.

No multiple of $x-2$ with a nonzero polynomial quotient can remove the constant $5$ without reintroducing a power of $x$.

Answer. Quotient $x^2+2x$, remainder $5$, so

Check. At $x=0$, the identity gives $5=(-2)\times0+5$. Expanding the product also recovers every coefficient of the dividend.

Worked example 4: a negative remainder

Question. Divide $x^3-4x-1$ by $x-2$, and express the answer as a fraction.

Plan. Use the same leading-term subtractions and retain the final constant with its sign.

Answer. Quotient $x^2+2x$, remainder $-1$, and

Check. At $x=3$, the original fraction is $14/1=14$; the answer gives $9+6-1=14$.

Watch out

A remainder $R$ becomes $R/(ax+b)$ in the fractional answer, not an extra constant $R$ added to the quotient. Multiplying back shows why.

You already know

Multiply every term in one bracket by every term in the other, then collect equal powers. Retain a zero coefficient when a power is absent.

Worked example 5: verifying an exact division

Question. Verify that dividing $x^3+x^2-4x-4$ by $x+1$ gives quotient $x^2-4$ and remainder $0$.

Plan. Reconstruct the dividend from divisor, quotient and remainder.

All four coefficients agree with the dividend.

Answer. The stated quotient and remainder are correct.

Check. At $x=2$, the dividend is $0$, and the product is $3\times0=0$, consistent with the full coefficient check.

Worked example 6: repairing a remainder

Question. A student says that $x^3+x^2-4x-2$ divided by $x+1$ has quotient $x^2-4$ and remainder $0$. Check the claim and give the correct remainder.

Plan. Expand the proposed product and compare its constant with the dividend's constant.

The variable coefficients agree, but the required constant is $-2$, which is $2$ greater than $-4$. Therefore

Answer. The quotient is correct; the remainder is $2$, not $0$.

Check. At $x=0$, the repaired identity gives $-2=1(-4)+2$. The proposed zero remainder would have given $-4$.

Exam tip

State both “quotient” and “remainder”, or write the complete identity. A bare quadratic leaves unclear whether it is the quotient or a claimed factorisation of the dividend.