Mathematics 9709/22 — October/November 2019

Walkthrough

The divisor is $x^2+2$, a quadratic. At each stage of long division we compare the leading term of what is left with the leading term $x^2$ of the divisor.

First $x^4 \div x^2 = x^2$. Multiplying $x^2(x^2+2)$ gives $x^4+2x^2$. Subtracting removes the $x^4$ and $x^2$ terms from the dividend, leaving $-3x^3 + 3x^2 - 6x + 11$.

Then $-3x^3 \div x^2 = -3x$. Multiplying by $-3x$ gives $-3x^3 - 6x$. Subtracting cancels both the $-3x^3$ and the $-6x$, leaving $3x^2+11$.

Finally $3x^2 \div x^2 = 3$. Multiplying by $3$ gives $3x^2+6$. Subtracting leaves remainder $5$, which has degree $0$, less than the degree of the divisor. The quotient is the sum of the quotient terms, $x^2 - 3x + 3$. So the answer satisfies

Key Takeaways

This question tests polynomial division by a quadratic divisor. The divisor is missing an $x$ term, so terms must be aligned by degree in the working. The final identity $f(x) = q(x)d(x) + r(x)$ is a useful check. It also reinforces that the degree of the remainder must be lower than the degree of the divisor.

Common Mistakes

• Subtracting only the first term of the product instead of the whole product.
• Losing the $x$ terms: when subtracting $-6x$, a common sign error gives $-12x$ instead of $0x$.
• Forgetting to carry the remaining $+11$ down when forming $3x^2 + 11$.
• Giving only a final answer without showing the division; the method mark requires dividing at least as far as the $x$ term in the quotient.

Things to Be Careful About

Align like terms so cancellations are visible. The divisor is $x^2+2$, not $x^2+0x+2$ in written form, but the zero $x$ term is still present when subtracting. The remainder has degree $0$, which is acceptable because it is lower than $2$. If you use the checking identity, expand the right-hand side to verify it equals $x^4 - 3x^3 + 5x^2 - 6x + 11$.