A polynomial in one variable $x$ is an expression of the form $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$Where $n \in \mathbb{N}_0$ $a_n, a_{n-1}, \ldots, a_0 \in \mathbb{R}$And $a_n \neq 0$. Polynomials and their manipulation Form a core part of the DSE compulsory mathematics syllabus, with applications ranging from Algebraic identities to combinatorial coefficient extraction.

Polynomial Basics

Definition and Terminology

A polynomial $f(x)$ of degree $n$ is written in standard form (descending powers of $x$):

$$F(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, \quad a_n \neq 0$$

• $a_n$ is the leading coefficient.
• $a_0$ is the constant term.
• The degree of $f(x)$ is the highest power of $x$ with a non-zero coefficient.
• A polynomial of degree 0 is a non-zero constant; the zero polynomial has undefined degree.

Polynomial Identities

A polynomial identity is an equality that holds for all values of the variable. Two polynomials $f(x)$ and $g(x)$ are identical (written $f(x) \equiv g(x)$) if and only if the coefficients of Corresponding powers of $x$ are equal.

Key identities at DSE level

- $(a+b)^2 = a^2 + 2ab + b^2$ - $(a-b)^2 = a^2 - 2ab + b^2$ - $(a+b)(a-b) = a^2 - b^2$ - $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ - $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ - $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ - $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ - $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Example: Using the method of undetermined coefficients

Find constants $A$, $B$, $C$ such that $x^2 + 4x + 6 \equiv A(x-1)^2 + B(x-1) + C$.

Expanding the right-hand side:

$$A(x^2 - 2x + 1) + Bx - B + C = Ax^2 + (-2A + B)x + (A - B + C)$$

Equating coefficients:

• $x^2$: $A = 1$
• $x$: $-2A + B = 4 \implies B = 6$
• constant: $A - B + C = 6 \implies 1 - 6 + C = 6 \implies C = 11$

Therefore $x^2 + 4x + 6 \equiv (x-1)^2 + 6(x-1) + 11$.

---

Binomial Theorem

Statement

For any positive integer $n$

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Where the binomial coefficient is

$$\binom{n}{k} = _n C_k = \frac{n!}{k!(n-k)!}$$

This is known as the Binomial Theorem. See also combinatorial notation.

Pascal”s Triangle

The binomial coefficients $\binom{n}{k}$ for successive values of $n$ form Pascal’s triangle:

\begin{array}{c@{\hspace{12pt}}c@{\hspace{12pt}}c@{\hspace{12pt}}c@{\hspace{12pt}}c@{\hspace{12pt}}c} & & & & 1 & \\ & & & 1 & & 1 \\ & & 1 & & 2 & & 1 \\ & 1 & & 3 & & 3 & & 1 \\ 1 & & 4 & & 6 & & 4 & & 1 \end{array}

Each entry is the sum of the two entries directly above it, reflecting the recurrence relation $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$.

Properties of Binomial Coefficients

1. Symmetry: $\displaystyle \binom{n}{k} = \binom{n}{n-k}$

2. Recurrence (Pascal’s identity): $\displaystyle \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

3. Sum of coefficients: Setting $a = b = 1$ in the binomial theorem,

$$\sum_{k=0}^{n} \binom{n}{k} = 2^n$$

4. Alternating sum: Setting $a = 1, b = -1$

$$\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$$

Expanding $(1+x)^n$

The expansion of $(1+x)^n$ is a frequently tested form:

$$(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n$$

The general term (the $(r+1)$-th term) is:

$$T_{r+1} = \binom{n}{r} x^r, \quad r = 0, 1, 2, \ldots, n$$

Expanding $(a + bx)^n$

For $(a + bx)^n$The general term is:

$$T_{r+1} = \binom{n}{r} a^{n-r} (bx)^r = \binom{n}{r} a^{n-r} b^r x^r$$

To find the coefficient of $x^k$Set $r = k$ and evaluate:

$$[\mathrm{coefficient of } x^k] = \binom{n}{k} a^{n-k} b^k$$

Example: Finding a specific coefficient

Find the coefficient of $x^3$ in the expansion of $(2 - 3x)^7$.

The general term is $T_{r+1} = \binom{7}{r} 2^{7-r}(-3x)^r$.

For the $x^3$ term, set $r = 3$:

$$\binom{7}{3} \cdot 2^4 \cdot (-3)^3 = 35 \cdot 16 \cdot (-27) = -15\,120$$

The coefficient of $x^3$ is $-15\,120$.

Example: Finding the constant term

Find the constant term in the expansion of $\left(x + \dfrac{2}{x}\right)^6$.

The general term is $T_{r+1} = \binom{6}{r} x^{6-r} \left(\dfrac{2}{x}\right)^r = \binom{6}{r} \cdot 2^r \cdot x^{6-2r}$.

For the constant term, $6 - 2r = 0 \implies r = 3$:

$$\binom{6}{3} \cdot 2^3 = 20 \cdot 8 = 160$$

The constant term is $160$.

Example: Finding the middle term

Find the middle term in the expansion of $\left(1 + \dfrac{x}{2}\right)^{10}$.

Since $n = 10$ (even), there is one middle term at position $\dfrac{n}{2} + 1 = 6$I.e. $r = 5$:

$$T_6 = \binom{10}{5} \left(\frac{x}{2}\right)^5 = 252 \cdot \frac{x^5}{32} = \frac{63}{8} x^5$$

---

Polynomial Division

Long Division

Given two polynomials $f(x)$ (dividend) and $g(x)$ (divisor) with $\deg g(x) \geq 1$Polynomial Long division yields a quotient $q(x)$ and a remainder $r(x)$ such that

$$F(x) = g(x) \cdot q(x) + r(x)$$

Where $\deg r(x) < \deg g(x)$ or $r(x) = 0$.

Example: Long division

Divide $f(x) = 2x^3 + 3x^2 - 5x + 7$ by $g(x) = x^2 - x + 2$. $$\begin{array}{r|l} X^2 - x + 2 & 2x^3 + 3x^2 - 5x + 7 \\ \hline & 2x \\ & 2x^3 - 2x^2 + 4x \\ \cline{2-2} & \phantom{2x^3} 5x^2 - 9x + 7 \\ & \phantom{2x^3} 5x^2 - 5x + 10 \\ & \cline{2-2} & \phantom{2x^3} \phantom{5x^2} -4x - 3 \end{array}$$

Quotient: $q(x) = 2x + 5$Remainder: $r(x) = -4x - 3$.

Verification: $(x^2 - x + 2)(2x + 5) + (-4x - 3) = 2x^3 + 3x^2 - 5x + 7$.

Remainder Theorem

When a polynomial $f(x)$ is divided by $(x - c)$The remainder is $f(c)$.

Proof. By the division algorithm, $f(x) = (x-c) \cdot q(x) + r$ where $r$ is a constant (since $\deg r < \deg(x-c) = 1$). Substituting $x = c$: $f(c) = 0 \cdot q(c) + r = r$.

For a divisor of the form $(ax - b)$Set $x = \dfrac{b}{a}$ to obtain the remainder $f\!\left(\dfrac{b}{a}\right)$.

Example: Remainder theorem

Find the remainder when $f(x) = 3x^4 - 2x^3 + x - 5$ is divided by $(x - 2)$.

By the Remainder Theorem, the remainder is $f(2)$:

$$F(2) = 3(16) - 2(8) + 2 - 5 = 48 - 16 + 2 - 5 = 29$$

Example: Remainder with a non-monic linear divisor

Find the remainder when $f(x) = 2x^3 - 5x + 3$ is divided by $(2x + 1)$.

Set $2x + 1 = 0 \implies x = -\dfrac{1}{2}$. The remainder is:

$$F\!\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 5\left(-\frac{1}{2}\right) + 3 = -\frac{1}{4} + \frac{5}{2} + 3 = \frac{21}{4}$$

Factor Theorem

$(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$.

This follows directly from the Remainder Theorem: if the remainder $f(c) = 0$Then $f(x) = (x-c) \cdot q(x)$So $(x-c)$ divides $f(x)$ exactly.

Corollary. $(ax - b)$ is a factor of $f(x)$ if and only if $f\!\left(\dfrac{b}{a}\right) = 0$.

Example: Factor theorem

Show that $(x - 3)$ is a factor of $f(x) = x^3 - 4x^2 + x + 6$ and hence factorize $f(x)$ completely.

$f(3) = 27 - 36 + 3 + 6 = 0$So $(x - 3)$ is a factor.

By division (or by comparing coefficients), $f(x) = (x - 3)(x^2 - x - 2) = (x-3)(x-2)(x+1)$.

Example: Finding an unknown constant

If $(x + 2)$ is a factor of $f(x) = x^3 + ax^2 - 3x + 10$Find $a$.

By the Factor Theorem, $f(-2) = 0$:

$$(-2)^3 + a(-2)^2 - 3(-2) + 10 = 0 \implies -8 + 4a + 6 + 10 = 0 \implies 4a + 8 = 0 \implies a = -2$$

---

Factorization of Polynomials

Common Techniques

HCF (highest common factor)

Factor out the greatest common factor from all terms. $$6x^3 - 9x^2 + 12x = 3x(2x^2 - 3x + 4)$$

Grouping

$$ X^3 + 2x^2 - 3x - 6 = x^2(x+2) - 3(x+2) = (x^2 - 3)(x+2) $$

Difference of squares

$$ A^2 - b^2 = (a+b)(a-b) $$ $$4x^2 - 25 = (2x+5)(2x-5)$$ $$9x^4 - 16y^2 = (3x^2 + 4y)(3x^2 - 4y)$$

Sum and difference of cubes

$$ A^3 + b^3 = (a+b)(a^2 - ab + b^2) $$ $$A^3 - b^3 = (a-b)(a^2 + ab + b^2)$$ $$8x^3 + 27 = (2x+3)(4x^2 - 6x + 9)$$ $$125x^3 - 8 = (5x-2)(25x^2 + 10x + 4)$$

Quadratic trinomials

For $ax^2 + bx + c$Find two numbers $p$ and $q$ such that $pq = ac$ and $p + q = b$. $$6x^2 - 7x + 2 = 6x^2 - 4x - 3x + 2 = 2x(3x - 2) - 1(3x - 2) = (2x - 1)(3x - 2)$$

If the discriminant $\Delta = b^2 - 4ac < 0$The quadratic cannot be factorized over $\mathbb{R}$.

Factorization by the Factor Theorem

For polynomials of degree 3 or higher, use the Factor Theorem to find linear factors by testing Integer roots (factors of the constant term), then factorize the resulting quotient.

Example: Complete factorization

Factorize $f(x) = 2x^3 + x^2 - 13x + 6$ completely.

Test integer factors of $6$: try $x = 1$.

$f(1) = 2 + 1 - 13 + 6 = -4 \neq 0$

Try $x = 2$:

$f(2) = 16 + 4 - 26 + 6 = 0$So $(x-2)$ is a factor.

Dividing: $f(x) = (x-2)(2x^2 + 5x - 3) = (x-2)(2x-1)(x+3)$.

---

Equations

Solving Polynomial Equations

To solve $f(x) = 0$:

1. Factorize $f(x)$ into linear (and possibly irreducible quadratic) factors.
2. Set each factor equal to zero and solve.

Example

Solve $x^3 - 3x^2 - 4x + 12 = 0$.

Factorizing by grouping: $x^2(x-3) - 4(x-3) = (x^2 - 4)(x-3) = (x-2)(x+2)(x-3) = 0$.

Solutions: $x = 2, -2, 3$.

Vieta’s Formulas (Quadratic)

For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

$$\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

These relationships between roots and coefficients are essential for DSE problems involving root Manipulation.

Example: Finding a new equation from roots

If $\alpha$ and $\beta$ are roots of $2x^2 - 5x + 1 = 0$Find the equation whose roots are $\alpha^2$ and $\beta^2$.

From Vieta: $\alpha + \beta = \dfrac{5}{2}$, $\alpha\beta = \dfrac{1}{2}$.

Sum of new roots:

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{25}{4} - 1 = \frac{21}{4}$$

Product of new roots:

$$\alpha^2 \beta^2 = (\alpha\beta)^2 = \frac{1}{4}$$

The required equation is $x^2 - \dfrac{21}{4}x + \dfrac{1}{4} = 0$Or equivalently $4x^2 - 21x + 1 = 0$.

Example: Symmetric expressions in roots

If $\alpha$ and $\beta$ are roots of $x^2 - 6x + 4 = 0$Find the value of $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$. $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{6}{4} = \frac{3}{2}$$

Extension: Vieta's formulas for cubic equations

For $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$: $$\alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}$$

---

Wrap-up Questions

1. **Question:** Expand $(1 + 2x)^6$ and find the coefficient of $x^4$. ### Details

Answer

Using the binomial theorem: $$(1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k}(2x)^k$$

The coefficient of $x^4$ corresponds to $k = 4$:

$$\binom{6}{4} \cdot 2^4 = 15 \cdot 16 = 240$$

The full expansion is $1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6$.

2. Question: Find the constant term in the expansion of $\left(x^2 + \dfrac{1}{x}\right)^9$.

Answer

The general term is $T_{r+1} = \binom{9}{r} (x^2)^{9-r} \cdot \left(\dfrac{1}{x}\right)^r = \binom{9}{r} x^{18 - 3r}$.

For the constant term: $18 - 3r = 0 \implies r = 6$.

$$\binom{9}{6} = \binom{9}{3} = 84$$

The constant term is $84$.

3. Question: When $f(x) = 2x^3 + ax^2 + bx - 6$ is divided by $(x-1)$The remainder is $-4$. When divided by $(x+2)$The remainder is $30$. Find $a$ and $b$.

Answer

By the Remainder Theorem:

• $f(1) = 2 + a + b - 6 = -4 \implies a + b = 0 \quad \mathrm{(i)}$
• $f(-2) = -16 + 4a - 2b - 6 = 30 \implies 4a - 2b = 52 \implies 2a - b = 26 \quad \mathrm{(ii)}$

Adding (i) and (ii): $3a = 26 \implies a = \dfrac{26}{3}$.

From (i): $b = -\dfrac{26}{3}$.

4. Question: Given that $(x - 3)$ and $(x + 1)$ are factors of $f(x) = x^3 + ax^2 + bx + c$ Find $a$, $b$And $c$. Hence factorize $f(x)$ completely.

Answer

By the Factor Theorem:

• $f(3) = 27 + 9a + 3b + c = 0 \quad \mathrm{(i)}$
• $f(-1) = -1 + a - b + c = 0 \quad \mathrm{(ii)}$

Since $(x-3)(x+1) = x^2 - 2x - 3$ is a factor, write $f(x) = (x^2 - 2x - 3)(x - d)$ for some Constant $d$.

Expanding: $f(x) = x^3 - (d+2)x^2 + (2d - 3)x + 3d$.

Comparing with $f(x) = x^3 + ax^2 + bx + c$:

• $a = -(d+2)$
• $b = 2d - 3$
• $c = 3d$

Also $f(3) = 0$ gives $27 + 9a + 3b + c = 0$.

Using $f(-1) = 0$: $-1 + a - b + c = 0$.

Subtracting (ii) from (i): $28 + 8a + 4b = 0 \implies 7 + 2a + b = 0 \quad \mathrm{(iii)}$.

From (ii): $a - b + c = 1$.

Substituting a = -(d+2)$$b = 2d-3$$c = 3d into (i):

$$27 + 9(-d-2) + 3(2d-3) + 3d = 27 - 9d - 18 + 6d - 9 + 3d = 0$$

This simplifies to $0 = 0$Which is consistent. From (ii):

$$-1 - d - 2 + 3 - 2d + 3d = 0 \implies 0 = 0$$

We need one more condition. Since the leading coefficient is $1$ and $f(x) = (x-3)(x+1)(x - d)$We Must have the constant term $c = 3d$. But $f(x)$ has constant term $c$. Comparing: $c = 3d$. We have One free parameter, so let us use $f(0) = c = 3d$But we need another constraint.

Let us equate the $x^2$ coefficient: $a = -(d+2)$. The $x$ coefficient: $b = 2d - 3$. Substituting Into (iii): $7 + 2(-d-2) + (2d-3) = 7 - 2d - 4 + 2d - 3 = 0$. Again automatically satisfied.

Without additional information, $d$ is undetermined. However, since $(x-3)$ and $(x+1)$ are the only stated factors, and the problem asks us to factorize completely, we observe that a cubic with Two known linear factors has a third linear factor. By Vieta, $\alpha + \beta + \gamma = -a$And $\alpha\beta\gamma = -c$. With $\alpha = 3, \beta = -1$:

$$3 + (-1) + \gamma = -a \implies 2 + \gamma = -a$$ $$3 \cdot (-1) \cdot \gamma = -c \implies -3\gamma = -c \implies c = 3\gamma$$

There are infinitely many cubics with $(x-3)$ and $(x+1)$ as factors. Assuming the problem intends a Monic cubic (which it is, with leading coefficient $1$), we write $f(x) = (x-3)(x+1)(x - d)$ where $d$ is the third root. Since no further condition is given, the general answer is:

a = -(d+2)$$b = 2d - 3$$c = 3dAnd $f(x) = (x-3)(x+1)(x-d)$ for any real $d$.

5. Question: Factorize $x^4 - 5x^2 + 4$ completely.

Answer

Let $u = x^2$: $$U^2 - 5u + 4 = (u-1)(u-4) = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)$$

6. Question: If $\alpha$ and $\beta$ are roots of $3x^2 - 8x + 2 = 0$Find the value of $\alpha^3 + \beta^3$ without solving the equation.

Answer

From Vieta: $\alpha + \beta = \dfrac{8}{3}$$\alpha\beta = \dfrac{2}{3}$. $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(\frac{8}{3}\right)^3 - 3 \cdot \frac{2}{3} \cdot \frac{8}{3}$$ $$= \frac{512}{27} - \frac{48}{9} = \frac{512}{27} - \frac{144}{27} = \frac{368}{27}$$

7. Question: Expand $(1 - 3x)^5$ in ascending powers of $x$ up to and including the term in $x^3$. Use the expansion to find an approximate value of $(0.97)^5$.

Answer

$$ (1-3x)^5 = \binom{5}{0} + \binom{5}{1}(-3x) + \binom{5}{2}(-3x)^2 + \binom{5}{3}(-3x)^3 + \cdots $$ $$= 1 - 15x + 90x^2 - 270x^3 + \cdots$$

Set $1 - 3x = 0.97 \implies x = 0.01$:

$$(0.97)^5 \approx 1 - 15(0.01) + 90(0.0001) - 270(0.000001) = 1 - 0.15 + 0.009 - 0.00027 = 0.85873$$

8. Question: The remainder when $f(x) = x^3 + px^2 + qx + 6$ is divided by $(x-1)$ is $12$. The Remainder when $f(x)$ is divided by $(x+1)$ is $18$. Find $p$ and $q$.

Answer

• $f(1) = 1 + p + q + 6 = 12 \implies p + q = 5 \quad \mathrm{(i)}$
• $f(-1) = -1 + p - q + 6 = 18 \implies p - q = 13 \quad \mathrm{(ii)}$

Adding: $2p = 18 \implies p = 9$.

From (i): $q = -4$.

9. Question: Prove that $\binom{n}{r} = \binom{n}{n-r}$ using the definition of binomial Coefficients.

Answer

$$ \binom{n}{n-r} = \frac{n!}{(n-r)!\,[n-(n-r)]!} = \frac{n!}{(n-r)!\,r!} = \binom{n}{r} $$

10. Question: Find the coefficient of $x^5$ in the expansion of $(1 + x)^8(1 - x)^6$.

Answer

Expand each factor using the binomial theorem and collect the $x^5$ terms.

From $(1+x)^8$The terms contributing to $x^5$ are $x^k$ where $k \leq 5$; from $(1-x)^6$The term $(-x)^{5-k}$.

The coefficient of $x^5$ is:

$$\sum_{k=0}^{5} \binom{8}{k}(-1)^{5-k}\binom{6}{5-k}$$

Evaluating each term:

• $k=0$: $\binom{8}{0}(-1)^5\binom{6}{5} = 1 \cdot (-1) \cdot 6 = -6$
• $k=1$: $\binom{8}{1}(-1)^4\binom{6}{4} = 8 \cdot 1 \cdot 15 = 120$
• $k=2$: $\binom{8}{2}(-1)^3\binom{6}{3} = 28 \cdot (-1) \cdot 20 = -560$
• $k=3$: $\binom{8}{3}(-1)^2\binom{6}{2} = 56 \cdot 1 \cdot 15 = 840$
• $k=4$: $\binom{8}{4}(-1)^1\binom{6}{1} = 70 \cdot (-1) \cdot 6 = -420$
• $k=5$: $\binom{8}{5}(-1)^0\binom{6}{0} = 56 \cdot 1 \cdot 1 = 56$

Sum: $-6 + 120 - 560 + 840 - 420 + 56 = 30$.

The coefficient of $x^5$ is $30$.

11. Question: Let $\alpha$ and $\beta$ be the roots of $x^2 - 7x + 3 = 0$. Form a quadratic Equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.

Answer

From Vieta: $\alpha + \beta = 7$$\alpha\beta = 3$.

Sum of new roots: $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{7}{3}$.

Product of new roots: $\dfrac{1}{\alpha} \cdot \dfrac{1}{\beta} = \dfrac{1}{3}$.

The equation is $x^2 - \dfrac{7}{3}x + \dfrac{1}{3} = 0$Or $3x^2 - 7x + 1 = 0$.

12. Question: Factorize $f(x) = x^3 - 3x^2 + 4$ completely.

Answer

Test integer factors of $4$: try $x = -1$.

$f(-1) = -1 - 3 + 4 = 0$So $(x+1)$ is a factor.

Dividing: $f(x) = (x+1)(x^2 - 4x + 4) = (x+1)(x-2)^2$.

---

:::tip Diagnostic Test Ready to test your understanding of Polynomials? The contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Polynomials with other DSE mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

---

DSE Exam Technique

Showing Working

For polynomial problems in DSE Paper 1:

1. When using the remainder theorem, write “By the Remainder Theorem, the remainder is $f(c)$” before computing.
2. When using the factor theorem, show that $f(c) = 0$ before stating that $(x - c)$ is a factor.
3. For polynomial division, show the division layout or state the quotient and remainder explicitly.
4. When finding unknown constants, set up a system of equations and solve step by step.
5. For binomial expansion, write the general term formula before substituting.

Significant Figures

Binomial coefficients and factorials are exact integers. Polynomial roots involving square roots should be left in exact form.

Common DSE Question Types

1. Remainder theorem with unknown constants.
2. Factor theorem to factorise cubics and quartics.
3. Binomial expansion (specific coefficient, constant term, approximation).
4. Vieta’s formulas for root manipulation.
5. Polynomial identities (equating coefficients).

---

Additional Worked Examples

Worked Example 13: Cubic with given conditions

The polynomial $f(x) = x^3 + ax^2 + bx - 12$ is divisible by $(x - 3)$ and $f(1) = -18$. Find a$$bAnd factorise $f(x)$ completely.

Solution

$f(3) = 0$: $27 + 9a + 3b - 12 = 0 \implies 9a + 3b = -15 \implies 3a + b = -5 \quad \text{(i)}$.

$f(1) = -18$: $1 + a + b - 12 = -18 \implies a + b = -7 \quad \text{(ii)}$.

(ii) from (i): $2a = 2 \implies a = 1$. From (ii): $b = -8$.

$f(x) = x^3 + x^2 - 8x - 12$.

Since $(x - 3)$ is a factor: $f(x) = (x - 3)(x^2 + 4x + 4) = (x - 3)(x + 2)^2$.

Worked Example 14: Sum of coefficients

Find the sum of all coefficients of $(2x - 3)^5$.

Solution

The sum of coefficients equals $f(1)$ where $f(x) = (2x - 3)^5$.

$f(1) = (2 - 3)^5 = (-1)^5 = -1$

Worked Example 15: Remainder when divided by a quadratic

Find the remainder when $f(x) = x^4 + 2x^3 - x^2 + 3$ is divided by $x^2 - x + 1$.

Solution

Since the divisor is degree 2, the remainder has degree at most 1: $r(x) = ax + b$.

The roots of $x^2 - x + 1 = 0$ are $\omega$ and $\omega^2$ (complex cube roots of unity, $\omega^3 = 1$).

By the remainder theorem for quadratic divisors:

$f(\omega) = a\omega + b \quad \text{and} \quad f(\omega^2) = a\omega^2 + b$

Since $\omega^2 + \omega + 1 = 0$ (i.e., $\omega^2 = -\omega - 1$) and $\omega^3 = 1$:

$f(\omega) = \omega^4 + 2\omega^3 - \omega^2 + 3 = \omega + 2 - (-\omega - 1) + 3 = \omega + 2 + \omega + 1 + 3 = 2\omega + 6$.

$f(\omega^2) = \omega^8 + 2\omega^6 - \omega^4 + 3 = \omega^2 + 2 - \omega + 3 = (-\omega - 1) + 2 - \omega + 3 = -2\omega + 4$.

From $a\omega + b = 2\omega + 6$: a = 2$$b = 6.

Check: $a\omega^2 + b = 2(-\omega - 1) + 6 = -2\omega + 4$. Consistent.

Remainder: $2x + 6$.

Worked Example 16: Vieta for cubic equations

If \alpha$$\beta$$\gamma are roots of $x^3 - 2x^2 + 3x - 4 = 0$Find $\alpha^2 + \beta^2 + \gamma^2$.

Solution

From Vieta: \alpha + \beta + \gamma = 2$$\alpha\beta + \beta\gamma + \gamma\alpha = 3$$\alpha\beta\gamma = 4.

$(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)$

$4 = \alpha^2 + \beta^2 + \gamma^2 + 6 \implies \alpha^2 + \beta^2 + \gamma^2 = -2$

Worked Example 17: Binomial coefficient ratio

If $\binom{n}{3} = 3\binom{n-1}{2}$Find $n$.

Solution

$\frac{n!}{3!(n-3)!} = 3 \cdot \frac{(n-1)!}{2!(n-3)!}$

$\frac{n}{6} = \frac{3}{2} \implies n = 9$

---

DSE Exam-Style Questions

DSE Practice 1. When $f(x) = x^3 + ax^2 + bx + c$ is divided by $(x - 1)$The remainder is $4$. When divided by $(x + 1)$The remainder is $-2$. When divided by $(x - 2)$The remainder is $14$. Find $a$, $b$And $c$.

Solution

$f(1) = 1 + a + b + c = 4 \implies a + b + c = 3 \quad \text{(i)}$

$f(-1) = -1 + a - b + c = -2 \implies a - b + c = -1 \quad \text{(ii)}$

$f(2) = 8 + 4a + 2b + c = 14 \implies 4a + 2b + c = 6 \quad \text{(iii)}$

(i) - (ii): $2b = 4 \implies b = 2$.

(iii) - (i): $3a + b = 3 \implies 3a + 2 = 3 \implies a = \dfrac{1}{3}$.

From (i): $\dfrac{1}{3} + 2 + c = 3 \implies c = \dfrac{2}{3}$.

DSE Practice 2. Find the coefficient of $x^3$ in the expansion of $(1 + 2x - x^2)^5$.

Solution

We need to find all ways to get $x^3$ from expanding $(1 + 2x - x^2)^5$ using the multinomial theorem.

The general term from choosing $a$ ones, $b$ copies of $2x$And $c$ copies of $-x^2$ where $a + b + c = 5$:

$\frac{5!}{a!\,b!\,c!} \cdot 1^a \cdot (2x)^b \cdot (-x^2)^c = \frac{5!}{a!\,b!\,c!} \cdot 2^b \cdot (-1)^c \cdot x^{b + 2c}$

For $x^3$: $b + 2c = 3$ with $a + b + c = 5$, $a, b, c \geq 0$.

Case $c = 0$: $b = 3$, $a = 2$. Coefficient: $\dfrac{120}{2! \cdot 3!} \cdot 8 = 10 \cdot 8 = 80$.

Case $c = 1$: $b = 1$, $a = 3$. Coefficient: $\dfrac{120}{3! \cdot 1! \cdot 1!} \cdot 2 \cdot (-1) = 20 \cdot (-2) = -40$.

Total coefficient of $x^3$: $80 + (-40) = 40$.

DSE Practice 3. If $(x + 1)$ and $(x - 2)$ are factors of $f(x) = 2x^3 + ax^2 + bx - 6$Find $a$ and $b$. Hence find the third factor.

Solution

$f(-1) = -2 + a - b - 6 = 0 \implies a - b = 8 \quad \text{(i)}$

$f(2) = 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5 \quad \text{(ii)}$

(i) + (ii): $3a = 3 \implies a = 1$. From (i): $b = -7$.

$f(x) = 2x^3 + x^2 - 7x - 6 = (x + 1)(x - 2)(2x + 3)$.

Verification: $(x + 1)(x - 2) = x^2 - x - 2$. $(x^2 - x - 2)(2x + 3) = 2x^3 + 3x^2 - 2x^2 - 3x - 4x - 6 = 2x^3 + x^2 - 7x - 6$. Correct.

Third factor: $(2x + 3)$.

DSE Practice 4. Expand $(1 + x)^{10}$ and use the expansion to find the value of $(1.01)^{10}$ correct to 5 decimal places.

Solution

$(1 + x)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^k = 1 + 10x + 45x^2 + 120x^3 + 210x^4 + 252x^5 + \cdots$

Set $x = 0.01$:

$(1.01)^{10} \approx 1 + 10(0.01) + 45(0.0001) + 120(0.000001) + 210(0.00000001)$

$= 1 + 0.1 + 0.0045 + 0.00012 + 0.0000021 = 1.1046221$

To 5 decimal places: $1.10462$.

DSE Practice 5. Prove that for positive integers $n \geq 2$, $n^n > 2^{n-1} \cdot n!$.

Solution

By the AM-GM inequality applied to the $n$ numbers $1, 2, 3, \ldots, n$:

$\frac{1 + 2 + \cdots + n}{n} \geq (1 \cdot 2 \cdots n)^{1/n}$

$\frac{n(n+1)}{2n} \geq (n!)^{1/n}$

$\frac{n+1}{2} \geq (n!)^{1/n}$

$\left(\frac{n+1}{2}\right)^n \geq n!$

We need to show $n^n > 2^{n-1} \cdot n!$I.e., $n^n / n! > 2^{n-1}$I.e., $\dfrac{n^n}{n!} > 2^{n-1}$.

Note $\dfrac{n^n}{n!} = \dfrac{n \cdot n \cdots n}{n \cdot (n-1) \cdots 1} = \prod_{k=1}^{n-1} \dfrac{n}{n - k}$.

Each factor $\dfrac{n}{n - k} \geq \dfrac{n}{n - 1} > 1$ for $n \geq 2$ and $k \geq 1$.

$\dfrac{n}{n-1} \cdot \dfrac{n}{n-2} \cdots \dfrac{n}{1} > 2 \cdot 2 \cdots 2 = 2^{n-1}$ when $n \geq 3$ (since $\dfrac{n}{n-k} \geq 2$ when $n - k \leq n/2$).

For $n = 2$: $4 > 2 \cdot 2 = 4$? No, $4 = 4$. For $n = 3$: $27 > 4 \cdot 6 = 24$. Yes.

The inequality holds strictly for $n \geq 3$. For $n = 2$Equality holds.

Common Pitfalls

• Forgetting the remainder theorem condition. The Remainder Theorem states $f(a)$ is the remainder when dividing by $(x - a)$, but only when the divisor is of the form $x - c$.

• Missing the factor theorem converse. $f(c) = 0$ implies $(x-c)$ is a factor, but students sometimes assume $(x+c)$ is a factor when $f(-c) = 0$.

• Arithmetic errors in polynomial long division. A single sign error propagates through all subsequent steps.

• Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’ — the former requires using previous work.

• Remainder Theorem: remainder when $f(x)$ is divided by $(x - a)$ equals $f(a)$.

• Factor Theorem: $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$.

• For polynomial equations, if one root is known, use polynomial division or synthetic division to reduce the degree.

• Vieta’s formulas relate coefficients to sums and products of roots for any degree polynomial.

Worked Examples

Example 1: Factor theorem application

Problem. Given that $(x - 2)$ is a factor of $f(x) = x^3 - 3x^2 + ax + 6$, find $a$ and factorise $f(x)$ completely.

Solution. By the Factor Theorem, $f(2) = 0$: $8 - 12 + 2a + 6 = 0 \implies 2a + 2 = 0 \implies a = -1$

So $f(x) = x^3 - 3x^2 - x + 6$. Dividing by $(x - 2)$:

$x^3 - 3x^2 - x + 6 = (x-2)(x^2 - x - 3)$

The quadratic $x^2 - x - 3 = 0$ has $\Delta = 1 + 12 = 13$, so: $f(x) = (x-2)\left(x - \frac{1+\sqrt{13}}{2}\right)\left(x - \frac{1-\sqrt{13}}{2}\right)$

$\blacksquare$

Example 2: Remainder theorem

Problem. When $f(x) = 2x^3 + px^2 - 5x + 3$ is divided by $(x - 1)$ the remainder is $4$. Find $p$.

Solution. By the Remainder Theorem: $f(1) = 4$. $2(1)^3 + p(1)^2 - 5(1) + 3 = 4 \implies 2 + p - 5 + 3 = 4 \implies p = 4$

$\blacksquare$

3. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

4. Dropping negative signs during algebraic manipulation — substitute back to verify your answer.

   Stashed changes:docs/docs_dse/Maths/compulsory/polynomials.md

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

:::