Polynomials | DSE - Wyatt's Notes

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.

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.

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

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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.

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 $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$.

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,

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

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

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

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:

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

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. $r = 5$:

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

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Polynomial Division

Section titled “Polynomial Division”

Long Division

Section titled “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: $(x^2 - x + 2)(2x + 5) + (-4x - 3) = 2x^3 + 3x^2 - 5x + 7$.

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:

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

$(x - c)$ 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, $f(-2) = 0$:

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

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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

Section titled “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: $f(x) = (x-2)(2x^2 + 5x - 3) = (x-2)(2x-1)(x+3)$.

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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)

Section titled “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 $\alpha, \beta, \gamma$: $$\alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}$$

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

1. 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 $84$.

1. 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): $b = -\dfrac{26}{3}$.

1. 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 = 3d And $f(x) = (x-3)(x+1)(x-d)$ for any real $d$.

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

Answer

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

1. 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}$$

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

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

1. 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): $q = -4$.

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

Answer

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

1. 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 $30$.

1. 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 $3x^2 - 7x + 1 = 0$.

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

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