Polynomials

A polynomial in $x$ is an expression involving the algebraic sum of powers of $x$.

The degree of the polynomial is the highest power.

The coefficient of a term is the number in front of that term.

The value of a polynomial is found by replacing each letter in the expression with a number.

Example

If $f(x) = 2x^3 + 3x^2 - 7$, find the value of the polynomial at $x = -2$.

\[ \begin{aligned} f(-2) &= 2(-2)^3 + 3(-2)^2 - 7 \\ &= 2(-8) + 3 \cdot 4 - 7 \\ &= -16 + 12 - 7 \\ &= -11 \end{aligned} \]

Evaluating a Polynomial in Nested Form

$f(x) = ax^3 + bx^2 + cx + d$

\[ \begin{aligned} f(x) &= (ax^2 + bx + c)x + d \\ &= ((ax + b)x + c)x + d \end{aligned} \]

Compare with:

Example

Evaluate $f(-2)$ when $f(x) = 2x^3 + 3x^2 + 5$.

Thus $f(-2) = 1$.

Division of Polynomials

Example

Divide $x^3 - 3x^2 + x + 8$ by $x - 2$.

Synthetic Division

Write the coefficients along the top.
Write the negative of the divisor on the side.
Carry the first number down.
Multiply the bottom number by the side number.
Place in the next column.
Add the column.
The degree of the quotient is one less than the original.

Example

Divide $x^3 - 3x^2 + x + 8$ by $x - 2$.

Divisor is $x - 2$ , so 2 goes on the side.

Quotient is $x^2 - x - 1$.

Remainder = 6.

\[ x^3 - 3x^2 + x + 8 = (x - 2)(x^2 - x - 1) + 6 \]

Example

Divide $3x^3 - 7x^2 + 5x + 4$ by $x + 3$.

\[ 3x^3 - 7x^2 + 5x + 4 = (x + 3)(3x^2 - 16x + 53) - 155 \]

And $f(-3) = -155$.

Division by $ax + b$

Example

Find the quotient and remainder when $4t^3 + 6t^2 - 2t - 1$ is divided by $2t + 1$.

$2t + 1 = 2(t + \tfrac12)$. Use synthetic division for $t + \tfrac12$.

$Synthetic division with fractional root$

So \[ 4t^3 + 6t^2 - 2t - 1 = (t + \tfrac12)(4t^2 + 4t - 4) + 1 \]

To return to the original divisor $2t + 1$: double $(t + \tfrac12)$ and halve the quotient.

\[ 4t^3 + 6t^2 - 2t - 1 = (2t + 1)(2t^2 + 2t - 2) + 1 \]

Quotient is $2t^2 + 2t - 2$, remainder is 1.

Interactive - Synthetic Division

The Remainder and Factor Theorems

\[ \textbf{Remainder Theorem:}\quad \text{If } f(x) \text{ is divided by } x - h,\text{ the remainder is } f(h). \]

\[ \textbf{Factor Theorem:}\quad f(h) = 0 \iff (x - h) \text{ is a factor of } f(x). \]

Example

Fully factorise $2t^3 - 5t^2 + 4t - 21$.

Try factors of 21 , the constant $ \pm1, \pm3, \pm7, \pm21$.

\[ 2t^3 - 5t^2 + 4t - 21 = (t - 3)(2t^2 + t + 7) \]

Solving Polynomial Equations

If the polynomial is of degree 3 or higher, factorise using synthetic division. Ensure the equation is written as $f(x) = 0$ before factorising.

Example

Solve the equation $x^4 - 5x^2 + 4 = 0$.

Polynomial equation factorisation step 1

Go again:

Polynomial equation factorisation step 2

Or combine:

\[ \begin{aligned} x^4 - 5x^2 + 4 &= 0 \\ (x - 1)(x^3 + x^2 - 4x - 4) &= 0 \\ (x - 1)(x + 1)(x^2 - 4) &= 0 \\ (x - 1)(x + 1)(x - 2)(x + 2) &= 0 \end{aligned} \]

So the roots are $x = 1, -1, 2, -2$.

Example

Given that $x + 2$ is a factor of $2x^3 + x^2 + kx + 2$, find the value of $k$.

Since $x + 2$ is a factor, the remainder must be zero.

\[ -2k - 10 = 0 \] \[ -2k = 10 \] \[ k = -5 \]

If the polynomial does not factorise, use an iterative method to approximate the solution.

Example

Show that $x^3 - 4x + 2 = 0$ has a root between 0.5 and 1, and find an approximation correct to 1 decimal place.

\[ f(x) = x^3 - 4x + 2 \] \[ f(0.5) = (0.5)^3 - 4(0.5) + 2 \] \[ f(0.5) = 0.125 - 2 + 2 = 0.125 \]

Since $f(0.5) > 0$ and $f(1) \lt 0$, a root lies between 0.5 and 1.

$f(0.7) = 0.343 - 2.8 + 2 = -0.457$
$f(0.6) = 0.216 - 2.4 + 2 = -0.184$

So the root lies between 0.5 and 0.6.

$f(0.55) = 0.166375 - 2.2 + 2 = -0.034$

Since the sign changes between 0.5 and 0.55, the root is $0.5$ correct to 1 decimal place.

Sketching the Graphs of Polynomials

To sketch the graph of a polynomial:

Identify the y‑intercept
Identify the x‑intercepts
Find stationary points
Identify the nature of stationary points
Investigate behaviour as $x \to \pm\infty$

Example

Sketch and fully annotate the graph of $y = x^3 - 12x - 16$.

Y‑intercept

When $x = 0$:
$y = 0^3 - 12 \cdot 0 - 16 = -16$
Point $(0, -16)$ is on the graph.

X‑intercepts

The graph cuts the x‑axis when $y = 0$. Use synthetic division to find the roots.

\[ \begin{aligned} x^3 - 12x - 16 &= (x - 4)(x^2 + 4x + 4) \\ &= (x - 4)(x + 2)(x + 2) \\ &= (x - 4)(x + 2)^2 \end{aligned} \]

$(x - 4)(x + 2)(x + 2) = 0$
Roots: $x = 4, -2$

Points $(4,0)$ and $(-2,0)$ are on the graph.

Find Stationary Points

\[ y = x^3 - 12x - 16 \] \[ \frac{dy}{dx} = 3x^2 - 12 \] \[ \text{Stationary points occur when } \frac{dy}{dx} = 0 \] \[ 3x^2 - 12 = 0 \] \[ 3(x^2 - 4) = 0 \] \[ 3(x + 2)(x - 2) = 0 \] \[ x = -2 \quad \text{or} \quad x = 2 \]

\[ \text{when } x = -2,\quad y = (-2)^3 - 12(-2) - 16 \] \[ \text{giving point } (-2, 0) \] \[ \text{when } x = 2,\quad y = 2^3 - 12(2) - 16 \] \[ \text{giving point } (2, -32) \]

Nature of Stationary Points

Final Sketch