Algebraic Operations

Two theorems do nearly all the work here. Both replace a page of long division with a single substitution, so knowing when they apply is worth more than being quick at dividing.

Polynomials

A polynomial in $x$ is a sum of terms $ax^n$ where $n$ is a non-negative integer:

$$f(x) = 3x^4 - 2x + 7$$

The degree is the highest power present — here $4$. The leading coefficient is the number attached to it, $3$.

$\sqrt{x}$, $\dfrac1x$ and $x^{-2}$ are not polynomial terms, so an expression containing them is not a polynomial.

Dividing polynomials

Dividing $f(x)$ by $d(x)$ gives a quotient $q(x)$ and a remainder $r(x)$:

$$f(x) = d(x)\,q(x) + r(x)$$

with the degree of $r$ strictly less than the degree of $d$. Dividing by a linear factor therefore leaves a constant remainder.

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

Working term by term, as with numerical long division:

• $x^3 \div x = x^2$. Multiply back: $x^3 - x^2$. Subtract: $-x^2 - 5x + 6$.
• $-x^2 \div x = -x$. Multiply back: $-x^2 + x$. Subtract: $-6x + 6$.
• $-6x \div x = -6$. Multiply back: $-6x + 6$. Subtract: $0$.

$$x^3 - 2x^2 - 5x + 6 = (x-1)(x^2 - x - 6)$$

The remainder is zero, so $x - 1$ is a factor.

The remainder theorem

$$\text{When } f(x) \text{ is divided by } (x - a), \text{ the remainder is } f(a).$$

No division needed — substitute and evaluate.

Find the remainder when $2x^3 + 3x^2 - 5x + 7$ is divided by $x - 2$.

$$f(2) = 2(8) + 3(4) - 5(2) + 7 = 16 + 12 - 10 + 7 = 25$$

And when it is divided by $x + 1$?

Write $x + 1$ as $x - (-1)$, so $a = -1$:

$$f(-1) = -2 + 3 + 5 + 7 = 13$$

Dividing by $(ax - b)$

The remainder is $f\!\left(\dfrac{b}{a}\right)$ — the value that makes the divisor zero.

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

Set $2x - 1 = 0$, so $x = \frac12$:

$$f\!\left(\tfrac12\right) = 2\left(\tfrac18\right) - \tfrac14 + 4 = \tfrac14 - \tfrac14 + 4 = 4$$

The factor theorem

The special case where the remainder is zero:

$$(x - a) \text{ is a factor of } f(x) \iff f(a) = 0$$

Is $x - 1$ a factor of $x^3 - 2x^2 - 5x + 6$?

$$f(1) = 1 - 2 - 5 + 6 = 0$$

Yes — and this took one line, where the long division above took several.

Finding an unknown constant

Find $k$ so that $x - 2$ is a factor of $x^3 + kx^2 - 4x + 4$.

By the factor theorem, $f(2) = 0$:

$$8 + 4k - 8 + 4 = 0$$ $$4k + 4 = 0$$ $$k = -1$$

Check: $f(x) = x^3 - x^2 - 4x + 4$ gives $f(2) = 8 - 4 - 8 + 4 = 0$ ✓

Factorising a cubic

The standard three-step method.

Factorise $x^3 - 2x^2 - 5x + 6$ completely.

Step 1 — find one root by trial. Try the factors of the constant term, $\pm1, \pm2, \pm3, \pm6$:

$$f(1) = 1 - 2 - 5 + 6 = 0 \ \checkmark$$

So $(x - 1)$ is a factor.

Step 2 — divide by $(x-1)$, giving the quadratic $x^2 - x - 6$.

Step 3 — factorise the quadratic:

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

$$f(x) = (x-1)(x-3)(x+2)$$

The roots are $x = 1$, $x = 3$ and $x = -2$ — three roots for a cubic, as expected.

Equating coefficients

An identity — written with $\equiv$ — holds for every value of $x$, so the coefficients of each power must match on both sides.

Find $A$ and $B$ if $5x + 1 \equiv A(x+2) + B(x-1)$.

Substitution method (faster): choose values of $x$ that kill one unknown.

Let $x = 1$: $\;6 = A(3) + B(0)$, so $A = 2$.

Let $x = -2$: $\;-9 = A(0) + B(-3)$, so $B = 3$.

Check: $2(x+2) + 3(x-1) = 2x + 4 + 3x - 3 = 5x + 1$ ✓

Comparing coefficients (also valid): expand the right side to $(A+B)x + (2A - B)$, then

$$A + B = 5 \qquad 2A - B = 1$$

Adding gives $3A = 6$, so $A = 2$ and $B = 3$ — the same answer.