Division and Factorisation of Polynomials Revision Notes for VCE SSCE Mathematical Methods

Division and Factorisation of Polynomials

Introduction to polynomial division

When we divide one polynomial by another, we get a quotient and possibly a remainder. This process is similar to dividing whole numbers.

The key relationship is:

P(x) = D(x)Q(x) + R(x)

where:

$P(x)$ is the dividend (the polynomial being divided)
$D(x)$ is the divisor (what we're dividing by)
$Q(x)$ is the quotient (the result)
$R(x)$ is the remainder

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The remainder $R(x)$ must either be zero or have a degree less than the divisor $D(x)$ . This is a fundamental constraint in polynomial division.

The long division method

Long division for polynomials follows the same pattern as numerical long division: divide, multiply, subtract, and repeat.

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Worked Example: Polynomial Long Division

Divide $x^3 + x^2 - 14x - 24$ by $x + 2$ .

Solution:

We set up the division:

x + 2 \overline{) x^3 + x^2 - 14x - 24}

Step 1: Divide $x$ into $x^3$ to get $x^2$

Step 2: Multiply $x^2$ by $(x + 2)$ to get $x^3 + 2x^2$

Step 3: Subtract from the dividend: $(x^3 + x^2 - 14x - 24) - (x^3 + 2x^2) = -x^2 - 14x - 24$

Step 4: Divide $x$ into $-x^2$ to get $-x$

Step 5: Multiply $-x$ by $(x + 2)$ to get $-x^2 - 2x$

Step 6: Subtract: $(-x^2 - 14x - 24) - (-x^2 - 2x) = -12x - 24$

Step 7: Divide $x$ into $-12x$ to get $-12$

Step 8: Multiply $-12$ by $(x + 2)$ to get $-12x - 24$

Step 9: Subtract: $(-12x - 24) - (-12x - 24) = 0$

The quotient is $x^2 - x - 12$ and the remainder is $0$ .

Since the remainder is zero, $x + 2$ is a factor of $x^3 + x^2 - 14x - 24$ .

Therefore:

\frac{x^3 + x^2 - 14x - 24}{x + 2} = x^2 - x - 12

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Worked Example: Division with Missing Terms

Divide $3x^4 - 9x^2 + 27x - 8$ by $x - 2$ .

Solution:

Notice that the dividend is missing an $x^3$ term, so we write it as $3x^4 + 0x^3 - 9x^2 + 27x - 8$ .

Following the division process:

x - 2 \overline{) 3x^4 + 0x^3 - 9x^2 + 27x - 8}

After completing the division steps, we get:

Quotient: $3x^3 + 6x^2 + 3x + 33$

Remainder: $58$

Therefore:

3x^4 - 9x^2 + 27x - 8 = (x - 2)(3x^3 + 6x^2 + 3x + 33) + 58

or equivalently:

\frac{3x^4 - 9x^2 + 27x - 8}{x - 2} = 3x^3 + 6x^2 + 3x + 33 + \frac{58}{x - 2}

infoNote

When a polynomial has missing terms, always include them with a coefficient of zero. This helps maintain proper alignment during the division process and prevents errors.

The equating coefficients method

An alternative to long division is to use equating coefficients. This algebraic method can be quicker in some cases, particularly when dealing with coefficients that are fractions.

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Worked Example: Equating Coefficients

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

Solution using equating coefficients:

First, write the identity:

3x^3 + 2x^2 - x - 2 = (2x + 1)(ax^2 + bx + c) + r

Now expand the right side and equate coefficients of like terms.

Equating coefficients of $x^3$ :

3 = 2a

Therefore $a = \frac{3}{2}$

Equating coefficients of $x^2$ :

2 = a + 2b

Therefore $b = \frac{1}{2}(2 - \frac{3}{2}) = \frac{1}{4}$

Equating coefficients of $x$ :

-1 = 2c + b

Therefore $c = \frac{1}{2}(-1 - \frac{1}{4}) = -\frac{5}{8}$

Equating constant terms:

-2 = c + r

Therefore $r = -2 + \frac{5}{8} = -\frac{11}{8}$

So the quotient is $\frac{3}{2}x^2 + \frac{1}{4}x - \frac{5}{8}$ and the remainder is $-\frac{11}{8}$ .

Dividing by a non-linear polynomial

The division technique works the same way when dividing by a polynomial of degree 2 or higher. The key difference is that the quotient will have a lower degree.

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Worked Example: Division by a Quadratic

Divide $3x^3 - 2x^2 + 3x - 4$ by $x^2 - 1$ .

Solution:

Write $x^2 - 1$ as $x^2 + 0x - 1$ to keep track of all terms.

Following the division steps:

The quotient is $3x - 2$ and the remainder is $6x - 6$ .

Therefore:

3x^3 - 2x^2 + 3x - 4 = (x^2 - 1)(3x - 2) + 6x - 6

or:

\frac{3x^3 - 2x^2 + 3x - 4}{x^2 - 1} = 3x - 2 + \frac{6x - 6}{x^2 - 1}

The remainder theorem

The remainder theorem provides a shortcut for finding remainders without performing full division.

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The Remainder Theorem

When a polynomial $P(x)$ is divided by $\beta x + \alpha$ , the remainder is $P\left(-\frac{\alpha}{\beta}\right)$ .

In the special case where we divide by $x - \alpha$ , the remainder is simply $P(\alpha)$ .

How it works:

If $P(x) = (x - \alpha)Q(x) + R$ , then substituting $x = \alpha$ gives:

P(\alpha) = (\alpha - \alpha)Q(\alpha) + R = R

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Worked Example: Using the Remainder Theorem

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

Solution:

By the remainder theorem, the remainder is:

P\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^3 + 2\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1

= -\frac{3}{8} + \frac{2}{4} - \frac{1}{2} + 1

= -\frac{3}{8} + \frac{4}{8} - \frac{4}{8} + \frac{8}{8}

= \frac{5}{8}

infoNote

Exam tip: The remainder theorem is much faster than long division when you only need the remainder, not the quotient. Use this strategically in exam situations to save time.

The factor theorem

The factor theorem is a special case of the remainder theorem that helps us identify factors of polynomials.

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The Factor Theorem

For a polynomial $P(x)$ :

If $P(\alpha) = 0$ , then $x - \alpha$ is a factor of $P(x)$
Conversely, if $x - \alpha$ is a factor of $P(x)$ , then $P(\alpha) = 0$

More generally:

If $\beta x + \alpha$ is a factor of $P(x)$ , then $P\left(-\frac{\alpha}{\beta}\right) = 0$
Conversely, if $P\left(-\frac{\alpha}{\beta}\right) = 0$ , then $\beta x + \alpha$ is a factor of $P(x)$

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Worked Example: Finding Unknown Coefficients

Given that $x + 1$ and $x - 2$ are factors of $6x^4 - x^3 + ax^2 - 6x + b$ , find the values of $a$ and $b$ .

Solution:

Let $P(x) = 6x^4 - x^3 + ax^2 - 6x + b$

By the factor theorem, $P(-1) = 0$ and $P(2) = 0$ .

For $P(-1) = 0$ :

6 + 1 + a + 6 + b = 0

a + b = -13 \text{ ... (1)}

For $P(2) = 0$ :

96 - 8 + 4a - 12 + b = 0

4a + b = -76 \text{ ... (2)}

Subtract equation (1) from equation (2):

3a = -63

a = -21

Substituting into equation (1):

-21 + b = -13

b = 8

Therefore $a = -21$ and $b = 8$ .

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Worked Example: Complete Factorisation

Show that $x + 1$ is a factor of $x^3 - 4x^2 + x + 6$ and hence find the other linear factors.

Solution:

Let $P(x) = x^3 - 4x^2 + x + 6$

P(-1) = (-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0

Therefore $x + 1$ is a factor by the factor theorem.

Now divide $x^3 - 4x^2 + x + 6$ by $x + 1$ :

The quotient is $x^2 - 5x + 6$

So:

x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)

Factorising the quadratic:

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

Therefore:

x^3 - 4x^2 + x + 6 = (x + 1)(x - 3)(x - 2)

The linear factors are $(x + 1)$ , $(x - 3)$ and $(x - 2)$ .

Alternative method:

Once we know $x + 1$ is a factor, we can write:

x^3 - 4x^2 + x + 6 = (x + 1)(x^2 + bx + c)

Equating constant terms: $6 = 1 \times c$ , so $c = 6$

Equating coefficients of $x^2$ : $-4 = 1 + b$ , so $b = -5$

Therefore:

x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)

Sums and differences of cubes

There are special factorisation formulas for the sum and difference of cubes that you should memorise.

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Cube Factorisation Formulas

Difference of cubes:

x^3 - a^3 = (x - a)(x^2 + ax + a^2)

Sum of cubes:

x^3 + a^3 = (x + a)(x^2 - ax + a^2)

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Memory aid: For the quadratic factor, the signs follow the pattern "same, opposite, always positive":

First sign matches the original expression
Second sign is opposite
Third term is always positive

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Worked Example: Factorising Cubes

Factorise:

a) $8x^3 + 64$

b) $125a^3 - b^3$

Solution:

a) $8x^3 + 64 = (2x)^3 + 4^3$

= (2x + 4)(4x^2 - 8x + 16)

b) $125a^3 - b^3 = (5a)^3 - b^3$

= (5a - b)(25a^2 + 5ab + b^2)

The rational-root theorem

The rational-root theorem helps us find rational solutions (fractions) to polynomial equations systematically by narrowing down the possible candidates.

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The Rational-Root Theorem

Let $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ be a polynomial with integer coefficients.

If $\beta x + \alpha$ is a factor of $P(x)$ , where $\alpha$ and $\beta$ are relatively prime integers (their highest common factor is 1), then:

$\beta$ divides the leading coefficient $a_n$
$\alpha$ divides the constant term $a_0$

In practical terms: If $-\frac{\alpha}{\beta}$ is a rational solution, then $\beta$ must divide the leading coefficient and $\alpha$ must divide the constant term.

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Worked Example: Using the Rational-Root Theorem

Use the rational-root theorem to help factorise $P(x) = 3x^3 + 8x^2 + 2x - 5$ .

Solution:

First, test for integer solutions: $\pm 1$ and $\pm 5$ (divisors of the constant term).

P(1) = 8 \neq 0

P(-1) = -2 \neq 0

P(5) = 580 \neq 0

P(-5) = -190 \neq 0

So there are no integer solutions. Now use the rational-root theorem.

If $-\frac{\alpha}{\beta}$ is a solution, then $\beta$ must divide 3 (the leading coefficient) and $\alpha$ must divide -5 (the constant term).

So $\beta \in \{1, 3\}$ and $\alpha \in \{-5, -1, 1, 5\}$

The only new possibility with $\beta = 3$ is $-\frac{\alpha}{\beta}$ where $\alpha = \pm 5$ or $\pm 1$ .

Testing: $P\left(-\frac{5}{3}\right) = 0$

Therefore $3x + 5$ is a factor.

Dividing gives:

3x^3 + 8x^2 + 2x - 5 = (3x + 5)(x^2 + x - 1)

To fully factorise, complete the square for $x^2 + x - 1$ :

x^2 + x - 1 = \left(x + \frac{1}{2}\right)^2 - \frac{5}{4}

= \left(x + \frac{1}{2} + \frac{\sqrt{5}}{2}\right)\left(x + \frac{1}{2} - \frac{\sqrt{5}}{2}\right)

Therefore:

P(x) = (3x + 5)\left(x + \frac{1 + \sqrt{5}}{2}\right)\left(x + \frac{1 - \sqrt{5}}{2}\right)

Solving polynomial equations

We can use factorisation to solve polynomial equations by applying the zero product property: if a product equals zero, at least one of the factors must be zero.

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Worked Example: Solving by Factorisation

Factorise $P(x) = x^3 - 4x^2 - 11x + 30$ and hence solve $x^3 - 4x^2 - 11x + 30 = 0$ .

Solution:

Test integer factors (divisors of 30):

P(1) = 1 - 4 - 11 + 30 \neq 0

P(-1) = -1 - 4 + 11 + 30 \neq 0

P(2) = 8 - 16 - 22 + 30 = 0

Therefore $x - 2$ is a factor.

Dividing $x^3 - 4x^2 - 11x + 30$ by $x - 2$ :

P(x) = (x - 2)(x^2 - 2x - 15)

Factorising the quadratic:

P(x) = (x - 2)(x - 5)(x + 3)

For $P(x) = 0$ :

$x - 2 = 0$ or $x - 5 = 0$ or $x + 3 = 0$

Therefore: $x = 2$ or $x = 5$ or $x = -3$

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Exam tip: Always check your factors by substituting back into the original polynomial. This helps catch any errors in your factorisation or division.

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Key Points to Remember:

The division formula is $P(x) = D(x)Q(x) + R(x)$ where the remainder has degree less than the divisor
The remainder theorem: when dividing by $\beta x + \alpha$ , the remainder equals $P\left(-\frac{\alpha}{\beta}\right)$
The factor theorem: $x - \alpha$ is a factor if and only if $P(\alpha) = 0$
Sum of cubes: $x^3 + a^3 = (x + a)(x^2 - ax + a^2)$
Difference of cubes: $x^3 - a^3 = (x - a)(x^2 + ax + a^2)$
The rational-root theorem narrows down possible rational solutions by limiting the numerator to divisors of the constant term and the denominator to divisors of the leading coefficient

Division and Factorisation of Polynomials (VCE SSCE Mathematical Methods): Revision Notes