01.2-Polynomials

# 1.2 Polynomials

A polynomial is an expression with

• only one variable (eg x)
• one or more terms
• each term has a non-negative, integer power of x
(1)
$$P(x) = a^nx_n + a^{n–1}x_{n–1} + ... + a^2x_2 + a^1x_1 + a^0$$

In the polynomial, P(x)

• an is the coefficient of xn, etc
• P(x) has a degree of n (the highest power of x)

Polynomials are usually named with consecutive capital letters, starting at P (ie P, Q, R, etc)

You are already familiar with many polynomials:

• a polynomial with degree 1 is Linear
• a polynomial with degree 2 is a Quadratic
• a polynomial with degree 3 is a Cubic
• a polynomial with degree 4 is a Quartic

Recall that we can factorise quadratics using the following techniques:

### 0. Common Factors

When factorising any polynomial, always take out any common factors first.

### 1. Perfect Squares

(2)
\begin{align} . \qquad a^2 \pm 2ab + b^2 = \big( a \pm b \big)^2 \end{align}

Example:

(3)
\begin{align} . \qquad 4x^2 + 12x + 9 = \big( 2x + 3 \big)^2 \end{align}

### 2. Difference of Perfect Squares

(4)
\begin{align} . \qquad a^2 - b^2 = \big(a + b \big) \big( a - b \big) \end{align}

Example:

(5)
\begin{align} . \qquad 9x^2 - 4y^2 = \big( 3x + 2y \big) \big( 3x - 2y \big) \end{align}

### 3. Shortcut Approach

For a full review of this approach when a = 1 go here

Example:

(6)
\begin{align} . \qquad x^2 + 2x - 24 \qquad \textit{look for factors of 24 that combine to give +2} \end{align}
(7)
\begin{align} . \qquad x^2 + 2x - 24 = \big( x + 6 \big) \big( x - 4 \big) \end{align}

For a full review of this approach when a <> 1, go here

Example:

(8)
\begin{align} . \qquad 3x^2 + 8x + 4 \qquad \textit {look for factors of } 3 \times 4 = 12 \textit{ that combine to give +8} \end{align}
(9)
\begin{align} . \qquad \quad 3x^2 + 8x + 4 \\ . \\ . \qquad = 3x^2 + 6x + 2x + 4 \\ . \\ . \qquad = 3x \big( x + 2 \big) + 2 \big( x + 2 \big) \\ . \\ . \qquad = \big( x + 2 \big) \big( 3x + 2 \big) \end{align}

### 4. Completing the Square

• This method will factorise any quadratic that can be factorised but use a quicker method if you can
• Note that it assumes a = 1, so if a <> 1 take that out as a common factor first
• For a full review of this approach, go here.

Example:

(10)
• Note that if you end up with a plus sign between the two terms in the 2nd last step,
• then the quadratic can not be factorised (in the Real number system)

## Factorising Cubics

We can factorise cubic polynomials using the following techniques

### 1. Perfect Cubes

(11)
\begin{align} . \qquad a^3 + 3a^2b + 3ab^2 + b^3 = \big( a + b \big)^3 \\ . \\ . \qquad a^3 - 3a^2b + 3ab^2 - b^3 = \big( a - b \big)^3 \end{align}

Example

(12)
\begin{align} . \qquad x^3 - 18x^2 + 27x - 27 = \big( x - 3 \big)^3 \end{align}

### 2. Sum and Difference of Two Cubes

(13)
\begin{align} . \qquad a^3 + b^3 = \big( a + b \big) \big( a^2 - ab + b^2 \big) \\.\\ . \qquad a^3 - b^3 = \big( a - b \big) \big( a^2 + ab + b^2 \big) \end{align}

Example

(14)
\begin{align} . \qquad x^3 + 8 = \big( x + 2 \big) \big( x^2 - 2x + 4 \big) \end{align}

### 3. Factorising Cubics by Grouping 2 and 2

• only some cubics can be factorised in this way

Example

(15)

## Factor Theorem

If factorising a higher degree polynomial, P(x), or the previous methods aren't effective, we can apply the factor theorem.

• If P(a) = 0 then the polynomial P(x) has a factor of (x – a)

NOTE: if (x – a) is a factor of P(x) then a must be a factor of the constant term in P(x).

To find a factor of P(x) use trial and error with different factors (positive and negative) of the constant term in P(x)

Example

(16)
\begin{align} . \qquad \text{Find a factor of } P(x) = x^3 - 2x^2 - 3x + 6 \end{align}

The constant term is 6, factors of 6 are: 1, –1, 2, –2, 3, –3, 6, –6

(17)
\begin{align} . \qquad P(1) = 1 - 2 - 3 + 6 = 2 \\ . \\ . \qquad P(-1) = -1 - 2 + 3 + 6 = 6 \\ . \\ . \qquad P(2) = 8 - 8 - 6 + 6 = 0 \end{align}
(18)

## Shortcut Method for Factorising Cubics when a factor is known

Once the factor theorem has revealed one factor, we can complete the factorisation using the following method

For a demonstration of this process in Powerpoint, download the following file (2.4 Mb)

Example

• Factorise $(x^3 – 2x^2 – 3x + 6)$ given that $(x – 2)$ is a factor

Solution:

By examination of the above, we can conclude that the other factor must be a quadratic.

Further the first term must be x2 and the last term must be –3 so we can write:

$x^3 – 2x^2 – 3x + 6 = (x – 2)(x^2 + ax – 3)$ where a is an unknown

If we expand these brackets we get:

(19)
$$x^3 – 2x^2 – 3x + 6 = x^3 + ax^2 – 3x – 2x^2 – 2ax + 6$$

Equate the coefficients of x2 (underlined above), we get:

–2 = a – 2

so a = 0

Thus

(20)
\begin{align} . \qquad x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x^2 - 3 \big) \end{align}

The quadratic in the second bracket can then be factorised
in this case using difference of two squares

Thus

(21)
\begin{align} . \qquad x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x - \sqrt{3} \big) \big( x + \sqrt{3} \big) \end{align}

## Solving Polynomial Equations

Once any type of polynomial equation has been factorised, it can be solved using the Null Factor Law.

Example

(22)

For any quadratic equation in the form:

(23)
\begin{align} . \qquad \qquad ax^2 + bx + c = 0 \end{align}

We can use the Quadratic Formula to solve it:

(24)

Recall that the Discriminant gives us information about the number and type of solutions:

(25)
• if discriminant less than zero, there are no real solutions
• if discriminant equal to zero, there is one real solution
• the solution is rational
• if discriminant greater than zero, there are two real solutions
• if discriminant is a perfect square, the solutions are rational
• if discriminant is not a perfect square, the solutions are irrational

## Equivalent Polynomials

Two polynomials are equivalent if and only if each of the respective coefficients are equal.

The notation to say that P(x) is equivalent to Q(x) is: $\; P(x) \equiv Q(x)$

Example
Find the value of k if:$3x^2 + kx - 5 \equiv 3x^2 - 4x - 5$

Solution: The equivalent coefficient of x is –4, so k = –4.

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