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)
\begin{equation} P(x) = a^nx_n + a^{n–1}x_{n–1} + ... + a^2x_2 + a^1x_1 + a^0 \end{equation}

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

Factorising Quadratics

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)
\begin{align} . \qquad x^2 + 8x + 2 = x^2 + 8x + \Big( \dfrac{8}{2} \Big)^2 + 2 - \Big( \dfrac{8}{2} \Big)^2 \\ . \\ . \qquad \qquad \qquad \quad = x^2 + 8x + 16 + 2 - 16 \\ . \\ . \qquad \qquad \qquad \quad = \big( x + 4 \big)^2 - 14 \\ . \\ . \qquad \qquad \qquad \quad = \big( x + 4 + \sqrt{14} \big) \big( x + 4 - \sqrt{14} \big) \end{align}
  • 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)
\begin{align} . \qquad x^3 - 5x^2 + 3x - 15 = \underline {x^3 - 5x^2 } + \underline{ 3x - 15 } \\ . \\ . \qquad \qquad \qquad \qquad \qquad = x^2 \big( x - 5 \big) + 3 \big( x - 5 \big) \\ . \\ . \qquad \qquad \qquad \qquad \qquad = \big( x - 5 \big) \big( x^2 + 3 \big) \end{align}

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
Start with the easiest values first

(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)
\begin{align} . \qquad P(2) = 0 \quad \text{ so } \quad \big( x - 2 \big) \quad \text{ is a factor} \end{align}

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)
\begin{equation} x^3 – 2x^2 – 3x + 6 = x^3 + ax^2 – 3x – 2x^2 – 2ax + 6 \end{equation}

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)
\begin{align} . \qquad \qquad x^3 - 2x^2 - 3x + 6 = 0 \\ . \\ . \qquad \qquad \big( x - 2 \big) \big( x - \sqrt{3} \big) \big( x + \sqrt{3} \big) = 0 \\ . \\ . \qquad \qquad x - 2 = 0 \qquad x - \sqrt{3} = 0 \qquad x + \sqrt{3} = 0 \\ . \\ . \qquad \qquad x = 2 \qquad \quad x = \sqrt{3} \qquad \quad x = -\sqrt{3} \end{align}

The Quadratic Formula

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)
\begin{align} .\qquad \qquad x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align}

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

(25)
\begin{align} . \qquad \qquad \Delta = b^2- 4ac \end{align}
  • 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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