A01.2-Polynomials

toc
 * 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

P(x) = a n x n + a n–1 x n–1 + ... + a 2 x 2 + a 1 x + a 0.

In the polynomial, P(x)
 * a n is the coefficient of x n, 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:

......... When factorising any polynomial, always take out any common factors first.
 * 0. Common Factors **

math . \qquad a^2 \pm 2ab + b^2 = \big( a \pm b \big)^2 \qquad. math
 * 1. Perfect Squares **

math . \qquad 4x^2 + 12x + 9 = \big( 2x + 3 \big)^2 \qquad. math
 * Example: **

math . \qquad a^2 - b^2 = \big(a + b \big) \big( a - b \big) \qquad. math
 * 2. Difference of Perfect Squares **

math . \qquad 9x^2 - 4y^2 = \big( 3x + 2y \big) \big( 3x - 2y \big) \qquad. math
 * Example: **


 * 3. Shortcut Approach **
 * For a full review of this approach when a = 1 go here

math . \qquad x^2 + 2x - 24 \qquad \textit{look for factors of 24 that combine to give +2} \qquad. math
 * Example: **

math . \qquad x^2 + 2x - 24 = \big( x + 6 \big) \big( x - 4 \big) \qquad. math


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

math . \qquad 3x^2 + 8x + 4 \qquad \textit {look for factors of } 3 \times 4 = 12 \textit{ that combine to give +8} \qquad. math
 * Example: **

math . \qquad \quad 3x^2 + 8x + 4 \\. \\ . \qquad = 3x^2 + 6x + 2x + 4 \\.\\ . \qquad = 3x \big( x + 2 \big) + 2 \big( x + 2 \big) \qquad. \\ . \\ . \qquad = \big( x + 2 \big) \big( 3x + 2 \big) math


 * 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.

math . \qquad x^2 + 8x + 2 = x^2 + 8x + \Big( \dfrac{8}{2} \Big)^2 + 2 - \Big( \dfrac{8}{2} \Big)^2 \qquad. \\ . \\ . \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) math
 * Example: **


 * 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

math . \qquad a^3 + 3a^2b + 3ab^2 + b^3 = \big( a + b \big)^3 \qquad. \\ . \\ . \qquad a^3 - 3a^2b + 3ab^2 - b^3 = \big( a - b \big)^3 math
 * 1. Perfect Cubes **

math . \qquad x^3 - 18x^2 + 27x - 27 = \big( x - 3 \big)^3 \qquad. math
 * Example **

math . \qquad a^3 + b^3 = \big( a + b \big) \big( a^2 - ab + b^2 \big) \qquad. \\.\\ . \qquad a^3 - b^3 = \big( a - b \big) \big( a^2 + ab + b^2 \big) math
 * 2. Sum and Difference of Two Cubes **

math . \qquad x^3 + 8 = \big( x + 2 \big) \big( x^2 - 2x + 4 \big) \qquad. math
 * Example **


 * 3. Factorising Cubics by Grouping 2 and 2 **
 * only some cubics can be factorised in this way

math . \qquad x^3 - 5x^2 + 3x - 15 = \underline {x^3 - 5x^2 } + \underline{ 3x - 15 } \qquad. \\ . \\ . \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) math
 * Example **

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)

math . \qquad \text{Find a factor of } P(x) = x^3 - 2x^2 - 3x + 6 \qquad. math
 * Example **


 * The constant term is 6, factors of 6 are: 1, –1, 2, –2, 3, –3, 6, –6
 * Start with the easiest values first

math . \qquad P(1) = 1 - 2 - 3 + 6 = 2 \qquad. \\ . \\ . \qquad P(-1) = -1 - 2 + 3 + 6 = 6 \\. \\ . \qquad P(2) = 8 - 8 - 6 + 6 = 0 math

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

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 x 2 and the last term must be –3 so we can write:

... ... x 3 – 2x 2 – 3x + 6 = (x – 2)(x 2 + ** a **x – 3) ... ... where ** a ** is an unknown


 * If we expand these brackets we get:

... ... x 3 __– 2__x 2 – 3x + 6 = x 3 __+ ** a **__x 2 – 3x __– 2__x 2 – 2** a **x + 6


 * Equate the coefficients of x 2 (underlined above), we get:

... ... –2 = ** a ** – 2

... ... so ** a ** = 0

Thus

math . \qquad x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x^2 - 3 \big) \qquad. math


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

Thus math . \qquad x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x - \sqrt{3} \big) \big( x + \sqrt{3} \big) \qquad. math

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

math . \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 \qquad x = 2 \qquad \quad x = \sqrt{3} \qquad \quad x = -\sqrt{3} math
 * Example **

The Quadratic Formula
For any quadratic equation in the form: math . \qquad \qquad ax^2 + bx + c = 0 \qquad. math

We can use the Quadratic Formula to solve it: math .\qquad \qquad x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \qquad. math

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

math . \qquad \qquad \Delta = b^2- 4ac \qquad. math


 * 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: math . \qquad \qquad P(x) \equiv Q(x) \qquad. math

Find the value of k if: math . \qquad \qquad 3x^2 + kx - 5 \equiv 3x^2 - 4x - 5 \qquad. math
 * Example **


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

.