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

… … $P(x) = a^nx_n + a^{n–1}x_{n–1} + ... + a^2x_2 + a^1x_1 + a^0$

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

… … $a^2 \pm 2ab + b^2 = \big( a \pm b \big)^2$

Example:

… … $4x^2 + 12x + 9 = \big( 2x + 3 \big)^2$
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2. Difference of Perfect Squares

… … $a^2 - b^2 = \big(a + b \big) \big( a - b \big)$

Example:

… … $9x^2 - 4y^2 = \big( 3x + 2y \big) \big( 3x - 2y \big)$
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3. Shortcut Approach

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

Example:

… … $x^2 + 2x - 24$ … … look for factors of 24 that combine to give +2

… … $x^2 + 2x - 24 = \big( x + 6 \big) \big( x - 4 \big)$
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For a full review of this approach when a <> 1, go here

Example:

… … $3x^2 + 8x + 4$ … … look for factors of $3 \times 4 = 12$ that combine to give +8

… … $3x^2 + 8x + 4$ … … desired factors of 12 are +6 and +2

… … $= 3x^2 + 6x + 2x + 4$ … … now factorise the 1st pair and the 2nd pair

… … $= 3x \big( x + 2 \big) + 2 \big( x + 2 \big)$

… … $= \big( x + 2 \big) \big( 3x + 2 \big)$
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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:

… … $x^2 + 8x - 2$

… … $= x^2 + 8x + \Big( \dfrac{8}{2} \Big)^2 + 2 - \Big( \dfrac{8}{2} \Big)^2$

… … $= \underline{x^2 + 8x + 16} \;\; \underline{ + 2 - 16}$

… … $= \big( x + 4 \big)^2 - 14$

… … $= \big( x + 4 + \sqrt{14} \big) \big( x + 4 - \sqrt{14} \big)$
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• 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)

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Factorising Cubics

We can factorise cubic polynomials using the following techniques

1. Perfect Cubes

… … $a^3 + 3a^2b + 3ab^2 + b^3 = \big( a + b \big)^3$

… … $a^3 - 3a^2b + 3ab^2 - b^3 = \big( a - b \big)^3$

Example

… … $x^3 - 18x^2 + 27x - 27 = \big( x - 3 \big)^3$
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2. Sum and Difference of Two Cubes

… … $a^3 + b^3 = \big( a + b \big) \big( a^2 - ab + b^2 \big)$

… … $a^3 - b^3 = \big( a - b \big) \big( a^2 + ab + b^2 \big)$

Example

… … $x^3+8 = \big(x+2\big)\big(x^2 - 2x + 4\big)$
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3. Factorising Cubics by Grouping 2 and 2

• only some cubics can be factorised in this way

Example

… … $x^3 - 5x^2 + 3x - 15$

… … $= \underline {x^3 - 5x^2 } \;\; \underline{+ 3x - 15 }$

… … $= x^2 \big( x - 5 \big) + 3 \big( x - 5 \big)$

… … $= \big( x - 5 \big) \big( x^2 + 3 \big)$
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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 $\big(x - a\big)$

NOTE: if $\big(x - a\big)$ 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

… … Find a factor of $P(x) = x^3 - 2x^2 - 3x + 6$

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

… … Start with the easiest values first

… … $P(1) = 1 - 2 - 3 + 6 = 2$

… … $P(-1) = -1 - 2 + 3 + 6 = 6$

… …. $P(2) = 8 - 8 - 6 + 6 = 0$
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… … $P(2) = 0$ so $\big( x - 2 \big)$ is a factor
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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

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 constant
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… … If we expand these brackets we get:

… … $x^3 \underline{- 2}x^2 - 3x + 6 = x^3 \underline{+ a}x^2 - 3x \underline{ - 2}x^2 - 2ax + 6$
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… … Equate the coefficients of x2 (underlined above), we get:

… … $-2 = a - 2$

… … so $a = 0$

… … Thus

… … $x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x^2 - 3 \big)$
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… … The quadratic in the second bracket can then be factorised

… … in this case using difference of two squares

… … Thus

… … $x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x - \sqrt{3} \big) \big( x + \sqrt{3} \big)$
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Solving Polynomial Equations

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

Example

… … x^3 - 2x^2 - 3x + 6 = 0

… … $\big( x - 2 \big) \big( x - \sqrt{3} \big) \big( x + \sqrt{3} \big) = 0$

… … $x - 2 = 0 \qquad x - \sqrt{3} = 0 \qquad x + \sqrt{3} = 0$

… … $x = 2 \qquad \quad x = \sqrt{3} \qquad \quad x = -\sqrt{3}$
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For any quadratic equation in the form:

… … $ax^2 + bx + c = 0$

We can use the Quadratic Formula to solve it:

… … $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

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

… … $\Delta = b^2- 4ac$

• 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

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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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