FACTORISING POLYNOMIALS - Maths Figured Out

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Section1

2  

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Factorising Polynomials TEXT

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Factorising Polynomials

Although this unit begins with a realistic example, much of the content has no immediate relevance to the outside world. Its purpose is to extend the algebra you already know.

This does not make

it a dull chapter. On the contrary, many people enjoy abstract mathematics as being interesting in its own right. Over the centuries, mathematical techniques have developed partly in response to problems that needed to be solved, but there have always been those who have pondered on the less applicable side of the subject purely out of interest and curiosity.

The Cubic Equation

Sup pose each smaller square has side x cm.

The dimensions of the box will then be

x by xx).

For the volume to be

500 cm

3 , x must satisfy the equation x x) 2 = 500, x When the expression has been multiplied out, the highest power of x in this equation is x 3 .

This is an example of a cubic equation.

As with a quadratic equation, it is good practice to reduce this equation to the form f (x) 0. x 80 x + 4 x 2 ) = 500 4 x 3 80 x
2 + 400 x 500 = 0 x 3 20 x 2 + x 25 = 0, 0 x Again by comparison with quadratic equations, the next thing you might want to do would be to try to factorise it. With a cubic function, however, this is not easy. Fortunately one solution to the problem can be seen 'by inspection'. A little experimentation reveals that x = 5 will make this equation true.

Is this the only solution?

Remember that for quadratic equations when there is one answer there is always another, except when there are equal roots.

How can an open-topped box with volume

500 cm3 be made from a square piece of

card 20 cm by 20 cm?

To make such a box, a net can be made by

cutting off equal squares from each corner, as shown in the diagram. x x x x x x x x ©

Factorising Polynomials TEXT

2 1 One method of solution is to draw the graph of the cubic function f(x) = x

3 20 x2 + x

and see where it crosses the x-axis. This is shown opposite. You can see that it does in fact have three real roots. You can use a graph calculator to find approximations for the other roots but we can use factorization to find values for these roots.

As x = 5

is a solution, we know that (x - 5) is a factor of the cubic equation x

3 20 x2 + x = 0

and can write x

3 20 x2 + x = (x - 5) × (quadratic formula)

Moreover, the quadratic must be of the form

x

2 + bx + 25

Finding b is not quite so obvious but can be done as follows. x

3 20 x2 x x 5) (x2 + bx + 25)

To make

x 2 in the cubic equation, b ֜ To make x in the cubic equation, b ֜

Hence x

3 20x2 x x 5) (x2 x + 25)

So we can now solve the cubic equation,

x

3 20x2 x - = 0

We can write (x 5) (x

2 x + 25) = 0

Either (x 5) = 0 or (x

2 x + 25) = 0

So x = 5 or, solving the quadratic, x =

ଵହ ±ξଶଶହିଵ଴଴ 6 = ଵହ ±ξଵଶହ 6 = ଵହ ±ଵଵ.ଵ଼ ଶ (working to 2 decimal places) y x ©

Factorising Polynomials TEXT

3 1

Worked Example

1

Find all solutions of x

3 x2 x + 35 = 0

Solution

You can see

that x is a root of this equation since f () = 0 .

Hence (x must be a factor, and you can write

x 3 x 2 x + 35 = (x )( x 2 + bx 35) Now to find b. The cubic equation contains x, so b ֜ ֜ 3 3x 2 33x + 35 = (x x
2 2 x 35) In this case, the quadratic factorises to (x 7)( x + 5).

Hence x

3 x 2 x + 35 = (x x x + 5)

The original equation can thus be re-written

(x x 7)( x + 5) = 0 giving the three solutions x = 5, x = and x = 7.

Worked Example

2

Solve x

3 x

Solution

x is a root of this equation. Hence (x + 3) is a factor. x

3 x x + 3)(x2 + bx 5)

Now to find b. The cubic equation has 0 x

2 , so

0 = b + 3

֜ ֜ 3 x = (x + 3)(x 2 3x 5)

The original

equation thus reads (x + 3)( x 2 x

So either x + 3 = 0 or x

2 3x 5 = 0 This gives x = x = or x = to 3 significant places. ©

Factorising Polynomials TEXT

4 2 1

Exercises

(a) Work out the missing quadratic factors (i)x 3 3x 2 6 x + 8 = ( x 4) ( ) (ii)x 3 + 8x 2 + x = ( x + 3) ( ) (iii)2 x 3 x 2 x 324 = (2 x + ( ) (b)Use your answers to (a) to find all the roots of (i)x 3 3x 2 6 x + 8 = 0 (ii)x 3 + 8x 2 x (iii)2x 3 x 2 x 324 = 0 2.Exp lain how you know that (a)(x 3) is a factor of x 3 2 x 2 + x (b)(x + 5) is a factor of 2 x 3 + 6 x 2 23x
(c)(2 x is a factor of 4 x 3 + 2 x 2 + 8x - 5 3.Fi nd all the roots of these equations (a)x 3 5x 2 + 6 x 2 = 0 (b)x 3 + 3x 2 46 x = 48
(c)2 x 3 x 2 x + = 0 4.Fo ur identical ‘square corners" are cut from a square piece of card measuring cm by make an open topped box with volume 64 cm3.

Find the

size of the squares that must be removed

Cubic Equation: Number of Roots

In the examples above, you have seen that cubic equations can have 3 real roots but this is not always the case and The other problem encountered above is knowing the value of one of the roots. There are methods that we can use to determine all the roots but this is rather more complex than the procedure for finding the roots of a quadratic equation.

The following graphs as the roots are given by

the x-value when the graph crosses the axis (and the value of the cubic is zero). ©

Factorising Polynomials TEXT

5 2 A practical way of finding the 3 roots is to consider a numerical or graphical approach.

For example, to solve the equation

ݔ 3 െ100 ݔ 2 +2000 ݔ െ1500= 0
it is a help to work out a few values of the function first. Let the function be labelled f (x).

A table of values is shown on the right

.

The arrows show where

f(x) changes from positive to negative or vice versa.

Hence the zeros of

f(x) must occur between 20 and 30 between 70 and 80 Trial and improvement, or graphs in the correct regions, give the answers of 0.78, 26.40 and 72.82, working to

2 decimal places.

Exercises

(a) Explain how you can tell that the function f (x) = x 3 + x 2 x + 5 has a zero between x x . (b)Find this value to 3 significant figures. x f(x) -20 - 0 20 30
40
50
60
70
80 -
-32500 - 6500
-4500 - -26500 -25500 -8500 30500

Note also that when the coefficient of the x

3 term is negative, then the curve looks like the one opposite; as before there will be ©

Factorising Polynomials TEXT

6 3 2

2.How many real roots does the equation f (x) = 0 have when

f (x) = x 3 െ 3x 2 + 3x െ 3.Sh ow that this cubic x 3 െ x 2 െ x െ 2 = 0 has just one real root. 4.Sho w that the cubic x 3 െ 4x 2 + 5x െ 2 = 0 has only two real roots.

Sketch a graph to illustrate this.

Factor Theorem

Quadratics and cubics are particular examples of

polynomial functions. a quadratic function is a polynomial of degree 2 a cubic has degree 3.

In general a

polynomial of degree n has the form an x n + a n-1 x n + ... + a 2 x 2 + a 1 x + a 0 where a 0 , a 1 , ... an are real numbers, and an . When solving cubic equations, the spotting of a root leads immediately to a simple linear factor. For example, 3 gives a zero value for the cubic x 3 x 2 x ֜ 3 5x 2 + x ֜ 3 + 5x 2 + x = (x 3) × (quadratic)

The same technique can be applied to polynomials

of any degree. For example, 2 gives a zero value for the function x 4 + 3x 2 x + 6 ֜ 4 + 3x 2 x + 6 ֜ 4 + 3x 2 x + 6 = (x 2) × (cubic)©

Factorising Polynomials TEXT

7 If P(x) is a polynomial of degree n and has a zero at x = a, that is P (a) = 0, then (x a) is a factor of P(x) and

P(x) = (x a) × Q(x)

where

Q(x) is a polynomial of degree (݊ െ1).

2 3

In general

Th is is known as the factor theorem.

Worked Example

1

Show that (x is a factor of the polynomial

x

5 4 x 4 x3 x2 + 25

Solution

Denote the polynomial by P(x) so

P(5) = 55 4 × 54 53 2 + 25

= 0 Hence 5 gives a zero value for P(x) and so by the factor theorem, (x is a factor of this polynomial.

Worked Example

2

Show that x =

ଵ ଶ gives a zero value to the 2x

3 3x2 3x + 2 .

Also show that 2x

3 3x2 3x + 2 has two linear factors other than x =

ଵ ଶ .

Solution

We write the function as

P(x) = 2x

3 3x2 3x + 2

We can see that ܲ

ଵ ଶ A = 2×ቀ ଵ ଶ A 7

F3 ×ቀ

ଵ ଶ A 6

F3 ×ቀ

ଵ ଶ A+ 2 = ଵ ସ െ ଷ ସ െ ଷ ଶ + 2 = 0

We know that ቀݔെ

ଵ ଶ A is a factor so (2x െ 1) is also a factor and we can write 2x

3 3x2 3x + 2 = (2x െ 1)

× (aݔ

ଶ ܾ+

Factorising Polynomials TEXT

8 3 Here a, b and c are constants and so a and c = 2. Comparing coefficients of x on each side of the identity, 3 = b + 2c, giving b =

You can check this by comparing coefficients of

ݔ ଶ .

So P(x) = (2x െ 1) × (ݔ

ଶ െ ݔെ 2) = (2x െ 1) × (ݔ െ2)× (ݔ+ 1), after factorising the quadratic.

Exercises

Show that

(a)(x is a factor of x 4 + 3x 3 2 x 2 + 5x - 7 (b)(x + is a factor of x 5 x 4 x 3 + 2 x 2 + x - (c)(x 3) is a factor of x 4 2 x 3 7 x - 6 (d) (x 2) is a factor of 2 x 6 + 5x 4 27 x
3 + 8 (e)(x + is a factor of 5x 5 + 23x 4 x 3 x + (f)(x 8) is a factor of x x + 8x 8 x 2 + x + 56 2.Th ree of the polynomials below have a linear factor from the list on the right.

Match the factors to the polynomials.

The 'odd one out' does have a simple linear factor, but not one in the list. Find the factor.

A: x

4 5x 2 + 3x 2 1: x +

B: x

5 4 x 4 + 2 x 3 8x 2 x + 4 2: x 2

C: x

3 4 x 2 6 x 3: x + 3 D: x 3 x 2 x + 6 3. Fi nd a linear factor for each of these polynomials. (a)x 3 + 4 x 2 + x 6 (b)x 4 x 4 (c)x 4 x 3 x 2 + 6 x + 7 (d)2 x 4 + x 3 x 2 x (e)3x 3 x 2 x + 2©

Factorising Polynomials TEXT

9

4Factors of Higher Order Polynomials

One method if we know one factor involves juggling with coefficients, and is best demonstrated by example.

Consider the quartic

x

4 + 5x3 + 7x2 + 6 x + 8

It is easy to check that

x gives a zero value for this function and hence (x + 2) is a factor.

The cubic factor can be found as follows.

x 4 + 5x 3 + 7x 2 + 6 x + 8 = x 3 (x + 2) + 3x 3 + 7x 2 + 6 x + 8 = x 3 (x + 2) + 3x 2 (x + 2) + x 2 + 6 x + 8 = (x 3 + 3x 2 ) (x + 2) + x(x + 2) + 4 x + 8 = (x 3 + 3x 2 + x) ( x + 2) + 4(x + 2) = (x 3 + 3x 2 + x + 4) (x + 2)

Hence the cubic factor is

(x 3 + 3x 2 + x + 4)

Worked Example

1

Find the missing

cubic expression here: x 4 3x 3 + 5x 2 + x 4 = (x ( cubic )

Solution

x 4 x 3 + 5x 2 + x 4 = x 3 (x x 3 + 5x 2 + x 4 = x 3 (x x 2 (x x 2 + x 4 = (x 3 x 2 ) (x x (x x = (x 3 x 2 + 3x) (x x = (x 3 x 2 + 3x + 4) (x

So the missing expression is

x

3 2 x2 + 3x + 4.

The second method is sometimes called 'long division of polynomials'.

The statement

x

4 + 5x3 + 7x2 + 6 x + 8 = (x +2) ( cubic )

can be written instead in this form ௫ ర ା ହ௫ య ା ଺௫ ା ଼ ௫ ା ଶ = cubic polynomial©

Factorising Polynomials TEXT

10 The cubic polynomial is thus the result of dividing the quartic by (x 2). Thi s division can be accomplished in a manner very similar to long division of numbers as we see below. x 3 + 3x 2 + x + 4 x + 2 x 4 + 5x 3 + 7x 2 + 6 x + 8 x 4 + 2 x 3 3x 3 + 7x 2 3x 3 + 6x 2 x 2 + 6 x x 2 + 2 x

4 x + 8

4 x + 8

0

So x

4 + 5x 5 + 7x 2 + 6 x + 8 = (x + 2) (x 3 + 3x 2 + x + 4) This is same answer is obtained as on the previous page by the 'juggling' method.

Worked Example

2

Evaluate the quotient

௫ ర ି ହ௫ య ି ଶ௫ మ ା ଶହ௫ ି ଷ ௫ି ଷ

Solution

x 3 െ 2x 2 െ 8x + 4 x െ 2 x 4 െ 5x 3 െ 2x 2 + 25 x െ 3 x 4 െ 3x 3 െ2x 3 െ 2x 2 െ2x 3 + 6 x 2 െ8x 2 + 25 x െ8x 2 + 24 x x െ 8 x െ 8 0 We can use these methods to help solve quartics or even higher order polynomials but in reality for practical work, numerical and graphical approaches are both easier and more ©

Factorising Polynomials TEXT

11 4 appropriate to the problem. Here though is one algebraic approach to solvin g a quartic equation.

Worked Example

3

Factorise the quartic

x

4 x3 x2 + 34 x as fully as possible and hence solve the

equation x

4 4 x3 7x2 + 34 x 24 = 0

Solution

x gives zero value for the quartic, so (x - ) is a factor.

Long division gives

x

4 4 x3 7x2 + 34 x 24 = (x (x3 3x2 x + 24)

Now factorise the cubic.

Since x = 2 is a zero of the cubic, (x 2) is a factor.

Long division now gives

x

3 3x2 x + 24 = (x 2) (x2

x

The quadratic

x

2 x factorises easily to give (x (x + 3).

The full factorisation of the original quartic is therefore (x ( x 2) ( x 4) ( x + 3) and the solutions to the equation are thus x

Worked Example

4

Solve the equation x 4 + 3x3 x2 x 6 = 0.

Solution

First, factorise the quartic as far as possible. x gives a zero value for the quartic so (x is a factor.

By long division

, x 4 + 3x 3 x 2 x -6 = (x x 3 + 2 x 2 x 6) Also x = 3 gives a zero value for the cubic so (x is a factor and x 3 + 2 x 2 x 6 = (x 3)( x 2 + 5x + 2) The quadratic x2 + 5x + 2 has no easy linear factors, but the equation x2 + 5x + 2 = 0 does have solutions, namely and . So the quartic thus factorises to ( x x x2 + 5x + 2 ) giving solutions x ©

Factorising Polynomials TEXT

12 5 4

Exercises

Use the 'juggling' method to find the missing factors (a)x 4 + 8x 3 + x 2 + x + = (x + 3) ( ) (b)x 4 5x 3 x 2 + 25 = (x 5) ( ) 2.Us e long division to find the missing factors (a)x 4 + 6 x 3 + x 2 + 5x + = ( x + () (b)x 4 + 7 x 3 x = (x + 6)( ) (c)2 x 3 4 x 2 7 x + = (x 2)( ) (d) x 5 + x 4 x 3 + x + 2 = (3x + 2)( )

3.Solve the following quartic equations:

(a)x 4 + 8x 3 + x 2 8x = 0 (b)x 4 + 8x 3 x 2 x + 36 = 0

4.Solve the quantic equation

x 5 x 3 + 6 x 2 + 7 x 6 = 0

Remainders

In the previous section you saw how to divide a polynomial by a factor. These ideas will be extended now to cover division of a polynomial by an expression of the form (x ). If (x ) is not a factor, there will be a remainder. As with numbers, long divisions of polynomials often leave remainders.

For example (x + 3) is not a factor of x

3 + 6 x2 + 7x , and so long division will yield a

remainder. x 2 + 3x - 2 x + 3 x 3 + 6 x 2 + 7x 4 x 3 + 3x 2 3x 2 + 7x 3x 2 + x

2 x 4

2 x 6

2 One way of expressing this might be to write ©

Factorising Polynomials TEXT

13 If P(x) is a polynomial of degree n and a is any number then

P(x) = (x െ a) Q(x) + R

where Q(x) is a polynomial of degree (n ) and R = P(a). 5 ௫ య ା଺௫ మ ା଻௫ିସ ௫ାଷ = x 2 + 3x rem 2 but the normal method is to write either x 3 6 x 2 7x 4 (x 2 3x 2) (x 3) 2 or ௫ య ା଺௫ మ ା଻௫ିସ ௫ାଷ = x 2 3x ଶ ௫ାଷ

The Factor Theorem can be extended to remainders

. The Remainder Theorem is given below. Wo rked Example 1

What is the remainder when x

3 7x + is divided by (x െ1)? Work it out by long division and check your answer using the Remainder Theorem.

Solution

x 2 + x - 6 x x 3 + 0 x 2 7x + x 3 x 2 x 2 7x x 2 x 6 x

6x + 6

4

Using the Remainder Theorem,

R = PP(x) = x 3 7x

So R =

3 7× 1 = 4 (as expected) ©

Factorising Polynomials TEXT

14 5 5

Exercises

.Wh at is the remainder when x 3 + 4 x 2 7 x + is divided by (x + 3)? 2.W ork out the remainder when (a)x 2 + 5x 7 is divided by ( x 2) ; (b)x 4 3x 2 + 7 is divided by (x + 3); (c)5x 3 + 6 x 2 + 2 x 3 is divided by (x + 5). 3.W hen the quadratic x 2 + px + is divided by (x the remainder is 5. Find p. 4.x 2 + px + q divides exactly by (x 5) and leaves remainder -6 when divided by (x + .

Find p and q.

5.Fi nd the linear expressions which give a remainder of 6 when divided into x 2 + x + 22.©