MEI C1 Study Resources Core1 Polynomials 1 Introduction

MEI Core 1 Polynomials Section 1 : Introducing polynomials Study plan Background You have been dealing with the addition and subtraction of polynomials since you first started with algebra and started collecting like terms. Multiplying polynomials follows the same procedure as for expanding brackets. Division of polynomials is probably new to you. Practice at this is important as you will meet this in other areas of the course.

Detailed work plan 1. Study the examples on pages 78 to 79 carefully. There are further worked examples in the Notes and Examples. 2. You can look at the Flash resources Polynomial addition, Polynomial subtraction and Polynomial multiplication. There are also some PowerPoint presentations on Multiplying polynomials and Multiplying three brackets, which help to show the thought processes involved in each step of these techniques. 3. Exercise 3A Try at least all of the odd numbered questions from 1 to 15 including 5*, 9* and 13*. Those with * beside them have worked solutions on the website for you to download. 4. Read the example on Polynomial division on pages 80 – 81 carefully. Many students find this technique difficult at first, but with plenty of practice you should be able to master it. 5. You can look at further examples using the Flash resources Polynomial division (long division), Polynomial division by inspection, Polynomial divison: the box method (no remainder) and Polynomial division: the box method (remainder). You can also look at the Geogebra resource Polynomial division, which shows the box method. There is also a PowerPoint presentation on Dividing polynomials, which looks at several different approaches. There is also a Polynomial division video, which concentrates on the long division approach. 6. Exercise 3A Try questions 16, 17, 18, 19* and 20. For additional practice in dividing polynomials, try the interactive resource Dividing polynomials.

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MEI C1 Polynomials Section 1 Study plan 7. Being able to visualise the shape of a curve by looking at its equation is a useful skill to develop. The notes on pages 82 – 86 highlight some of the things to look for and remember. Also read the additional notes in the Notes and Examples – some important points are given which will help you gain maximum marks for curve sketching in examinations. 8. Look at the Flash resource Sketching factorised cubics. In each case, before clicking on “Show graph”, think about where the graph will cut the axes and what the graph will look like. You may also find the Mathcentre video Polynomial functions useful. 9. Exercise 3B Try the odd numbered questions including 5* and 9*. 10. For additional practice, try the interactive resources Sketching polynomial curves and Finding a polynomial from its roots.

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MEI Core 1 Polynomials Section 1: Introducing Polynomials Notes and Examples These notes contain subsections on Adding and subtracting polynomials Multiplying polynomials Dividing polynomials Graphs of polynomial functions Sketching graphs of polynomials in factorised form Finding the equation of a curve

This section involves quite a lot of algebraic manipulation. It may look a bit daunting, but you will probably find that it is easier than it looks once you have got the hang of it! There are a lot of possible different approaches to the examples in this section. You may also see different variations in your textbook or in class, or you may find your own method. In these notes, the examples are done using fairly traditional methods, as these are easy to show on a printed page. Other approaches are difficult to show on a printed page because of the thought processes involved, so you will also find links to PowerPoint presentations that demonstrate these methods step-by-step. You will find several “in your head” methods in the PowerPoint presentations. You may find these approaches difficult at first. If you do, then leave them and choose a different method, but it is well worth returning to them and trying again later in your course as you become more confident.

Adding and subtracting polynomials Example 1 For the polynomials f(x) = 2x³ – 3x² + 1 g(x) = x³ + x² – 3x – 4 find (i) f(x) + g(x) (ii) f(x) – g(x) Solution (i) (2x³ – 3x² + 1) + (x³ + x² – 3x – 4) = (2x³ + x³) + (-3x² + x²) + (-3x) + (1 – 4) = 3x³ – 2x² – 3x – 3 Add the terms in x³, the terms in x², the terms in x and the constant terms separately. Notice that f(x) does not have a term in x. It is not essential to put in the brackets as shown here, but it can be helpful.

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MEI C1 Polynomials Section 1 Notes and Examples +

2 x3 x3

3x 2 x2

3x

1 4

3x3

2 x2

3x

3

Alternatively you can write it out like an addition sum:

(ii) (2x³ – 3x² + 1) – (x³ + x² – 3x – 4) = (2x³ – x³) + (-3x² – x²) + (3x) + (1 + 4) = x³ – 4x² + 3x + 5

Do this in a similar way, but be careful about signs.

–

2 x3 x3

3x 2 x2

3

2

x

4x

3x

1 4

3x

5

Alternatively (again be careful about signs):

You can see more examples like this using the Flash resources Polynomial addition and Polynomial subtraction.

Multiplying polynomials You are already familiar with multiplying out two linear expressions to obtain a quadratic expression. You now need to be able to deal with more complicated multiplications. The principle is the same – for a multiplication involving two brackets, each term in one bracket needs to be multiplied by each term in the other bracket. First of all, it is helpful to think about how many terms there will be in the result (before simplifying). Each of the 3 terms in the first bracket must be multiplied by each of the terms in the second bracket, so there should be 9 terms altogether. There are a number of different ways of setting out this multiplication. One way is shown in Example 2 below. You can also view the animated PowerPoint presentation Multiplying polynomials, which shows two other methods: (1) using a table, and (2) doing it “in your head”.

Example 2 Multiply x² + 3x – 2 by 2x² – x + 4

Each term in the first bracket is multiplied by the whole of the second bracket

Solution (x² + 3x – 2)(2x² – x + 4) = x²(2x² – x + 4) + 3x(2x² – x + 4) – 2(2x² – x + 4) = (2x4 – x³ + 4x²) + (6x³ – 3x² + 12x) + (-4x² + 2x – 8)

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MEI C1 Polynomials Section 1 Notes and Examples = 2x4 + (–x³ + 6x³) + (4x² – 3x² – 4x²) + (12x + 2x) – 8 = 2x4 + 5x³ – 3x² + 14x – 8

Collect like terms together

You can see more examples using the Flash resource Multiplying polynomials.

You also need to be able to multiply out expressions involving more than one bracket. This is shown in the next example.

Example 3 Multiply out (x – 2)(2x + 3)(3x – 1) Solution It is often easiest to multiply out one pair of brackets, and then multiply the result by the third bracket. (x – 2)(2x + 3) = 2x² + 3x – 4x – 6 = 2x² – x – 6 (2x² – x – 6)(3x – 1)

Simplify here or the next step will be more complicated!

= 2x²(3x – 1) – x(3x – 1) – 6(3x – 1) = 6x³ – 2x² – 3x² + x – 18x + 6 = 6x³ – 5x² – 17x + 6

You can of course use another approach, such as using a table, or doing it in your head, to multiply the quadratic by the third bracket. You can also do the whole thing “in your head”. The PowerPoint presentation Multiplying three brackets gives an animated example.

Dividing polynomials When you divide one polynomial by another, it may divide exactly, or there may be a remainder, just as in arithmetic. For example:

26

6 = 4 remainder 2

You could rewrite this statement as: 26 = 6

26 is called the dividend, 6 is called the divisor, 4 is the quotient and 2 is the remainder.

4+2

This rearrangement helps to give one method of dividing polynomials. When working with polynomials, remember the following points: If you are dividing by a linear expression, the quotient is of order one less than the dividend (e.g. for a quartic, the quotient is cubic) and the remainder, if any, is a constant term.

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MEI C1 Polynomials Section 1 Notes and Examples If you are dividing by a quadratic term, the quotient is of order two less than the dividend (e.g. for a quartic, the quotient is quadratic) and the remainder, if any, could be linear or a constant term. This idea can be extended to a polynomial divisor of any order. Most, if not all, of the examples you meet will only involve dividing by a linear expression.

Example 4 Divide 2x³ + 3x² - x + 1 by x + 2

When you divide a cubic expression by a linear expression, as in this example, the quotient is a quadratic expression and the remainder, if any, is a constant term.

Solution

Let the quotient be ax² + bx + c and the remainder be d. 2x³ + 3x² – x + 1

= (x + 2)(ax² + bx + c) + d = x(ax² + bx + c) + 2(ax² + bx + 2) + d = ax³ + bx² + cx + 2ax² + 2bx + 2c + d = ax³ + (b + 2a)x² + (c + 2b)x + 2c + d

Equating coefficients of x³ Equating coefficients of x² Equating coefficients of x Equating constant terms

a=2 b + 2a = 3 c + 2b = -1 2c + d = 1

b + 4 = 3 b = -1 c – 2 = -1 c = 1 2 + d = 1 d = -1

2x³ + 3x² – x + 1 = (x + 2)(2x² – x + 1) – 1 The quotient is 2x² – x + 1 and the remainder is -1. Don’t be put off by a negative remainder, as in this example – this is quite acceptable in polynomial division!

There are other approaches to polynomial division. The PowerPoint presentation Dividing polynomials shows (1) algebraic long division and (2) polynomial division by inspection (i.e. in your head). The Geogebra resource Polynomial division shows the box method. You can also look at more examples using the Flash resources Polynomial division (long division), Polynomial division by inspection, Polynomial division: the box method (no remainder) and Polynomial division: the box method (remainder). There is also a Polynomial division video – this deals with the long division method only. For some extra practice in examples like the one above, try the interactive resource Dividing polynomials.

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MEI C1 Polynomials Section 1 Notes and Examples Graphs of polynomial functions There are two simple rules about the graphs of polynomial functions: a polynomial of order n meets the x-axis at most n times. a polynomial of order n has at most n – 1 turning points. For example, a quadratic, which has order 2, has 1 turning point and meets the x-axis at most twice.

This quadratic crosses the x-axis twice.

This quadratic touches the x-axis once.

This quadratic does not meet the x-axis at all.

A cubic, which has order 3, has at most 2 turning points and meets the x-axis at most three times and at least once. Here are some examples.

This cubic graph has two turning points and crosses the x-axis three times.

This cubic graph has two turning points, but only crosses the x-axis once.

The simplest cubic graph of all, y = x³, has just one turning point, a point of inflection. You can think of this as two turning points in the same place. It crosses the x-axis just once.

A quartic, which has order 4, has at most 3 turning points and meets the x-axis at most four times. Here are some examples.

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MEI C1 Polynomials Section 1 Notes and Examples

This quartic graph has three turning points and crosses the x-axis four times.

This quartic graph has three turning points and crosses the x-axis twice.

This quartic graph has just one turning point and crosses the x-axis twice. Notice that the turning point is much flatter than usual – it is actually three turning points all in the same place. This makes the shape different from a quadratic graph.

Sketching graphs of polynomials in factorised form You have already done some work on factorising quadratic expressions. As you know, a quadratic expression can sometimes be factorised into two linear factors. These factors can be used to tell you where the graph of the quadratic cuts the x-axis. For example, the quadratic expression can be written in factorised form as

y = x² + x – 2 y = (x – 1)(x + 2)

When the graph cuts the x-axis, the value of y = 0.

(x – 1)(x + 2) = 0 x = 1 or x = -2.

The graph therefore cuts the x-axis at (1, 0) and (-2, 0). You can also find out where the graph cuts the y-axis by substituting x = 0. In this case, when x = 0, y = -2, so the graph cuts the y-axis at (0, -2). This information allows you to sketch the graph. y = x² + x – 2

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MEI C1 Polynomials Section 1 Notes and Examples It is also useful to think about the behaviour of the graph as x → ( i.e. when x is very large and positive) and as x → - ( i.e. when x is very large and negative). For the quadratic graph above, for values of x which are numerically large the term x² is the most significant term (the dominant term). So you just need to think about whether this term is positive or negative. In this case it is positive as x → and as x → - . So the graph disappears off the top of the page at both the left and right of the graph. The same ideas can be extended to any polynomial. You will learn how to factorise cubics and higher order polynomials in Section 2. For now, you will look at polynomials that are given in factorised form. Examination questions commonly ask you to sketch the graph of a function which you have already factorised. This should be an easy source of marks (provided you have managed the factorising!) but a lot of candidates throw marks away. Here are some tips: A sketch means a sketch! Do it in the answer booklet, not on graph paper (just because graph paper is provided does not mean that you are expected to use it!). You should certainly not be calculating and plotting points – this is very time-consuming, unnecessary and may result in only part of the graph. Show the points at which the graph cuts BOTH axes. The points at which the graph cuts the x-axis can be found from the factorised equation. Find the point at which the graph cuts the y-axis by substituting x = 0, and mark on this point – one mark is often given for this. Make sure that your graph is the right way up. If you have marked on the point where the graph cuts the y-axis then it should be clear how the graph goes. Also think about whether y is positive or negative when x is very large and positive, and when x is very large and negative. Do not stop the graph when you reach an axis! You may lose marks for this. The graph should go beyond each of the points marked.

Example 5 Sketch the following graphs, showing the points where the graphs meet the coordinate axes. (i) y ( x 1)( x 3)(2 x 1) (ii) y x( x 2) 2 ( x 2) Solution (i) y ( x 1)( x 3)(2 x 1) When y = 0, x = 1, x = -3 or x =

1 2

.

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MEI C1 Polynomials Section 1 Notes and Examples When x = 0, y = -1 3 1 = -3 The graph crosses the x-axis at (1, 0), (-3, 0) and ( The graph crosses the y-axis at (0, -3).

1 2

y

, 0).

( x 1)( x 3)(2 x 1)

As x → 1

-3

, y is positive. As x → - , y is negative.

1

2

-3

(ii) y x( x 2) 2 ( x 2) When y = 0, x = 0, x = 2 (repeated) or x = -2.

When a root is repeated, the graph touches the x-axis

There is no need to look at where the graph crosses the y-axis, since you already know it passes through the origin.

The graph crosses the x-axis at (0, 0) and (-2, 0), and touches the x-axis at (2, 0).

As x →

, y is positive. As x → - , y is positive. -2

0

2

For some examples, look at the Flash resource Sketching factorised cubics. For each example, look at the equation in its factorised form and work out the points at which the graph will cross both axes, and decide which way up it will be. Then press “Show curve” to check. You may also find the Mathcentre video Polynomial functions useful.

For some extra practice in recognising the shapes of quadratic and curves, try the interactive questions Sketching polynomial curves. (If you look at the worked solutions, don’t worry about the mention of differentiation which is

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MEI C1 Polynomials Section 1 Notes and Examples covered in Core 2 – all you need to know at this stage is that quadratics have one turning point and that cubics can have up to two).

Finding the equation of a curve If you know where a polynomial curve cuts the axes (or alternatively, if you are given the roots of a polynomial equation), you can deduce the equation of the curve by writing it in factorised form and then multiplying out the brackets.

Example 6 Find a polynomial equation which has roots x = 1, x = -3 and x = 0.5. Solution A polynomial equation with these roots is ( x 1)( x 3)(2 x 1) 0 . Multiplying out: ( x 1)( x 3)(2 x 1) ( x 2 2 x 3)(2 x 1) 2 x3 4 x 2 6 x x 2 2 x 3 2 x3 3x 2 8 x 3

An equation with these roots is 2 x3 3x2 8x 3 0

Note that any multiple of this polynomial (e.g. 4 x3 6 x 2 18x 6 0 ) would have the same roots. This is the simplest possible equation.

For some extra practice in examples like the one above, try the interactive resource Finding a polynomial from its roots.

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Core 1 Polynomials Section 1: Introducing polynomials Crucial points 1. Always check your work carefully It is very easy to make silly mistakes when manipulating algebraic expressions. 2. When multiplying out brackets, make sure that you know how many terms there should be You can find out how many by multiplying together the number of terms in each bracket, so that you know that you have not missed any. 3. Make sure that you can divide polynomials with confidence Try the different methods (long division, inspection or using a table) and choose the one that you feel most comfortable with. Get lots of practice until you feel really confident. Remember that you can check your answer by multiplying 4. Make sure that you know the basic rules about polynomial graphs A polynomial of degree n crosses the x axis at most n times and has at most n – 1 turning points. A repeated root means that the graph touches the x-axis at this point. 5. Don’t lose easy marks when sketching polynomials You will often be asked to sketch a polynomial which you have already factorised. Remember you are being asked for a sketch, so you should do this in your answer booklet and NOT on graph paper. You should certainly not be wasting time plotting points. Make sure that your graph does not stop at the axes, that you have shown all the points at which the graph cuts the axes (including the y-axis), and that your graph is the correct way up.

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MEI Core 1 Polynomials Section 1: Introducing polynomials Exercise Do not use a calculator or graph-drawing package for this exercise. 1. Given that f ( x) = x3 + 2 x 2 − 5 x + 4 and g( x) = x3 − 3 x 2 + 1 , find (i) f(x) + g(x) (ii) f(x) – g(x) 2. Given that p( x) = 2 x3 − 5 x 2 + 3x − 2 and q( x) = x3 − 2 x 2 + 1 , find (i) q(x) – p(x) (ii) 2p(x) + 3q(x) 3. Given that f ( x) = x3 + 5 x 2 − 3 and g( x) = 3 x 4 − 2 x3 + x , find (i) g(x) – 3f(x) (ii) (2x + 1)f(x) 4. Expand the brackets and simplify the following as far as possible: (i) ( x − 2)(2 x 2 − 3x + 1) (ii) (3 x − 2)( x3 − 2 x + 4) (iii) (2 x + 1)( x3 + 2 x 2 − 3x − 5) (iv) (3x 2 − x + 2)(2 x 2 + 5 x − 1) (v) ( x + 3)(2 x − 1)( x − 4) (vi) (2 x + 3)( x − 2)( x 2 + 1) 5. Given that in each case there is no remainder, divide (i) 2 x3 − x 2 + 7 x + 4 by 2x + 1. (ii) x3 + 2 x 2 − 3 by x – 1. (iii) 6 x 4 − 4 x3 + 3x 2 + 4 x − 4 by 3x – 2. 6. Sketch the following graphs: (i) y = ( x + 1)( x − 3)( x + 4) (ii) y = ( x + 2) 2 (2 x − 1) (iii) y = x(3 − x)(2 x + 3) (iv) y = x 2 ( x − 2)( x + 3) (v) y = ( x − 2) 2 (3 x + 4) 2

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MEI Core 1 Polynomials Section 1 : Introduction to polynomials Solutions to Exercise 1. (i)

f( x )  g( x )   x 3  2 x 2  5 x  4    x 3  3 x 2  1   x 3  2 x 2  5 x  4  x 3  3x 2  1  x 3  x 3  2 x 2  3x 2  5 x  4  1  2x3  x2  5 x  5

(ii) f( x )  g( x )   x 3  2 x 2  5 x  4    x 3  3 x 2  1 

 x 3  2 x 2  5 x  4  x 3  3x 2  1  x 3  x 3  2 x 2  3x 2  5 x  4  1  5 x2  5 x  3

2. (i)

q( x )  p( x )   x 3  2 x 2  1    2 x 3  5 x 2  3 x  2   x 3  2 x 2  1  2 x 3  5 x 2  3x  2  x 3  2 x 3  2 x 2  5 x 2  3x  1  2  x 3  3x 2  3x  3

(ii) 2 p( x )  3q( x )  2  2 x 3  5 x 2  3 x  2   3  x 3  2 x 2  1 

 4 x 3  10 x 2  6 x  4  3 x 3  6 x 2  3  4 x 3  3 x 3  10 x 2  6 x 2  6 x  4  3  7 x 3  16 x 2  6 x  1

3. (i)

g( x )  3f( x )  3 x 4  2 x 3  x  3  x 3  5 x 2  3   3 x 4  2 x 3  x  3 x 3  15 x 2  9  3 x 4  2 x 3  3 x 3  15 x 2  x  9  3 x 4  5 x 3  15 x 2  x  9

(ii) (2 x  1)f( x )   2 x  1   x 3  5 x 2  3 

 2 x  x 3  5 x 2  3   x 3  5 x 2  3  2 x 4  10 x 3  6 x  x 3  5 x 2  3  2 x 4  10 x 3  x 3  5 x 2  6 x  3  2 x 4  11 x 3  5 x 2  6 x  3

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MEI C1 Polynomials Section 1 Exercise solutions 4. (i) ( x  2)(2 x 2  3 x  1)  x(2 x 2  3 x  1)  2(2 x 2  3 x  1)

 2 x 3  3x 2  x  4x 2  6x  2  2x3  7x2  7x  2 (ii) (3 x  2)( x 3  2 x  4)  3 x( x 3  2 x  4)  2( x 3  2 x  4)

 3 x 4  6 x 2  12 x  2 x 3  4 x  8  3 x 4  2 x 3  6 x 2  16 x  8 (iii) (2 x  1)( x 3  2 x 2  3 x  5 )

 2 x( x 3  2 x 2  3 x  5 )  ( x 3  2 x 2  3 x  5 )  2 x 4  4 x 3  6 x 2  10 x  x 3  2 x 2  3 x  5  2 x 4  5 x 3  4 x 2  13 x  5 (iv) (3 x 2  x  2)(2 x 2  5 x  1)

 3 x 2(2 x 2  5 x  1)  x(2 x 2  5 x  1)  2(2 x 2  5 x  1)  6 x 4  15 x 3  3 x 2  2 x 3  5 x 2  x  4 x 2  10 x  2  6 x 4  13 x 3  4 x 2  11 x  2 (v) ( x  3)(2 x  1)( x  4)  ( x  3)(2 x 2  8 x  x  4)

 ( x  3)(2 x 2  9 x  4)  x(2 x 2  9 x  4)  3(2 x 2  9 x  4)  2 x 3  9 x 2  4 x  6 x 2  27 x  12  2 x 3  3 x 2  23 x  12 (vi) (2 x  3)( x  2)( x 2  1)  (2 x  3)( x 3  2 x 2  x  2)

 2 x( x 3  2 x 2  x  2)  3( x 3  2 x 2  x  2)  2 x 4  4x 3  2 x 2  4x  3x 3  6 x 2  3x  6  2 x 4  x 3  4x 2  x  6

5. (i)

2 x 3  x 2  7 x  4  (2 x  1)(ax 2  bx  c )  2ax 3  ax 2  2bx 2  bx  2cx  c  2ax 3  (a  2b )x 2  (b  2c )x  c Equating coefficients of x³: 2  2a  a  1 Equating coefficients of x²: 1  a  2b  1  1  2b Equating constant terms: 4  c Check: coefficient of x  b  2c  1  8  7 2x3  x2  7x  4 2x  1  x2  x  4

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MEI C1 Polynomials Section 1 Exercise solutions (ii) x 3  2 x 2  3  ( x  1)(ax 2  bx  c )

 ax 3  ax 2  bx 2  bx  cx  c  ax 3  (b  a )x 2  (c  b )x  c Equating coefficients of x³: 1  a  a  1 Equating coefficients of x²: 2  b  a  2  b  1  b  3 Equating constant terms: 3  c  c  3 Check: coefficient of x  c  b  3  3  0 x 3  2 x 2  3  x  1  x 2  3x  3 (iii) 6 x 4  4 x 3  3 x 2  4 x  4  (3 x  2)(ax 3  bx 2  cx  d )

 3ax 4  2ax 3  3bx 3  2bx 2  3cx 2  2cx  3dx  2d  3ax 4  (3b  2a )x 3  (3c  2b )x 2  (3d  2c )x  2d Equating coefficients of x4: 6  3a  a  2 Equating coefficients of x³: 4  3b  2a  4  3b  4  b  0 Equating coefficients of x²: 3  3c  2b  3  3c  0  c  1 Equating constant terms: 4  2d  d  2 Check: coefficient of x  3d  2c  6  2  4 6 x 4  4x 3  3 x 2  4x  4  3 x  2  2 x 3  x  2

6. (i)

y  ( x  1)( x  3)( x  4) This is a cubic graph which cuts the x-axis at (-1, 0), (3, 0) and (-4, 0). When x = 0, y  1  3  4  12 When x is large and positive, y is positive. When x is large and negative, y is negative.

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(ii) y  ( x  2)2(2 x  1) This is a cubic graph which touches the x-axis at (-2, 0) and cuts the x-axis at  21 ,0  . When x = 0, y  2 2  1  4 When x is large and positive, y is positive. When x is large and negative, y is negative. (iii) y  x(3  x )(2 x  3) This is a cubic graph which cuts the x-axis at (0, 0), (3, 0) and   23 ,0  . When x = 0, y  0 When x is large and positive, y is negative. When x is large and negative, y is positive.

© MEI, 13/09/10

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MEI C1 Polynomials Section 1 Exercise solutions (iv) y  x 2( x  2)( x  3) This is a quartic graph which touches the x-axis at (0, 0) and cuts the x-axis at (2, 0) and (-3, 0). When x = 0, y  0 When x is large and positive, y is positive. When x is large and negative, y is positive. (v) y  ( x  2)2(3 x  4)2 This is a quartic graph which touches the x-axis at (2, 0) and   43 ,0  . When x = 0, y  2 2  42  64 When x is large and positive, y is positive. When x is large and negative, y is positive.

© MEI, 13/09/10

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MEI Core 1 Polynomials Section 1: Introducing polynomials Multiple Choice Test Do not use a calculator for this test. Questions 1 – 6 are about the polynomials f(x) = x4 – 2x³ + 2x – 1 g(x) = 3x³ + 4x² – 2x + 5

1) f(x) + g(x) = (a) x4 + x³ + 6x² + 4 (c) x4 + x³ + 4x² + 4 (e) I don’t know

(b) 4x4 + 2x³ + 4 (d) x4 + 3x³ + 2x² + 2x + 4

2) f(x) – g(x) = (a) x4 + x³ + 4x² + 4 (c) x4 + 5x³ + 4x² – 4x – 6 (e) I don’t know

(b) x4 – 5x³ – 4x² + 4x – 6 (d) x4 – 3x³ – 6x² – 4x – 6

3) (x – 2)f(x) = (a) x5 – 4x4 + 6x³ – 5x + 2 (c) x5 – 4x4 + 6x² – 5x + 2 (e) I don’t know

(b) x5 – 4x4 + 4x³ + 2x² – 5x + 2 (d) x5 + 4x³ + 2x² + 2

4) (2x + 1)g(x) = (a) 6x4 + 5x³ + 8x² + 8x + 5 (c) 6x4 + 11x³ + 8x² + 2x + 5 (e) I don’t know

(b) 6x4 + 7x² + 2x² + 3x + 5 (d) 6x4 + 11x³ + 8x + 5

5) f(x) ÷ (x – 1) = (a) x³ + x² – x + 1 (c) x³ – x² + x + 3 remainder 2 (e) I don’t know

(b) x³ + x² + x + 3 remainder 2 (d) x³ – x² – x + 1

© MEI, 10/01/09

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MEI C1 Polynomials Section 1 MC test 6) g(x) ÷ (x + 3) = (a) 3x² + x – 5 remainder 20 (c) 3x² + x + 2 (e) I don’t know

(b) 3x² – 5x + 13 remainder -34 (d) 2x² – x + 2

Questions 7 to 9 are about the graphs shown below.

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7) One of these graphs is of the form y = ax³ + bx² + cx + d, where a > 0. This is graph (a) P (c) R (e) I don’t know

(b) Q (d) S

8) One of these graphs is of the form y = ax² + bx + c, where a ≠ 0. This is graph (a) P (c) R (e) I don’t know

(b) Q (d) S

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MEI C1 Polynomials Section 1 MC test 9) One of these graphs is of the form y = ax4 + bx³ + cx + d, where a ≠ 0. This is graph (a) P (c) R (e) I don’t know

(b) Q (d) S

10) The graph of y = ( x − 2)(2 x − 3)( x + 1) cuts the coordinate axes at (a) (2, 0), (1.5, 0), (-1, 0) and (0, 6) (c) (-2, 0), (-1.5, 0), (1, 0) and (0, -6) (e) I don’t know

(b) (2, 0), (1.5, 0) and (-1, 0) (d) (2, 0), (3, 0), (-1, 0) and (0, 6)

© MEI, 10/01/09

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