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*for Edexcel*

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**Linear simultaneous equations**

**Linear simultaneous equations**

**Simultaneous equations involving one linear and one quadratic equation**

**Linear inequalities**

**Quadratic inequalities**

**Polynomials**

**Examination-style questions**

**Linear simultaneous equations**

An equation with two unknowns has an infinite number of solution pairs. For example:

*x*

+

*y*

= 3 is true when

*x*

= 1 and

*y*

= 2

*x*

=

–4 and

*y*

= 7

*x*

= 6.4 and

*y*

=

–3.4 and so on.

We can represent the set of solutions graphically.

The coordinates of every point on the line satisfy the equation.

3

*y x*

+

*y*

= 3

0

3

*x*

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**Linear simultaneous equations**

Similarly, an infinite number of solution pairs exist for the equation

*y*

–

*x*

= 1

Again, we can represent the set of solutions graphically.

*y*

There is one pair of values that satisfies

*both*

these equations simultaneously.

1

*y*

–

*x*

= 1

This pair of values corresponds to the point where the lines and

*y*

–

*x*

= 1 intersect:

*x*

+

*y*

= 3

-1

0

*x*

+

*y*

= 3

This is the point (1, 2). At this point

*x*

= 1 and

*y*

= 2.

*x*

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**Linear simultaneous equations**

Two linear equations with two unknowns, such as

*x*

and

*y*

, can be solved simultaneously to give a single pair of solutions.

When will a pair of linear simultaneous equations have no solutions?

In the case where the lines corresponding to the equations are parallel, they will never intersect and so there are no solutions.

Linear simultaneous equations can be solved algebraically using:

The elimination method,

*or*

The substitution method.

The solution to the equations can be illustrated graphically by finding the points where the two lines representing the equations intersect.

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**The elimination method**

If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example:

Solve the simultaneous equations

3

*x*

+ 7

*y*

= 22 and 3

*x*

+ 4

*y*

= 10.

Subtracting gives:

–

3

*x*

+ 7

*y*

= 22

3

*x*

+ 4

*y*

= 10

3

*y y*

= 12

= 4

The terms in

*x*

have been eliminated.

Substituting

*y*

= 4 into the first equation gives:

3

*x*

+ 28 = 22

3

*x*

=

–6

*x*

=

–2

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**The elimination method**

We can check whether

*x*

=

–2 and

*y*

= 4 solves both

3

*x*

+ 7

*y*

= 22

3

*x*

+ 4

*y*

= 10 by substituting them into the second equation.

LHS = 3

× –2 + 4 × 4

=

–6 + 16

= 10

= RHS

So the solution is

*x*

=

–2

,

*y*

= 4 .

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**The elimination method**

Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example:

Solve: 5

*x*

–

2

*y*

= 31

4

*x*

+ 3

*y*

= 11

**1**

**2**

We need to have the same number in front of either the

*x*

or the

*y*

terms before adding or subtracting the equations.

Call these equations

**1**

and

**2**

.

3

×

2

×

**3**

**1**

**2**

+

**4**

+

15

*x*

– 6

*y*

= 93

8

*x*

+ 6

*y*

= 22

23

*x *

=

115

*x*

= 5

**3**

**4**

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**The elimination method**

Substitute

*x*

= 5 in

**1 :**

5

× 5 – 2

*y*

= 31

25

– 2

*y*

= 31

–2

*y*

= 6

*x*

=

–3

Check by substituting

*x*

= 5 and

*y*

=

–3 into

LHS = 4

× 5 + 3 × –3

= 20

– 9

**2**

:

= 11

= RHS

So the solution is

*x*

= 5 ,

*y*

=

–3

.

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**The elimination method**

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**The substitution method**

Two simultaneous equations can also be solved by substituting one equation into the other. For example:

Solve:

= 2

*x*

2

*x*

+ 3

*y*

= 23

Call these equations

**1**

and

**2**

.

Substitute

**1**

into

**2**

:

2

*x*

+ 3( 2

*x*

– 3

) = 23

2

*x*

+ 6

*x*

– 9 = 23

8

*x*

– 9 = 23

8

*x*

= 32

*x*

= 4

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**1**

**2**

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**The substitution method**

Substituting

*x*

= 4 into

**1**

gives

*y*

= 2

× 4 – 3

*y*

= 5

Check by substituting

*x*

= 4 and

*y*

= 5 into

**2**

:

LHS = 2

× 4 + 3 × 5

= 8 + 15

= 23

= RHS

So the solution is

*x*

= 4 ,

*y*

= 5 .

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**The substitution method**

Solve:

3

*x*

–

*y*

= 9

8

*x*

+ 5

*y*

= 1

**1**

**2**

Call these equations

**1**

and

**2**

.

One of the equations needs to be arranged in the form

*x*

or

*y*

= …

= … before it can be substituted into the other equation.

Rearrange equation

**1**

:

3

*x*

–

*y*

= 9

–

*y*

= 9

– 3

*x y*

= 3

*x*

– 9

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**The substitution method**

Now substitute

*y*

= 3

*x*

– 9 into equation

**2**

:

8

*x*

8

*x*

+ 5( 3

*x*

+ 15

*x*

– 9

) = 1

– 45 = 1

23

*x*

– 45 = 1

23

*x*

= 46

*x*

= 2

Substitute

*x*

= 2 into equation

**1**

to find the value of

*y*

:

6

–

*y*

–

*y y*

= 9

= 3

=

–3

So the solution is

*x*

= 2 ,

*y*

=

–3

.

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**One linear and one quadratic equation**

**Linear simultaneous equations**

**Simultaneous equations involving one linear and one quadratic equation**

**Linear inequalities**

**Quadratic inequalities**

**Polynomials**

**Examination-style questions**

**One linear and one quadratic equation**

Suppose one of the equations in a pair of simultaneous equations is linear and the other is a quadratic of the form

*y*

=

*ax*

2

+

*bx*

+

*c*

.

By considering the points where the graphs of the two equations might intersect we can see that there could be two, one or no pairs of solutions.

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**One linear and one quadratic equation**

If the second equation contains terms in

*xy*

or

*y*

2 the shape of the corresponding graph will not be a parabola but a circle, a hyperbola or an ellipse:

**A line and a circle**

**A line and a hyperbola**

**A line and an ellipse**

Again we can have two, one or no pairs of solutions.

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**One linear and one quadratic equation**

When a pair of simultaneous equations contains one linear and one quadratic equation, we usually solve them by substitution. For example:

Solve:

*y*

=

*x*

2

+ 1

*y*

=

*x*

+ 3

**1**

**2**

Substituting equation

**1**

into equation

**2**

gives

*x*

2

+ 1 =

*x*

+ 3

Rearranging to give a quadratic equation of the form

*ax*

2

+

*bx*

+

*c*

= 0 gives

*x*

2

–

*x*

– 2 = 0

(

*x*

+ 1)(

*x*

– 2) = 0

*x*

=

–1 or

*x*

= 2

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**One linear and one quadratic equation**

We can substitute these values of

*x*

into one of the equations

*y*

=

*x*

2

+ 1

*y*

=

*x*

+ 3 to find the corresponding values of

*y*

.

**1**

**2**

It is easiest to substitute into equation

**2**

because it is linear.

When

*x*

=

–1:

*y y*

=

–1 + 3

= 2

When

*x*

= 2:

*y*

= 2 + 3

*y*

= 5

The solutions are

*x*

=

–1

,

*y*

= 2 and

*x*

= 2 ,

*y*

= 5 .

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**One linear and one quadratic equation**

*y*

*y*

=

*x*

2

+ 1

*y*

=

*x *

+ 3 We can demonstrate the solutions to

*y*

=

*x*

2

+ 1

*y*

=

*x*

+ 3 using a graph.

(

–1, 2)

(2, 5)

0

*x*

It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs, particularly when the solutions are not integers.

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**One linear and one quadratic equation**

Look at this pair of simultaneous equations:

*y*

–

*x*

= 1

*x*

2

+

*y*

2

= 13

**1**

**2**

What shape is the graph given by

*x*

2

+

*y*

2

= 13?

The graph of

*x*

2

+

*y*

2

= 13 is a circular graph with its centre at

We can solve this pair of simultaneous equations algebraically using substitution.

We can then sketch the graphs of the equations to demonstrate where they intersect.

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**One linear and one quadratic equation**

*y*

–

*x*

= 1

*x*

2

+

*y*

2

= 13

**1**

**2**

Rearranging

**1**

gives

*y*

=

*x*

+ 1

Substituting into

**2**

gives

*x*

2

+ (

*x*

+ 1)

2

= 13

*x*

2

+

*x*

2

+ 2

*x*

+ 1 = 13

2

*x*

2

+ 2

*x*

– 12 = 0

*x*

2

+

*x*

– 6 = 0

(

*x*

+ 3)(

*x*

– 2) = 0

*x*

=

–3 or

*x*

= 2

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**One linear and one quadratic equation**

We can substitute these values of

*x*

into one of the equations

*y*

=

*x*

+ 1

*x*

2

+

*y*

2

= 13 to find the corresponding values of

*y*

.

**1**

**2**

It is easiest to substitute into equation

**1**

because it is linear.

When

*x*

=

–3:

When

*x*

= 2:

*y*

=

–3 + 1

*y*

=

–2

*y*

= 2 + 1

*y*

= 3

The solutions are

*x*

=

–3

,

*y*

=

–2 and

*x*

= 2 ,

*y*

= 3 .

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**One linear and one quadratic equation**

Demonstrating these solutions graphically gives:

*x*

2

+

*y*

2

= 13

*y*

(2, 3)

*y*

=

*x *

+ 1

0

*x*

(

–3, –2)

The graphs intersect at the points (

–3, –2) and (2, 3).

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**Using the discriminant**

In summary, to solve a pair of simultaneous equations where one equation is linear and the other is quadratic:

Rearrange the linear equation so that one of the variables is written in terms of the other.

Substitute the linear equation into the quadratic equation to give a single equation of the form

*ax*

2

+

*bx*

+

*c*

= 0.

We can find the determinant of this equation to find how many times the line and the curve will intersect. When

*b*

2

– 4

*ac*

> 0, there are two distinct points of intersection.

*b*

2

– 4

*ac*

= 0, there is one point of intersection (or two coincident points). The line is a

**tangent**

to the curve.

*b*

2

– 4

*ac*

< 0, there are no points of intersection.

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**Using the discriminant**

Show that the line

*y*

– 4 tangent to the curve

*y x*

=

*x*

+ 7 = 0 is a

2

– 2

*x*

+ 2.

Call these equations

**1**

and

**2**

.

*y*

– 4

*x*

+ 7 = 0

*y*

=

*x*

2

– 2

*x*

+ 2

**1**

**2**

Rearranging

**1**

gives

Substituting into

**2**

gives

4

*x y*

= 4

*x*

– 7

– 7 =

*x x*

2

– 6

*x*

+ 9 = 0

2

– 2

*x*

+ 2

The discriminant =

*b*

2

– 4

*ac*

= (

–6)

2

– 4(9)

= 36

– 36

= 0

*b*

2

– 4

*ac *

= 0 and so the line is a tangent to the curve.

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**Linear inequalities**

**Linear simultaneous equations**

**Simultaneous equations involving one linear and one quadratic equation**

**Linear inequalities**

**Quadratic inequalities**

**Polynomials**

**Examination-style questions**

**Linear inequalities**

An

**inequality**

the symbols: links two or more expressions with

<, >,

≤ or ≥.

Inequalities are linear if the expressions they contain can be written in the form

*ax*

+

*b*

where

*a*

and

*b*

are constants.

For example:

3

*x*

+ 2 > 5

Solving this inequality involves finding the values of

*x*

that make the inequality true.

In this example, the inequality is true when

*x*

> 1.

The solution can be illustrated using a number line as follows:

–3

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–2 –1

0 1 2 3 4 5

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**Solving linear inequalities**

Like an equation, we can solve an inequality by adding or subtracting the same value to both sides of the inequality sign.

We can also multiply or divide both sides of the inequality by a

*positive*

value. For example:

Add 7 to both sides:

Add 2

*x*

to both sides:

Divide both sides by 6:

Solve 4

*x*

– 7 > 11 – 2

*x.*

4

*x*

> 18

– 2

*x*

6

*x*

> 18

*x*

> 3

How could we check this solution?

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**Checking solutions**

To verify that is the solution to

*x*

> 3

4

*x*

– 7 > 11 – 2

*x*

substitute a value just above 3 into the inequality and then substitute a value just below 3.

Substituting

*x*

= 4 into the inequality gives

4

× 4 – 7 > 11 – 2 × 4

16

– 7 > 11 – 8

9 > 3 This is true.

Substituting

*x*

= 2 into the inequality gives

4

× 2 – 7 > 11 – 2 × 2

8

– 7 > 11 – 4

1 > 7 This is not true.

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**Multiplying or dividing by negatives**

Although most inequalities can be solved like equations we have to take great care when multiplying or dividing both sides of an inequality by a negative value.

The following simple inequality is true:

–3 < 5

Look what happens if we multiply both sides by

–1:

–3 × –1

< 5

× –1

3 <

–5

This is not true.

To keep the inequality true we have to reverse the inequality sign:

3 >

–5

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**Multiplying or dividing by negatives**

**Remember:**

when both sides of an inequality are multiplied or divided by a negative number the inequality is reversed.

For example:

4

– 3

*x*

≤ 10

–3

*x*

≤ 6

*x*

≥ –2

The inequality sign is reversed.

We could also solve this type of inequality by collecting

*x*

terms on the right and reversing the inequality sign at the end.

4

– 3

*x*

≤ 10

4

≤ 10 + 3

*x*

–6 ≤ 3

*x*

–2 ≤

*x x*

≥ –2

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**Solving combined linear inequalities**

The two inequalities 4

*x*

+ 3

≥ 5 and 4

*x*

as a single combined inequality:

+ 3 < 15 can be written

5

≤ 4

*x*

+ 3 < 15

We can solve this inequality as follows:

Subtract 3 from each part:

2

≤ 4

*x*

< 12

Divide each part by 4:

0.5

≤

*x*

< 3

We can illustrate this solution on a number line as follows:

–1 –0.5

0 0.5

1 1.5

2 2.5

3 3.5

4

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**Solving combined linear inequalities**

Some combined inequalities contain variables in more than one part. For example:

*x*

– 2 ≤ 3

*x*

+ 2

≤ 2

*x*

+ 7

Treat this as two separate inequalities:

*x*

– 2 ≤ 3

*x*

+ 2 and 3

*x*

– 2 ≤ 2

*x*

– 4 ≤ 2

*x*

+ 2

*x*

+ 2

*x*

≤ 2

+ 2

≤ 7

≤ 5

*x*

+ 7

– 2 ≤

*x*

We can write the complete solution as

–2 ≤

*x *

≤ 5 and illustrate it on a number line as follows:

–3 –2 –1

0 1 2 3 4 5 6 7

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**Overlapping solutions**

Solve the following inequality and illustrate the solution on a number line:

2

*x*

– 1 ≤

*x*

+ 2 < 7

Treating as two separate inequalities gives

2

*x*

– 1 ≤

*x*

+ 2 and

*x*

+ 2 < 7

*x*

– 1 ≤ 2

*x*

≤ 3

*x*

< 5

If

*x*

< 5 then it is also

≤ 3. The whole solution set is therefore given by

*x*

< 5. This can be seen on the number line:

–3 –2 –1

0 1 2 3 4 5 6 7

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**Solutions in two parts**

Solve the following inequality and illustrate the solution on a number line:

4

*x*

+ 5 < 3

*x*

+ 5

≤ 4

*x*

+ 3

Treating as two separate inequalities gives

4

*x*

+ 5 < 3

4

*x x*

< 3

< 0

*x x*

+ 5 and 3

*x*

+ 5

≤ 4

*x*

+ 3

5

≤

2

≤

*x x x *

≥ 2

+ 3

We cannot write these solutions as a single combined inequality. The solution has two parts.

–3 –2 –1

0 1 2 3 4 5 6 7

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**Quadratic inequalities**

**Linear simultaneous equations**

**Simultaneous equations involving one linear and one quadratic equation**

**Linear inequalities**

**Quadratic inequalities**

**Polynomials**

**Examination-style questions**

**Quadratic inequalities**

**Quadratic inequalities**

contain terms in both

*x*

2 example:

*x*

2

+

*x*

– 6 ≥ 0 and

*x*

. For

Factorizing gives

(

*x*

+ 3)(

*x*

– 2) ≥ 0

*x*

2

+

*x*

– 6 is equal to 0 when:

*x*

+ 3 = 0

*x*

=

–3 and

*x*

– 2 = 0

*x*

= 2

These values give the end points of the solution set:

–5 –4 –3 –2 –1

0 1 2 3 4 5

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**Quadratic inequalities**

To find the solution set we can substitute a value from each of the following three regions:

**region 2 region 3**

–5 –4 –3 –2 –1

0 1 into the original inequality

*x*

2

+

*x*

– 6 ≥ 0.

2 3 4

When

*x*

=

–4: –4

2

+

–4 – 6 ≥ 0

16

– 4 – 6 ≥ 0

6

≥ 0

This is true and so values in region 1 satisfy the inequality.

5

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**Quadratic inequalities**

When

*x*

= 0 0

2

+ 0

– 6 ≥ 0

–6 ≥ 0

This is not true and so values in region 2 do not satisfy the inequality.

**region 1**

–5 –4

When

*x*

= 3

–3 –2 –1

0 1

3

2

+ 3

–

6

≥ 0

9 + 3

–

6

≥ 0

6

≥ 0

2 3 4

This is true and so values in region 3 satisfy the inequality.

5

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**Quadratic inequalities**

We have shown that values in

**region 1**

inequality

*x*

2

+

*x*

–

6

≥ 0.

and

**region 3**

satisfy the

**region 1 region 2 region 3**

–5 –4 –3 –2 –1

0 1 2 3

We can show the complete solution set as follows:

4 5

–5 –4 –3 –2 –1

0

So the solution to

*x*

2

+

*x*

– 6 ≥ 0 is:

*x*

≤ –3 or

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1

*x*

≥ 2

2 3 4

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5

**Quadratic inequalities**

An alternative method for solving inequalities involves using graphs. For example:

Solve

*x*

2

+

*x*

– 3 > 4

*x*

+ 1.

The first step is to rearrange the inequality so that all the terms are on one side and 0 is on the other.

*x*

2

– 3

*x*

– 4 > 0

Sketching the graph of

*y*

inequality.

=

*x*

2

– 3

*x*

– 4 will help us to solve this

The coefficient of

*x*

2

> 0 and so the graph will be

-shaped.

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**Quadratic inequalities**

Next, we find the roots by solving

*x*

2

– 3

*x*

– 4 = 0.

Factorizing gives (

*x*

+ 1)(

*x*

– 4) = 0

*x*

=

–1 or

*x*

= 4

We can now sketch the graph.

The inequality

*x*

2

– 3

*x*

– 4 > 0 is true for the parts of the curve that lie above the

*x*

-axis.

( (

–1, 0)

So, the solution to

*x*

2

+

*x*

– 3 > 4

*x*

+ 1 is

*x*

<

–1 or

*x*

> 4

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**Solving quadratic inequalities using graphs**

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**Polynomials**

**Linear simultaneous equations**

**Simultaneous equations involving one linear and one quadratic equation**

**Linear inequalities**

**Quadratic inequalities**

**Polynomials**

**Examination-style questions**

**Polynomials**

A

**polynomial in **

*x*

*ax n*

is an expression of the form

+

*bx n*

1

+

*cx n*

2

+...+

*px*

2

+ where

*a*

,

*b*

,

*c*

, … are constant coefficients and

*n*

is a positive integer.

The value of

*a*

is called the

**leading coefficient**

.

Examples of polynomials include:

3

*x*

7

+ 4

*x*

3

–

*x*

+ 8

*x*

11

– 2

*x*

8

+ 9

*x*

and 5 + 3

*x*

2

– 2

*x*

3

.

Polynomials are usually written in

**descending **

powers of

*x*

.

They can also be written in

**ascending **

powers of

*x*

, especially when the leading coefficient is negative, as in the last example.

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**Polynomials**

The

**degree**

, or

**order**

, of a polynomial is given by the highest power of the variable.

A polynomial of degree 1 is called

**linear**

and has the general form

*ax*

+

*b*

.

A polynomial of degree 2 is called

**quadratic**

general form

*ax*

2

+

*bx *

+

*c*

.

and has the

A polynomial of degree 3 is called

**cubic**

general form

*ax*

3

+

*bx*

2

+

*cx *

+

*d*

.

and has the

A polynomial of degree 4 is called

**quartic**

general form

*ax*

4

+

*bx*

3

+

*cx*

2

+

*dx *

+

*e*

.

and has the

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**Using function notation**

Polynomials are often expressed using function notation.

For example, consider the polynomial function:

*f*

(

*x*

) = 2

*x*

2

– 7

We can use this notation to substitute given values of

*x*

.

For example:

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Find

*f*

(

*x*

) when a)

*f*

(

–2) = 2(–2)

2

– 7

= 8

– 7

= 1 a)

*x*

=

–2 b)

*x*

=

*t*

+ 1 b)

*f*

(

*t*

+ 1) = 2(

*t*

+ 1)

2

– 7

= 2(

*t*

2

+ 2

*t*

+ 1)

– 7

= 2

*t*

2

+ 4

*t*

+ 2

– 7

= 2

*t*

2

+ 4

*t*

– 5

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**Adding and subtracting polynomials**

When two or more polynomials are added, subtracted or multiplied, the result is another polynomial.

Polynomials are added and subtracted by

**collecting like terms**

.

For example:

*f*

(

*x*

) = 2

*x*

3

– 5

*x*

+ 4 and

*g*

(

*x*

) = 2

*x*

– 4

Find a)

*f*

(

*x*

) +

*g*

(

*x*

) b)

*f*

(

*x*

)

–

*g*

(

*x*

) a)

*f*

(

*x*

) +

= 2

*x*

3

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*g*

(

*x*

– 3

)

= 2

*x*

3

– 5

*x x*

+ 4 + 2

*x*

– 4 b)

*f*

(

*x*

)

–

*g*

(

*x*

)

= 2

*x*

3

– 5

*x*

+ 4

– (2

*x*

– 4)

= 2

*x*

3

– 5

*x*

+ 4

– 2

*x*

+ 4

= 2

*x*

3

– 7

*x*

+ 8

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**Multiplying polynomials**

When two polynomials are multiplied together every term in the first polynomial must by multiplied by every term in the second polynomial. For example:

*f*

(

*x*

) = 3

*x*

3

– 2 and

*g*

(

*x*

) =

*x*

3

+ 5

*x*

– 1

*f*

(

*x*

)

*g*

(

*x*

) = (3

*x*

3

– 2)(

*x*

3

+ 5

*x*

– 1)

= 3

*x*

6

+ 15

*x*

4

– 3

*x*

3

– 2

*x*

3

– 10

*x*

+ 2

= 3

*x*

6

+ 15

*x*

4

– 5

*x*

3

– 10

*x *

+ 2

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**Multiplying polynomials**

Sometimes we only need to find the coefficient of a single term.

For example:

Find the coefficient of

*x*

2 is multiplied by 2

*x*

3 when

*x*

3

+ 5

*x*

2

–

*x*

– 4

*x*

2

– 6.

+ 2

*x*

We don’t need to multiply this out in full. We only need to decide which terms will multiply together to give terms in

*x*

2

.

(

*x*

3

– 4

*x*

2

+ 2

*x*

)(2

*x*

3

+ 5

*x*

2

–

*x*

– 6)

We have: 24

*x*

2

– 2

*x*

2

= 22

*x*

2

So, the coefficient of

*x*

2 is 22 .

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**Examination-style questions**

**Linear simultaneous equations**

**Simultaneous equations involving one linear and one quadratic equation**

**Linear inequalities**

**Quadratic inequalities**

**Polynomials**

**Examination-style questions**

**Examination-style question 1**

a) Solve the simultaneous equations

*x*

– 2

*y*

= 2

*x*

2

+ 4

*y*

2

= 100 b) Interpret your solution geometrically.

a) Label the equations

*x*

– 2

*y*

= 2

*x*

2

+ 4

*y*

2

= 100

**1**

**2**

Rearranging equation

**1**

Substituting into equation

**2**

*x*

= 2 + 2

*y*

(2 + 2

*y*

)

2

+ 4

*y*

2

= 100

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**Examination-style question 1**

4 + 8

*y*

+ 4

*y*

2

+ 4

*y*

2

= 100

8

*y*

2

+ 8

*y*

– 96 = 0

*y*

2

+

*y*

– 12 = 0

(

*y *

+ 4)(

*y*

– 3) = 0

*y*

=

–4 or

*y*

= 3

Substituting into equation

**1**

When

*y*

=

–4,

*x*

=

–6

When

*y*

= 3,

*x*

= 8 b) The line

*x*

– 2

*y*

= 2 crosses the curve

*x*

2 points (

–4, –6) and (3, 8).

+ 4

*y*

2

= 100 at the

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**Examination-style question 2**

a) Write an expression for the area

*A*

of the following rectangle:

*x*

– 2

*x*

+ 2 b) If the area satisfies the inequality

5 <

*A*

< 12 find the range of possible values for

*x*

.

a)

*A*

= (

*x*

=

*x*

2

+ 2)(

*x*

– 4

– 2) b) The range of possible values for

*x*

is given by

5 <

*x*

2

– 4 < 12

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**Examination-style question 2**

We have to solve 5 <

*x*

2

– 4 and then solve

*x*

2

– 4 < 12

5 <

*x*

2

– 4

*x*

2

– 9 > 0

*x x*

2

2

– 4 < 12

– 16 < 0

(

*x*

+ 3)(

*x*

– 3) > 0

Sketching

*y*

=

*x*

2

– 9

(

*x*

+ 4)(

*x*

– 4) < 0

Sketching

*y*

=

*x*

2

– 16

*y y*

(

–3, 0)

0

(3, 0)

*x*

(

–4, 0)

0

(4, 0)

*x*

*x*

2

– 9 > 0 when

*x*

<

–3 or

*x*

> 3

*x*

2

– 16 < 0 when –4 <

*x*

So the range of possible value for

*x*

is

< 4

3 <

*x*

< 4 (ignoring negative solutions)

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