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2\textcolor{blue}{-3}x+\textcolor{red}{2}

Step 1: First we can write two brackets with an x placed in each bracket.

(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)

Step 2: We can identify that this is a sub-type (b) quadratic, meaning both brackets will contain \large{-}

(x\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,)

Step 3: We have to find two numbers which multiply to make \textcolor{red}{2} and when added together make \textcolor{blue}{-3}.

We know both numbers will be negative.

-2 \times -1 = \textcolor{red}{2}

-2 + -1 = \textcolor{blue}{-3}

Finally add these numbers to the brackets.

(x-2) (x-1)

Level 4-5GCSEAQAEdexcelOCRWJECEdexcel iGCSE

Level 6-7GCSEAQAEdexcelOCRWJECEdexcel iGCSE

Type 2: Factorising quadratics (a> 1)

In this instance the general form of the equation is ax^2+bx+c where a>1. 2+\textcolor{blue}{3x}\textcolor{red}{-1}

Step 1: When a>1 it makes things more complicated. It is not immediately obvious what the coefficient of each x term should be. There are two possible options,

(4x \kern{1 cm} ) (x  \kern{1 cm} ) or (2x \kern{1 cm} ) (2x  \kern{1 cm} )

Step 2: We can identify that this quadratic is part of sub-type (c) meaning it can contain + and —

This is most important for quadratic pairs which are non-symmetrically creating a third option, all three are shown below.

\begin{aligned}(4x \kern{0.4 cm} +\kern{0.4 cm} )&(x  \kern{0.4 cm}-\kern{0.4 cm} )\\ (4x \kern{0.4 cm} -\kern{0.4 cm} )&(x  \kern{0.4 cm}+\kern{0.4 cm} )\\(2x \kern{0.4 cm}+\kern{0.4 cm} )&(2x \kern{0.4 cm}-\kern{0.4 cm} )\end{aligned}

Step 3: We need to find two numbers which when multiplied make \textcolor{red}{-1}

\textcolor{red}{-1} has only one factor. 2 — x — 12.

[2 marks]

Step 1: Draw empty brackets

(x \kern{1 cm} ) (x  \kern{1 cm} )

Step 2: Identify sub-type (b)

(x \kern{0.4cm} + \kern{0.4cm} ) (x  \kern{0.4 cm} — \kern{0.4cm} )

Step 3: We are looking for two numbers which multiply to make \textcolor{red}{-12} and add to make \textcolor{blue}{-1}. Let’s consider some factor pairs of -12.

\begin{aligned}(-1)\times12&=-12  \,\,\text{    and  } -1 + 12 = 11\\(-2)\times6&=-12 \,\,\text{    and  } -2 + 6 = 4\\ (-6)\times2&=-12 \,\,\text{  and  } -6 + 2 = -4\\ (-3)\times4&=-12 \,\,\text{    and  } -3 +4 = 1\\ \textcolor{red}{(-4)\times3}&\textcolor{red}{=-12 }\,\,\text{    and  } \textcolor{blue}{-4 + 3 = -1}\end{aligned}

We could keep going, but there’s no need because the last pair, -4 and 3, add to make -1. This pair fills both criteria, (as highlighted above) so the factorisation of x^2 — x — 12 is

(x — 4)(x + 3)

Note: You can try expanding the double brackets to check your answer is correct. 2 -5m -6

We can see that last option with +3 and -2 is the correct combination.

This gives the final answer to be:

(4m+3)(m-2)

MME

Level 1-3GCSEKS3

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MME

Level 4-5GCSEKS3

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MME

Level 1-3GCSEKS3

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MME

Level 4-5GCSEKS3

Revise

## Factorising Quadratics — GCSE Maths

Introduction

Common misconceptions

Learning checklist

Next lessons

Still stuck?

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Introduction

Common misconceptions

Learning checklist

Next lessons

Still stuck?

Here we will learn about factorising quadratics; we will explore what quadratic expressions are and the steps needed to factorise into double brackets. {2}\color{#00BC89}{+3}x\color{#7C4DFF}{-2}\]

We factorise quadratic expressions of this sort using double brackets. There are different methods we can use depending on whether the coefficient of x2 is greater than 1.

Factorising, or factoring quadratic equations is the opposite of expanding brackets and is used to solve quadratic equations.

For example, in the form of x2 + bx + c requires two brackets (x + d) (x + e).

1. Write out the factor pairs of the last number (c).
2. Find a pair of factors that + to give the middle number (b) and multiply to give the last number (c).
3. Write two brackets and put the variable at the start of each one.
4. Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, but the signs of the factors are.

If you’re looking for a summary of all the different ways you can factorise expressions then you may find it helpful to start with our main factorising lesson or look in detail at the other lessons in this section. • Factorising
• Factorising single brackets
• Difference of two squares

A quadratic equation is a quadratic expression that is equal to something. We can solve quadratic equations by using factorisation (or factoring), the quadratic formula or by completing the square.

Step by step guide: Quadratic equations

x

### Factorising quadratics in the form x

2 + bx + c

To factorise a quadratic expression in the form x2 + bx + c we need double brackets. Factorisation into double brackets is the reverse process of expanding double brackets.

In this case, the coefficient (number in front) of the x2 term is 1 (a=1). These are known as monic quadratic.

### How to factorise quadratics: x

2 + bx + c (double brackets)

In order to factorise a quadratic algebraic expression in the form x2 + bx + c into double brackets:

1. Write out the factor pairs of the last number (c).
2. Find a pair of factors that + to give the middle number (b) and ✕ to give the last number (c).
3. Write two brackets and put the variable at the start of each one.
4. Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, the signs of the factors are.

#### Explain how to factorise quadratics: x² + bx + c (double brackets)

2 + bx + c (double brackets)

#### Example 1: with +x coefficient and a +constant

Fully factorise:

$x^2 \color{#00BC89}{+ 6x}\color{#7C4DFF} {+ 5}$

1. Write out the factor pairs of the last number (5) in order. 2 – 2x – 24\]

Write out the factor pairs of the last number (24) in order

x2 – 2x – 24

Factors of 24:

1, 24
2, 12
3, 8
4, 6

We need a pair of factors that + to give the middle number (-2) and ✕ to give the last number (-24).

x2 – 2x – 24

Factors of 24:
1, 24
2, 12
3, 8
4, 6

-6 + 4 = -2 ✔

-6 ✕ 4 = -24 ✔

(It’s a good idea to do a quick check that we have the correct numbers)

Remember: to multiply two values together to give a negative answer, the signs must be the different.

Write two brackets and put the variable at the start of each one (x in this case).

(x       )(x       )

Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, the signs of the factors are.

(x – 6)(x + 4)

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

(x – 6)(x + 4)  = x2 – 2x – 24

#### Example 3: with +x coefficient and a –constant

Fully factorise:

x2 + x – 20

Write out the factor pairs of last number (20) in order. 2 – 8x + 15\]

Write out the factor pairs of the last number (15) in order.

Factors of 15:

1, 15

3, 5

We need a pair of factors that + to give the middle number (-8) and ✕ to give the last number (15).

x2 – 8x + 15

Factors of 15:
1, 15
3, 5

-3 + -5 = -8 ✔
-3 ✕ -5 = 15 ✔

It’s a good idea to do a quick check that we have the correct numbers.

Remember: to ✕ two values together to give a positive answer, the signs must be the same

Write two brackets and put the variable at the start of each one (x in this case)

(x       )(x       )

Write one factor in the first bracket and the other factor in the second bracket.

$(x – 3)(x – 5)$

We have now fully factorised the quadratic expression.

We can check the answer by multiplying out the brackets!

(x – 3)(x – 5) = x2 – 8x + 15

### Practice factorising quadratics questions: x

2 + bx + c (double brackets)

(x+5)(x+6)

(x+3)(x+2)

(x+1)(x+6)

(x+3)(x+3)

We need numbers that have a product of 6 and a sum of 5 . {2}-10x+24=(x-4)(x-6) .

### Factorising quadratics GCSE questions: x

2 + bx + c (double brackets)

1. Factorise: x2 + 3x – 10

(x – 2)(x + 5)

(2 marks)

2. Factorise: y2 – 10y + 16

(y – 2)(y – 8)

(2 marks)

3. Factorise: x2 – 12x + 27

(x – 3)(x – 9)

(2 marks)

### Factorising quadratics in the form ax

2 + bx + c

To factorise a quadratic expression in the form ax2 + bx + c we need double brackets. Factorising into double brackets is the reverse process of expanding double brackets.

In this case the coefficient (number in front) of the x2 term is greater than 1 (a > 1). These are known as non-monic quadratics.

### How to factorise quadratics: ax

2 + bx + c (double brackets)

In order to factorise a quadratic algebraic expression in the form ax2 + bx + c into double brackets:

1. Multiply the end numbers together (a and c) then write out the factor pairs of this new number in order. 2 + 5x + 3\]

1. Multiply the end numbers together (2 and 3) then write out the factor pairs of this new number in order.

2x2 + 5x + 3

2 × 3 = 6

Factors of 6:
1, 6
2, 3

2We need a pair of factors that + to give the middle number (5) and ✕ to give this new number (6).

2x2 + 5x + 3

2 × 3 = 6

Factors of 6:
1, 6
2, 3

+ 5
✕ 6

2 + 3 = 5 ✔
2 x 3 = 6 ✔

Remember: to x two values together to give a positive answer, the signs must be the same.

3Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.

2x2 + 5x + 3
2x2 + 2x + 3x + 3

4Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!

2x2 + 5x + 3
2x2 + 2x + 3x + 3
2x(x + 1) + 3(x + 1)

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

5Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. 2 + 3x – 2\]

Multiply the the end numbers together (2 and -2) then write out the factor pairs of this new number in order.

2x2 + 3x – 2

2 ✕ -2 = -4

Factors of 4:
1, 4
2, 2

We need a pair of factors that + to give the middle number (3) and ✕ to give this new number (-4)

2x2 + 3x – 2

2 ✕ -2 = -4

Factors of 4:
1, 4
2, 2

⊕ 3
✕ -4

-1 + 4 = 3 ✔
-1 ✕ 4 = -4 ✔

Remember: to x two values together to give a negative answer, the signs must be different

Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.

2x2 + 3x - 2
2x2 - x + 4x - 2

Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!

2x2 + 3x - 2
2x2 - x + 4x - 2
x(2x + 1) + 2(2x - 1)

Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. 2 – 2x – 8\]

Multiply the the end numbers together (3 and -8) then write out the factor pairs of this new number in order.

3x2 – 2x – 8

3 ✕ -8 = -24

Factors of 24:
1, 24
2, 12
3, 8
4, 6

We need a pair of factors that + to give the middle number (-2) and to ✕ give this new number (-24)

3x2 – 2x – 8

3 ✕ -8 = -24

Factors of 24:
1, 24
2, 12
3, 8
4, 6

⊕ -2
✕ -24

-6 + 4 = -2 ✔
-6 ✕ 4 = -24 ✔

Remember: to ✕ two values together to give a negative answer, the signs must be different

Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.

3x2 - 2x - 8
3x2 - 6x + 4x - 8

Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!

3x2 - 2x - 8
3x2 - 6x + 4x - 8
3x(x - 2) + 4(x - 2)

Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. 2 – 7x + 2 \]

Multiply the the end numbers together (6 and 2) then write out the factor pairs of this new number in order.

6x2 – 7x + 2

6 ✕ 2 = 12

Factors of 12:
1, 12
2, 6
3, 4

We need a pair of factors that + to give the middle number (-7) and ✕ to give this new number (12)

6x2 – 7x + 2

6 ✕ 2 = 12

Factors of 12:
1, 12
2, 6
3, 4

+ -7
✕ -24

-3 + -4 = -7 ✔
-3 ✕ -4 = 12 ✔

Remember: to ✕ two values together to give a positive answer, the signs must be the same

Go back to the original equation and rewrite it this time splitting the middle term into the two factors we found in step 2 – the order of these factors doesn’t matter, the signs do.

6x2 - 7x + 2
6x2 - 3x - 4x + 2

Split the equation down the middle into two halves and fully factorise each half – the expressions in the brackets must be the same!

6x2 - 7x + 2
6x2 - 3x - 4x + 2
3x(2x - 1) - 2(2x - 1)

Now factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket. {2}-8x-x+4] or 2[2x(x-4)-(x-4)] so that the fully factorised expression is 2(2x-1)(x-4) .

### Factorising quadratics GCSE questions: ax

2 + bx + c (double brackets)

1. Factorise: 2x2 + 9x + 4

(2x + 1)(x + 4)

(2 marks)

2. Factorise: 2y2 – y – 3

(2y – 3)(y + 1)

(2 marks)

3. Factorise: 2x2 – x – 10

(2x – 5)(x + 2)

(2 marks)

### Common misconceptions

• The order of the brackets

When we multiply two values the order doesn’t matter. This is true for the brackets when factorising quadratics

e.g. 2 ✕3 = 3 ✕2

It is exactly the same here.

(x – 6)(x + 4) means (x – 6)(x + 4)

So,

(x – 6)(x + 4)=(x + 4)(x – 6)

+ ✕ + = +

e.g 2 ✕ 3 = 6

4 ✕ 5=20

– ✕ – = +

e.g -2 ✕ -3 = 6

-4 ✕ -5= 20

+ ✕ – = –

e.g 2 ✕ -3= -6

4 ✕ -5= -20

– ✕ + = –

e. {2} + bx + c (H)

2. ### Still stuck?

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## Possibilities of the computer program «GeoGebra» in preparation for the OGE in mathematics.

900 02

Opportunities
computer program «
GeoGebra »
in preparation for the OGE in mathematics.

Work completed:

Teacher
Mathematics MBOU «Secondary School No. 114″

Mullakhmetov Ilnaz Rashitovich

Kazan 2019

Contents

…………………………………………………………..5

1.1 Computer Algebra System (CAS —
Computer Algebra System) ………6

1.3. Symbolic calculations.
functions containing the module……..11

2. Geometric tasks in the GeoGebra program………………………………..14

2.1. Geometric constructions………………………………………………..14

2.2 Geometric

Conclusion………………………………………………………………………. ….24

References……………………………………………………………………25

900 22

Introduction

Computer technology is taking over
more and more trust and sympathy of schoolchildren and teachers of mathematics. On lessons
technology in the 7th grade, we got acquainted with the mathematical program «GeoGebra».
Initial ideas about the capabilities of this program, its availability for
any person has given me the opportunity to develop the application of this program to
studying the school course of algebra in the classroom. In algebra lessons, the teacher
showed the plotting of function graphs using the GeoGebra program. Me too
wanted to explore the possibilities of this program.On the Internet by keywords
«geogebra online» you can easily find a video that shows
some features of this application. This program was created in 2012 and
is developing very rapidly. The program was written by Markus Hohenwarter in Java, which means it runs on
a large number of operating systems. Translated to 39languages ​​and currently
is being actively developed. Translated into Russian in 2013.

Probably the program
there is no one specific purpose, it is a package of tools «for all occasions
life.» When working with this program as a result of computer simulation
many mathematical concepts and theorems become «visible» and «tangible» for students. The interface of the GeoGebra program (GeoGebra) resembles a blackboard on which
you can draw graphs, create geometric shapes, etc. In the program window
the changes being made will be visually displayed: if I change the equation,
the curve will be rebuilt, the scale or its position in space will change,
the equation written next to the curve will automatically be adjusted,
according to the new values.

The GeoGebra program is widely
used in the world by millions of users to teach algebra and geometry.
The learning process is visual due to the visual form of using the application.

The purpose of the work is to develop a methodology for using computer algebra systems
when solving algebraic and geometric problems in mathematics lessons.

To achieve the set goal, the following tasks were formulated

:

1. Analyze educational, methodological and scientific literature

for this topic.

2. Learn the basics of working in the GeoGebra system. 3. Analyze the methodological features of the application

of this system in the educational process.

4. Highlight guidelines for the use of GeoGebra

in the study of mathematics.

Methods used in the work — analysis of educational and methodological literature

comparison, generalization of pedagogical experience in
use of

computer algebra systems at school.

Object of study : the process of teaching mathematics to schoolchildren with the help of the

GeoGebra system.

Subject of research: the use of GeoGebra in the learning process

solving algebraic and geometric problems.

The relevance of the work is associated with the demand for methods

of using the GeoGebra computer algebra system in mathematics lessons

9000 3

1. Theoretical section

Mathematics programs are not limited to plotting and
computing. But when preparing for the OGE to solve problems of increased complexity (
task 23 OGE) you need to know

— Computer Algebra System (CAS)

symbolic calculations,

solution of equations,

factorization,

construction
function graphs, use of sliders,

plotting the function containing the module.

Starting the program

After starting GeoGebra
a window appears as shown below (Figure 1). With the help of drawing
tools (models) that are selected on the toolbar, you can
build drawings in notepad using the mouse. At the same time, the corresponding
coordinates and equations are displayed in the algebra window. Text input field
used for direct input of coordinates, equations, commands, functions;
they are immediately displayed in Notepad after pressing the Enter key. 9

1. 1 Computer algebra system (CAS — Computer
Algebra System)

The Computer Algebra System (CAS) in GeoGebra allows
perform symbolic calculations. The CAS window can be opened by left-clicking
clicking the arrow on the right side of the GeoGebra main window and selecting the CAS term. (Drawing
2). This window contains cells with an input line at the top and an output field just below
input field. The input line can be used just like the normal input line at the bottom
GeoGebra main screen with the following differences: 9)
in expressions,

— the equal sign (=) is used in equations, and the sign
:= for assigning values. This means that if you enter in the input line a=2 ,
then the variable a will be equal to 2 .

— Multiplication must be marked with *. If in a normal input line
at the bottom of the GeoGebra main screen, the correct entries are a(b+c) and a*(b+c) ,
then only entry 9 is correct in the CAS input line0061 a*(b+c) . CAS window toolbar

compute, decimal, freeze input, factorization,

expand brackets, replace, solve, delete.

Below the toolbar of the CAS window is the style bar (figure
3).

This panel contains the following buttons:

format button
text, allows you to change the style and color of the text: button
includes bold font, works after pressing the Enter key; button
includes italic font, works after pressing the Enter key; button displays the virtual keyboard. 92+6x-12=0
and press the x= button on the top
panels

3.
We got the result.

1.3 Character
calculations.

Plotting a quadratic function, using sliders.

GeoGebra is plotting functions, including those with parameters.
Consider examples and make explanations.

The GeoGebra program window looks like
(Fig. 4):

Fig.4

Coordinate axes and grid can be
show (or hide) if you right-click on the canvas and in
In the context menu that opens (Fig. 2), select Axes or Grid.

Fig.2

parameter, first define the parameter using the Slider tool.
Perform a left mouse click at that location in the construction area

92-12x+3a+9 and press Enter.

Now when moving the slider
the location of the function graph changes, i.e. the drawing became dynamic and the model
interactive.

You can also use the GeoGebra program
use when solving tasks of the OGE and the Unified State Examination. 1.5

Plotting a quadratic function containing a modulo.

Consider the solution of task 23 from
open form OGE 20192-3|x|-x and press Enter.

We get the graph of this function (Fig. 2).

Fig.2

3. Create a slider for our second
functions y=c
.

Next, we set the parameters for our
slider: the name will be denoted by the letter «c», since our function is y=c,
set the interval [-10;10], press ok

4. In the input field, type our second function
y=c
for which we made a slider in the previous step, press Enter.

92-3|x|-x
and the line y=c have 3 points of intersection at c=0 and c=-1 (Fig. 5, Fig. 6)

Fig. 5.
6

2.
GeoGebra .

2.1.
Geometric constructions

Point:
1.
Select the «Place a dot» tool.
2. Left click where we want
put a point.

Cut:
1.
In the tool «Straight» click on the white triangle.
2. Select «Segment» from the list.
3. Put 2 points — vertices of the segment.
Beam:

triangle. 2. Select «Beam» from the list.
3. Select two points on the canvas 2 points: the first
— the beginning of the beam, the second — the point through which the beam will be drawn.
Line:
1. Select the Line tool.

2. Specify 2 points through which the line will pass.
Perpendicular:
1. Select the Perpendicular tool.
2. Choose a line, ray or segment to which
we want to draw a perpendicular.
3. Choose a point through which it will pass
(point can lie on this line/ray/segment)
Line parallel to this line:
1. In the Perpendicular tool, click on
white triangle.
2. From the pop-up list, select «Parallel
straight».
3. Choose a line, ray or segment to which
a parallel line will be drawn.
4. Let’s choose a point through which it will pass.
Perpendicular bisector of the line:
1. In the Perpendicular tool, click on
white triangle.
2. From the pop-up list, select «Middle
perpendicular». 3. Let’s choose a segment or 2 points denoting
the segment through which the perpendicular bisector will be drawn.

Tangent line to circle:
1.
In the Perpendicular tool, click on the white triangle.
2. In the pop-up list, select «Tangent».
3. Select the circle to which the
tangent.
4. Select the point through which the
tangent. Two tangents are drawn. If only 1 tangent is needed,
you can hide one of them by right-clicking on the tangent and removing
checkmark in front of «Show object»

Polygon:
1.
Select the Polygon tool.
2. Let’s choose some points denoting
vertices, ending with the first point. For example, a triangle and immediately construct the described
circle around this triangle:

— we draw the perpendicular bisectors to two
sides,

— find the point of intersection of the middle
perpendiculars,

— draw a circle along the center and point

Inscribed circle in a triangle:

— draw the bisectors of the spirit of the angles of the triangle

— find the intersection points of these bisectors

— draw a perpendicular line to one of
sides

— find the point of intersection of the side of the triangle with
this line

— draw a circle in the center and the found point

Regular polygon:
1. In the Polygon tool, click on the white triangle.
2. From the pop-up list, select «Correct
polygon»
3. Select or put 2 points.
4. From the pop-up window, select how many vertices
will have a regular polygon.

Diagonal intersection points
polygon:

1.
To draw diagonals, we will use the Segment tool.
2. After carrying out two (or more) necessary
diagonals in the Point tool, click on the white triangle.
3. Select «Intersection» from the pop-up list.
4. Select 2 intersecting diagonals.
Points by coordinates:
1. Click on the entry line.
2. Write the name of the point and its
coordinates (for example A=(1,1) )

Geometric
GeoGebra

In a right triangle ABC with a right
the angle C knows the legs: AC=6, BC=8. Find the median CK of this triangle.

1. Take points A(0.6), B(8.0), C(0.0)

2. Use the button to connect the polygon
our three points A, B, C. We get triangle ABC.

3. Using the middle or center command
find the midpoint of the segment AB

4. Next, use the button to connect the segment
vertex C and D and get the segment CD
which on the command line is denoted by the letter f — the length of which is 5 cm

Task. Divide the prism into three pyramids.

1. Open
windows 2d
and 3d
in window 2 d we will construct three points
A (-3; 2), B (3; 3), C (1; -1)

2. Take the polygon tool and
build triangle ABC

straight prism, take the height 4.

we need them to understand how to break our prism.

5. Let’s remove everything superfluous on two canvases,
leave only our prism.

6. We begin to highlight our three pyramids on
which we will break our prism. We will give each pyramid its own different colors.

7. Now we need to animate our
spreading of the prism into 3 pyramids, for this we will create a slider, denote it n
and set some parameters that are shown in Figure

we need to make our pyramids move, for this we will set the vectors
DF,FE,BA

. 9. Next to make our pyramids
move in the input line, type : Move(