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# Factorisation

This might sound like a silly/basic question to many, but I'm anything but bright in maths.

As much as I try, and despite the fact that I've tried reading from various books, I cannot for the love of God understand perfect square factorisation.

Apparently, there's an easy way of solving it with the GDC. Can someone please explain both methods, with and without GDC?

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you mean that, for instance, you do not understand the calculation of (x+5)2 or you mean the completion of square process?

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Nope, it's completion of square. The rule stated in my book is (a+b)^2=a^2+2ab+b^2

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the rule that you state above is not the completion of square. Anyway, can you explain a bit more the problem you are facing?

When completing the square by the way, it means that any quadratic expression y = ax^2 + bx + c can be expressed into the form y = a(x-k)^2 + I

but from what I understand this is not your problem.

If you have (x+5)^2 this can be written as x^2+2(5)x+5^2 = x^2 + 10x + 25. Is it the reverse that you don't get?

*sorry if I'm not very helpful

Edited by Jirashimosu
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The perfect square is in the form of:

(ax(square)+2abx+b(square))

By normal factorisation you will get

(ax + b) square

example:

x (square) +2x + 1

first of all look to the third term is it a square (or you can find the two dividends of it that can when added giv you the middle number)

you know that 1= 1*1

and you know that 1+1=2

so you can make two brakets

(x+1) (x+1)

This can be written in a form of (x+1)square

Look for the third term and its dividends

Note if the third term is a square, it should not be a perfect square:

Example :

x(square)+ 10 x +16

16= 4*4

but in this case 4+4 =/10

but 8+2 = 10

Other perfect square root is when (ax-b) square

then the exansion is ax(square) -2axb+b(square)

I Do not know how to solve it by GDP, but i will search for

Is that what you want

If you have an example in specific it will be better

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obviously I have failed to understand that such extensive answer was needed.

anyway, if you are using the casio calc (fx - 9860GII). you go to equation mode (from the main menu, you press A or else it is the square saying "EQUA axn+ ...= 0), then you press F2 (it is for polynomial eqs) and if it is a quadratic you press F1 (the respective button for a second degree eq). You want to have the equation written in its expanded form, for example: (x+2)^2 you first need to expand the square and write it in the form:

x2 + 4x + 4 (it is the expansion of (x+2)2) and then at the first bracket (above which there is an "a") you write the coefficient of x2 which in this case is 1. Then by pressing the EXE button (at the bottom right of the calc), it moves you to a block above which there is the letter "b" and here you must write the coefficient of x, which in the above example is 4. Following the procedure explained above, you press again EXE and in the next block write the 4 (which is the final part of the equation). After this press again EXE and the calc will show you the double root the equation has (since the discriminant is 0) which is -2.

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inm, nope, that's not my question.

Jirashimosu, my apologies! The formula I gave is the wrong one; I got confused. The one you gave is the one I don't understand. For further elaboration, here's the worked solution they gave in the book:

x^2 + 10x + 25

= x^2 + (2)(x)(5) + 5^2

= (x+5)^2

wut?

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Work it to understand it;

(a+b)^2= (a+b) (a+b)

then using simple foill methode

To use foil you mulitiply

First symbol of each bracke= a.a= a^2

Outer symbol= a. b= ab

then inner=a.b=ab

then last= b.b= b^2

a^2+ab+ab+b^2

so it equals to:

a^2+2ab+b^2

which is the same as what your book says:

(a+b)^2=a^2+2ab+b^2

In the example:

x^2 + 10x + 25

= x^2 + (2)(x)(5) + 5^2

= (x+5)^2

You found that 25 is the squae of 5

and 5+ 5= 10

then you can write it as

= x^2 + (2)(x)(5) + 5^2 (here a=x , b= 5)

the you will change it from a^2+2ab+b^2 shape to

(a+b)^2 (you know that a=x and b=5)

so (x+5)^2

is this what are searching for..

Edited by inm

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No worries but the formula you gave was correct.

Assuming that you have the example: x2 + 10x + 25. Consider the formula (a + b)2 = a2 + 2ab + b2. You should be aware that it works in both ways which means that if you have a2 + 2ab + b2 it can be converted to (a+b)2.

In this example, x^2 + 10x + 25, you should find which represents the factor "a" and which one the factor "b". The second step your book gives you is so that it is clearer for you which is "a" and which is b" as you can see from the initial formula, these two factors are squared and their product multiplied by two gives you 10. Once you understand which one is "a" and which one is "b", you can write it in the form (a + b)2 but instead of "a" writing "x" and instead of "b" writing 5.

Better?

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I think I have a slight idea. Is it like...

x^2 + 6x + 9

= x^2 + 2x3 + 3^2

...Um, now I'm lost. I don't really understand the next step. Did I even do that first step correctly?

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yes!! it is correct! so, in your example, which one is represented by "a" and which one by "b" (shown in the formula: (a + b)2 = a2 + 2ab + b2)?

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You reached a good step, now you should ask you self what is:

a and b

a = is the first term which is under the square

examples:

4x^2= 2^2x^2 then a=

= 2x

x^4= (x^2)^2 then a = x^2

b= the square root of the third term

if it is 9, then b=3

if it is 16, then b= 4

if it is y^2, then b=y

(a+b)^2

a=x, b= 3

(x+3)^2

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Um, so...

x^2 + 3x2 + 3^2 = (2 + 3)^2? That's it? I'm sorry but I don't really get it Can you explain more? Like...in detail; I don't really understand from just equations.

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First of all)) how can you factorise any quadratic formula:

example: x^2+ 7X+ 6

You should look at third term and its dividends, in the example it is 6:

6= 2.3=6.1

you will choose the one in which when added give you the coffiecient of the second term (7)

2+3=5

6+1= 7

you will choose the second one

so open two brakets

( ) ( )

since you have x^2

(x + ) (x+ )

since 6 and 1, satisfy you formula

it will be

(x+1) (x+6)

check it

Using foil

((x+1) (x+6)= x.x + 6.x + 1.x+ 1.6

first outer inner last

(x+1) (x+6)= x^2+ 6x +1x +6

since you have 6x and 1 x with the same degree

x^2+ 7x+6

in perfect square you will have the same thin but the two brakets are equal to each other

to have (a+b) (a+b) = (a+b)^2

Do you understand that

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Yes! Thank you very much just to make sure, this also applies for when x^2 has a coefficient, or no?

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Yes, diiffently, it seems that you understand , to make sure solve this:

1) x^2 + 20 x + 64

and

2) x^2 + 16 x + 64

even if you do not know, i will find in which step you have a problem

Edited by inm

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x^2 + 10x + 25.

which is like

a^2 + bx + c

the way i'd put it is take the square root of "c"

which means, the square root of 25 = 5

and now take the square root of "a"

which in this case, "a" is 1x2

so the root of 1x2 is 1x which is the same as x

therefore you have x and 5

since the 5 is positive you put a + in front of it = + 5

now combine (x + 5)

and then all you have to do is square it.

therefore you get (x + 5)2

when you expand you will see (x + 5)(x + 5)

which through normal expansion comes to x2 + 10x + 25

thus i have proved my steps

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I think these two guidelines will help a lot (you may consider them tricks but they are not tricks)

when you have a quadratic polynomial x^2 + bx + c, you can factorise it into (x + k) (x + n) by:

1) looking at the bx. The bx is always the sum of the k and n

2) c is always the multiplication of k and n.

Thus, what you can do, is to look for all the possible values that can give k+n=b and all the values that can give kn=c. Then, choose the correct one (i.e, the one that matches in both operations). That's a quick way to solve factorisations and I always do like this

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