Math, asked by zaira92, 1 year ago

⭐⭐ formula ⭐⭐

★polynomial ★

Answers

Answered by aman3495
6
polynomial identities(short multiplication formulas):
(x + y)^2 = x^2 + 2xy + y^2
(x - y)^2 = x^2 - 2xy + y^2

Example 1: If x = 10, y = 5a
(10 + 5a)2 = 10^2 + 2·10·5a + (5a)2 = 100 +100a + 25a^2

The opposite is also true:
25 + 20a + 4a2 = 52 + 2·2·5 + (2a)2 = (5 + 2a)2

Consequences of the above formulas:

(-x + y)^2 = (y - x)^2 = y^2 - 2xy + x^2
(-x - y)^2 = (-(x + y))^2 = (x + y)^2 = x^2 + 2xy + y^2
Formulas for 3rd degree:

(x + y)^3 = x^3 + 3x2y + 3xy2 + y^3
(x - y)^3 = x^3 - 3x2y + 3xy2 - y^3

Example: (1 + a2)^3 = 1^3 + 3.12.a2 + 3.1.(a2)2 + (a2)^3 = 1 + 3a^2 + 3a^4 + a^6

(x + y + z)2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz
(x - y - z)2 = x^2 + y^2 + z^2 - 2xy - 2xz + 2yz
Factor Rules

x^2 - y^2 = (x - y)(x + y)

x^2 + y^2 = (x + y)2 - 2xy
or
x^2 + y^2 = (x - y)2 + 2xy

Example: 9a2 - 25b2 = (3a)^2 - (5b)^2 = (3a - 5b)(3a + 5b)

x^3 - y^3 = (x - y)(x^2 + xy + y^2)
x^3 + y^3 = (x + y)(x2 - xy + y2)

If n is a natural number

x^n - y^n = (x - y)(x^n-1 + x^n-2y +...+ y^n-2x + y^n-1)
If n is even (n = 2k)

x^n + y^n = (x + y)(x^n-1 - x^n-2y +...+ y^n-2x - y^n-1)

If n is odd (n = 2k + 1)

x^n + y^n = (x + y)(x^n-1 - x^n-2y +...- y^n-2x + y^n-1)
More Algebraic Formulas

2(a^2 + b^2) = (a + b)^2 + (a - b)^2
(a + b)^2 - (a - b)^2 = 4ab
(a - b)^2 = (a + b)^2 - 4ab
a^4 + b^4 = (a + b)(a - b)[(a + b)^2 - 2ab]

i hope it help you
Answered by Anonymous
8

Formulas involving squares


1) (a+b)²


=(a+b)(a+b)


=a²+ab+ab+b²


=a²+2ab+b²


Hence:


(a+b)²=a²+b²+2ab


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2) (a-b)²


=(a-b)(a-b)


=a²-ab-ab-b²


=a²-2ab+b²


Hence:


(a-b)²=a²+b²-2ab


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3) (a+b)²+(a-b)²


=a²+2ab+b²+a²-2ab+b²


=2a²+2b²


=2(a²+b²)


So : (a+b)²+(a-b)²=2(a²+b²)


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4) (a+b)²-(a-b)²


=a²+b²+2ab-(a²+b²-2ab)


=a²+b²+2ab-a²-b²+2ab


=4ab


(a+b)²-(a-b)²=4ab


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5) (a+b+c)²


=(a+b+c)(a+b+c)


=a(a+b+c)+b(a+b+c)+c(a+b+c)


=a²+ab+ac+ab+b²+bc+ac+bc+c²


=a²+b²+c²+2ab+2ac+2bc


=a²+b²+c²+2(ab+bc+ac)


a²+b²+c²+2(ab+bc+ac)


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6)(a+b)(a-b)


=a(a-b)+b(a-b)


=a²-ab+ab-b²


=a²-b²


(a+b)(a-b)=a²-b²


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7)(a-b-c)²


(a-b-c)(a-b-c)


=a(a-b-c)-b(a-b-c)-c(a-b-c)


=a²-ab-ac-ab+b²+bc-ac+bc+c²


=a²+b²+c²-2ab-2ac+2bc


=a²+b²+c²-2(ab-bc+ac)


(a-b-c)²=a²+b²+c²-2(ab-bc+ac)


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Hey I am a human being how can I prove every formula in a day:


Other such important formulas are as follows:-


(x+a)(x+b)=x²+x(a+b)+ab


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(x+a)(x-b)=x²+x(a-b)-ab


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(x-a)(x+b)=x²-x(a-b)-ab


---------------------------------------------------------


(x-a)(x-b)=x²-x(a+b)+ab


--------------------------------------------------------


a(x+a)(x+b)=x²+x(a+b)+ab


------------------------------------------------------------------


a²+b²+c²-ab-bc-ac=1/2*[(a-b)²+(b-c)²+(c-a)²]


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Formulas involving cubes


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(a+b)³=a³+b³+3ab(a+b)


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(a-b)³=a³-b³-3ab(a-b)


-------------------------------------------------------------------


a³-b³=(a-b)(a²+b²+ab)


----------------------------------------------------------------


a³+b³=(a+b)(a²-ab+b²)


------------------------------------------------------------


If (a+b+c)=0


then:


a³+b³+c³=3abc


----------------------------------------------------------------


(a+b+c)(a²+b²+c²-ab-bc-ac)=a³+b³+c³-3abc


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(x+a)(x+b)(x+c)=x³+ax²+bx²+cx²+abx+bcx+acx+abc


==> (x+a)(x+b)(x+c)=x³+x(a+b+c)+(ab+bc+ac)


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(a+b+c)³=(a³+b³+c³)+3[(a+b+c)(ab+ac+bc)−abc]


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Other important formulas:


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For : ax²+bx+c



where x is the root of the equation


x satisfies the equation and makes the equation zero.


There may be more than one root in an equation.


Sum of roots of a quadratic equation is -b/a


Product of roots is c/a


For : ax³+bx²+cx+d


Let the roots be x,y and z


x+y+z=-b/a


xyz=-d/a


xy+yz+xz=c/a


For any polynomial


sum of roots=-b/a


product of roots=last term/first term


If the degree of polynomials are odd then the product is -last term/first term


If it is even then the product is last term/first term.


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Hope these formulas help you:-)



Please ask me if you have any doubts.





Anonymous: took very long to write the answer :(
zaira92: tmne itne jaldi se likha kese ???
Anonymous: hehe guess how?
zaira92: i don't know
zaira92: bolo na kese
Anonymous: i had already written one answer like this so some formulas i have taken from there
zaira92: kaha likh ke rakhe the
zaira92: but i don't understand
Anonymous: mein ekh aur answer brainly mein diya tha
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