Question

Give me formulas of polynomial. .......

Answer 1

polynomial identities(short multiplication formulas):

(x + y)2 = x2 + 2xy + y2

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

Example 1: If x = 10, y = 5a

(10 + 5a)2 = 102 + 2·10·5a + (5a)2 = 100 + 100a + 25a2

Example 2: if x = 10 and y is 4

(10 - 4)2 = 102 - 2·10·4 + 42 = 100 - 80 + 16 = 36

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 = y2 - 2xy + x2

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

Formulas for 3rd degree:

(x + y)3 = x3 + 3x2y + 3xy2 + y3

(x - y)3 = x3 - 3x2y + 3xy2 - y3

Example: (1 + a2)3 = 13 + 3.12.a2 + 3.1.(a2)2 + (a2)3 = 1 + 3a2 + 3a4 + a6

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2xz + 2yz

(x - y - z)2 = x2 + y2 + z2 - 2xy - 2xz + 2yz

Factor Rules

x2 - y2 = (x - y)(x + y)

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

or

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

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

x3 - y3 = (x - y)(x2 + xy + y2)

x3 + y3 = (x + y)(x2 - xy + y2)

If n is a natural number

xn - yn = (x - y)(xn-1 + xn-2y +...+ yn-2x + yn-1)

If n is even (n = 2k)

xn + yn = (x + y)(xn-1 - xn-2y +...+ yn-2x - yn-1)

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

xn + yn = (x + y)(xn-1 - xn-2y +...- yn-2x + yn-1)

More Algebraic Formulas

2(a2 + b2) = (a + b)2 + (a - b)2

(a + b)2 - (a - b)2 = 4ab

(a - b)2 = (a + b)2 - 4ab

a4 + b4 = (a + b)(a - b)[(a + b)2 - 2ab]

Problems Involving Polynomial Identities

1) Solve the equation: x2 - 25 = 0

Solution: x2 - 25 = (x - 5)(x + 5)

=> we have to solve the following 2 equations:

x - 5 = 0 or x + 5 = 0

so the equation have two decisions: x = 5 and x = -5

Answer 2

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

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(x-a)(x-b)=x²-x(a+b)+ab

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a(x+a)(x+b)=x²+x(a+b)+ab

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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)

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a³-b³=(a-b)(a²+b²+ab)

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a³+b³=(a+b)(a²-ab+b²)

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If (a+b+c)=0

then:

a³+b³+c³=3abc

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(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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If n is a natural number,

$a^n - b^n = (a - b)(a^(n-1 )+ a^(n-2)b +.........+ b^(n-2)a + b^(n-1))$

If n is even :(n = 2k),

$a^n + b^n = (a +b)(a^(n-1) - a^(n-2)b +........+ b^(n-2)a - b^(n-1))$

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

$a^n + b^n = (a + b)(a^(n-1) -a^(n-2)b ........- b^(n-2)a + b^(n-1))$

Also see these are important:

For : ax²+bx+c

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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.