Math, asked by Anonymous, 1 day ago

Polynomials Class 9

Explain
➝ Dividend , Divisor , Quotient , Remainder
➝ Explain factorization of polynomials
➝ Explain all 8 algebraic identities with explanation

You can answer this question with examples for better explanation. ​

Answers

Answered by rajpurohitlalit1000
17

Step-by-step explanation:

The dividend divisor quotient remainder formula helps in verifying the answer obtained by the division method. Usually, when we divide a number by another number, it results in an answer, such that;

a/b = c where a is the dividend, b is the divisor, and c is the quotient. In other words, it can be written as:

Dividend/Divisor = Quotient

Dividend = Divisor × Quotient

And if any remainder is left, after the division process, then it is written as:

Dividend = Divisor × Quotient + Remainder

Therefore, the dividend divisor quotient remainder formula is Dividend = Divisor x Quotient + Remainder

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Examples Using Dividend Divisor Quotient Remainder Formula

Example 1: Find the remainder when the dividend is 75, the divisor is 5 and the quotient is 15. Use the dividend divisor quotient remainder formula.

Solution: Given, dividend = 75, divisor = 5, quotient = 15 and let the remainder be x

Marking them in the formula:

Dividend = Divisor × Quotient + Remainder

75 = 5 × 15 + x

75 = 75 + x

x = 75 - 75

x = 0

Therefore, by using the formula we obtained the remainder which is 0. Remainder = 0

Example 2: Find the dividend when the remainder is 1, the divisor is 3, and the quotient is 31.

Solution: Given, remainder = 1, divisor = 3, quotient = 31 and let the dividend be x

Using the dividend divisor quotient remainder formula:

Dividend = Divisor × Quotient + Remainder

Dividend = 3 × 31 + 1

Dividend = 94

Therefore, the dividend is 94

Example 3: Divide 120 by 5 using the division method and verify it with the dividend divisor quotient remainder formula.

Solution: First let's divide 120 by 5 using the simple division method:

120/5 = 24

Here, 120 is the dividend, 5 is the divisor, 24 is the quotient, and 0 is the remainder.

Let us verify this answer by using the dividend divisor quotient remainder formula:

Dividend = Divisor × Quotient + Remainder

120 = 5 × 24 + 0

120 = 120

As we can see that the LHS = RHS, hence the division is correct.

FAQs on Dividend Divisor Quotient Remainder Formula

What is Meant by the Dividend Divisor Quotient Remainder Formula?

What Happens when Zero is Used in the Dividend Divisor Quotient Remainder Formula?

Find the value of x in 70 = 5 × x + 0 using the Dividend Divisor Quotient Remainder Formula.

How can we apply Dividend Divisor Quotient Remainder Formula?

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Book A

Answered by tennetiraj86
41

Step-by-step explanation:

Dividend :-

The number or term or polynomial which is divided by another number or term or polynomial is called dividend .

Ex:-

In 12÷3 , 12 is called dividend

In polynomials generally the dividend is denoted by P(x).

Divisor :-

The number or term or polynomial which divides is called a divisor.

Ex:-

In 12÷3 , 3 is called divisor .

In polynomials the divisor generally denoted by g(x).

Quotient :-

The number or term or polynomial which is obtained by dividing the dividend by the divisor is called Quotient.

Quotient is the result in the process of division.

Ex:-

In 12÷3 = 4 , 4 is called the quotient .

Remainder:-

The number or term or polynomial which the part remaining after division process.

Ex:- On dividing 12 by 3 then we get the remainder 0

3) 12 ( 4

12

(-)

_____

0

_____

Ex:-

Dividend = 12

Divisor = 3

Quotient = 4

Remainder = 0

x+1 ) x²+4x+3 ( x+3

x²+x

(-) (-)

________

3x+3

3x+3

(-) (-)

_________

0

___________

Where,

Dividend = x²+4x+3

Divisor = x+1

Quotient = x+3

Remainder = 0

Factorization of polynomials:-

The polynomial q(x) is said to have divided a polynomial p(x) exactly if the remainder is zero.

In case q(x) is a factor of p(x)

Ex:-

x+1 ) x²+4x+3 ( x+3

x²+x

(-) (-)

________

3x+3

3x+3

(-) (-)

_________

0

___________

In the above example , x²+4x+3 is divided by x+1 then we get the remainder is zero.

So x+1 is a factor of x²+4x+3

We know that

The division rule

Dividend = Divisor x Quotient+Remainder

P(x) = g(x).q(x)+r(x)

Where,p(x) is the dividend

g(x) is the divisor

q(x) is the quotient

r(x) is the remainder

We use Factor Theorem for finding the factors of the given polynomial.

Factor Theorem:-

If P(x) is a polynomial of the degree n≥1 and a is any real number, then

i) x-a is a factor of P(x) ,if P(a) = 0

ii) and its converse ,if x-a is a factor of P(x) then P(a) = 0.

Algebraic Identities :-

I) (x+y)² = x²+2xy+y²

Proof:-

(x+y)²

=> (x+y)(x+y)

=> x(x+y)+y(x+y)

=> x²+xy+yx+y²

=>x²+2xy+y²

Ex:-

(t+5)²

it is in the form of (x+y)²

Where, x = t and y = 5

We know that

(x+y)² = x²+2xy+y²

=> (t+5)² = t²+2(t)(5)+5²

=> (t+5)² = t²+10t+25

ii) (x-y)² = x²-2xy+y²

Proof:-

(x-y)²

=> (x-y)(x-y)

=> x(x-y)-y(x-y)

=> x²-xy-yx+y²

=>x²-2xy+y²

Ex:-

(t-5)²

it is in the form of (x-y)²

Where, x = t and y = 5

We know that

(x-y)² = x²-2xy+y²

=> (t-5)² = t²-2(t)(5)+5²

=> (t-5)² = t²-10t+25

iii)(x+y)(x-y) = x²-y²

Proof:-

(x+y)(x-y)

=>x(x-y)+y(x-y)

=>x²-xy+yx-y²

=>x²-y²

Ex:-

(t+5)(t-5)

It is in the form of (x+y)(x-y)

Where, x = t and y = 5

=> (t+5)(t-5) = t²-5²

=> (t+5)(t-5) = t²-25

iv) (x+a)(x+b) = x²+(a+b)x+ab

Proof:-

(x+a)(x+b)

=> x(x+b)+a(x+b)

=> x²+bx+ax+ab

=> x²+(a+b)x+ab

Ex:-

(t+2)(t+4)

It is in the form of (x+a)(x+b)

Where, x = t, a = 2 , b = 4

We know that

(x+a)(x+b) = x²+(a+b)x+ab

=> (t+2)(t+4) = t²+(2+4)t+(2×4)

=> (t+2)(t+4) = t²+6t+8

v)(x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Proof:-

(x+y+z)²

=>[(x+y)+z]²

=> (x+y)²+2(x+y)(z)+z²

=> x²+2xy+y²+2xz+2yz+z²

=> x²+y²+z²+2xy+2yz+2zx

Ex:-

(2a+3b+4c)²

=> (2a)²+(3b)²+(4c)²+2(2a)(3b)+2(3b)(4c)+2(4c)(2a)

=> 4a²+9b²+16c²+12ab+24bc+16ac

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

Proof:-

(x+y)³

=> (x+y)(x+y)(x+y)

=> (x+y)(x+y)²

=> x(x+y)²+y(x+y)²

=> x(x²+2xy+y²)+y(x²+2xy+y²)

=> x³+2x²y+xy²+x²y+2xy²+y³

=> x³+3x²y+3xy²+y³

=> x³+3xy(x+y)+y³

Ex:-

(2a+3b)³

=> (2a)³+3(2a)²(3b)+3(2a)(3b)²+(3b)³

=> 8a³+3(4a²)(3b)+3(2a)(9b²)+27b³

=> 8a³+36a²b+54ab²+27b³

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

Proof:-

(x-y)³

=> (x-y)(x-y)(x-y)

=> (x-y)(x-y)²

=> x(x-y)²-y(x-y)²

=> x(x²-2xy+y²)-y(x²-2xy+y²)

=> x³-2x²y+xy²-x²y+2xy²-y³

=> x³-3x²y+3xy²+y³

=> x³-3xy(x-y)-y³

Ex:-

(2a-3b)³

=> (2a)³-3(2a)²(3b)+3(2a)(3b)²-(3b)³

=> 8a³-3(4a²)(3b)+3(2a)(9b²)+27b³

=> 8a³-36a²b+54ab²-27b³

viii) (x+y+z)(x²+y²+z²-xy-yz-zx) = x³+y³+z³-3xyz

Proof:-

(x+y+z)(x²+y²+z²-xy-yz-zx)

x(x²+y²+z²-xy-yz-zx)+y(x²+y²+z²-xy-yz-zx)+z(x²+y²+z²-xy-yz-zx))

=>x³+xy²+xz²-x²y-xyz-zx²+yx²+y³+yz²-xy²-y²z-xyz+x²z+zy²+z³-xyz-yz²-z²x

=>x³+y³+z³-xyz-xyz-xyz

=> x³+y³+z³-3xyz

Ex:-

(2a+b+c)(4a²+b²+c²-2ab-bc-2ca)

=> (2a+b+c)[(3a)²+b²+c²-(2a)(b)-(b)(c)-(c)(2a)]

=> It is in the form of (x+y+z)(x²+y²+z²-xy-yz-zx)

Where, x = 2a , y = b and z = c

We know that

(x+y+z)(x²+y²+z²-xy-yz-zx) = x³+y³+z³-3xyz

=> (2a)³+b³+c³-3(2a)(b)(c)

=> 8a³+b³+c³-6abc

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