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
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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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