In division of algebraic expression if x is a variable and m, n are positive integers such that m > n then (xᵐ ÷ xⁿ) = x\(^{m  n}\).
Quotient of two monomials is a monomial which is equal to the quotient of their numerical coefficients, multiplied by the quotient of their literal coefficients.
Rule:
Quotient of two monomials
= (quotient of their numerical coefficients) x (quotient of their variables)
We may proceed according to the steps given below:
(i) Arrange the terms of the dividend and divisor in descending order of their degrees.
(ii) Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
(iii) Multiply all the terms of the divisor by the first term of the quotient and subtract the result from the dividend.
(iv) Consider the remainder (if any) as a new dividend and proceed as before.
(v) Repeat this process till we obtain a remainder which is either 0 or a polynomial of degree less than that of the divisor.
Let us understand it through some examples.
Solution:
12 – 14a² – 13a by (3 + 2a).
Write the terms of the polynomial (dividend and divisor both) in decreasing order of exponents of variables.
So, dividend becomes – 14a² – 13a + 12 and divisor becomes 2a + 3.
Divide the first term of the dividend by the first term of the divisor which gives first term of the quotient.
Multiply the divisor by the first term of the quotient and subtract the product from the dividend which gives the remainder.
Now, this remainder is treated as, new dividend but the divisor remains the same.
Now, we divide the first term of the new dividend by the first term of the divisor which gives second term of the quotient.
Now, multiply the divisor by the term of the quotient just obtained and subtracts the product from the dividend.
Thus, we conclude that divisor and quotient are the factors of dividend if the remainder is zero.
Quotient = 7a + 4
Remainder = 0
= (2a + 3)(7a + 4) + 0
= 2a(7a + 4) +3(7a + 4) + 0
= – 14a² + 8a – 21a + 12 + 0
= – 14a² – 13a + 12
Solution:
Therefore, quotient = (2x + 1) and remainder = 0.
Solution:
Therefore, Dividend = x² + 6x + 8
Divisor = x + 4
Quotient = x + 2 and
Remainder = 0.
Solution:
Arranging the terms of the dividend and divisor in descending order and then dividing,
Therefore, quotient = (x²  4x + 1) and remainder = 0.
Solution:
Arranging the terms of the dividend and divisor in descending order and then dividing,
Therefore, (29x  6x²  28) ÷ (3x  4) = (2x + 7).
Solution:
The terms of the dividend are in descending order.
Arranging the terms of the divisor in descending order and then dividing,
Therefore, 5x³4x² + 3x  18) ÷ (x²  2x + 3) = (5x + 6).
Solution:
(x  1) completely divides (x³  1).
Hence, (x  1) is a factor of(x³ 1).
Solution:
Arranging the terms of dividend and divisor in descending order and then dividing,
Therefore, quotient is (5x + 2) and remainder is (x  1).
Solution:
The terms of the dividend and that of the divisor are in descending order. So, we divide them as;
(10x⁴ + 17x³  62x² + 30x  3) ÷ (2x² + 7x  1) = (5x²  9x + 3).
● Algebraic Expression
Algebraic Expression
Addition of Algebraic Expressions
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Division of Algebraic Expressions
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