Divide polynomials using long division :
Let p(x) and g(x) be two polynomials such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0. Then there exists unique polynomials q(x) and r (x) such that
p(x) = g(x) q(x) + r (x) ... (1)
where r (x) = 0 or degree of r (x) < degree of g(x) .
The polynomial p(x) is the dividend, g(x) is the divisor, q(x) is the quotient and r (x) is the remainder.
(1) ==> Dividend = (Divisor x Quotient) + Remainder
Let us see some examples based on the above concept.
Example 1 :
Find the quotient and the remainder when (10- 4x + 3x²) is divided by x - 2.
Let us first write the terms of each polynomial in descending order ( or ascending order).
Thus, the given problem becomes (10- 4x + 3x²) ÷ (x - 2)
f(x) = 10- 4x + 3x²
= 3x² - 4x + 10
g (x) = x - 2
Step 1 :
In the first step, we are going to divide the first term of the dividend by the first first term of the divisor.
After changing the signs, +3x² and - 3x² will get canceled. By simplifying we get 2x + 10
Step 2 :
In the second step again we are going to divide the first term that is 2x by the first term of divisor that is x.
Quotient = 3x + 2
Remainder = 14
Example 2 :
Find the quotient and the remainder when (4x³ + 6x² - 23 x - 15) is divided by 3 + x
Step 1 :
Divide the first term of the dividend by the first term of the divisor. That is 4x³/x = 4x². Now we have to multiply this 4x² by the divisor. So we get 4x³ + 12x².
In order to subtract 4x³ + 12x², we changed the sign. After simplifying we get - 6x² - 23x - 15.
Step 2 :
Divide the first term of the dividend by the first term of the divisor. That is -6x²/x = -6x. Now we have to multiply this -6x by the divisor(x + 3). So we get -6x² - 18 x.
Subtract -6x² - 18 x from -6x² - 23 x - 15. By subtracting these two polynomials we get -5x - 15.
Step 3 :
-5x/x = -5. Now we have to multiply this -5 by the divisor(x + 3). So we get -5x - 15.
By subtracting these two polynomials, we get 0.
Hence the quotient = 4 x²-6 x - 5 and the remainder = 0.
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