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Next, subtract this new polynomial –6a – 12a Now, divide the first term of new dividend
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from the original dividend –6a + 18a + 5a – 110. 30a + 5a – 110 by the first term of the divisor
Pull down the remaining terms to obtain a new a + 2. We get the second term of the quotient,
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dividend 30a + 5a – 110. i.e., 30a and repeat the same process.
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–6a 2 –6a + 30a
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a + 2 –6a + 18a + 5a – 110 a + 2 –6a + 18a + 5a – 110
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–6a – 12a 2 –6a – 12a 2
+ + + +
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2
30a + 5a – 110 30a + 5a – 110
30a + 60a
2
– –
– 55a – 110
Again, divide the first term of the new dividend –55a –110 by the first term of the divisor a + 2. We
get the third term of the quotient, i.e., –55 and repeat the above process.
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–6a + 30a – 55 Since, we get the remainder zero. Thus, we can
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a + 2 –6a + 18a + 5a – 110 say that
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–6a – 12a 2 (–6a + 18a + 5a – 110) ÷ (a + 2) = –6a + 30a –55
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+ + \ Quotient = –6a + 30a – 55 and remainder = 0.
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30a + 5a – 110
2
30a + 60a
– –
– 55a – 110
– 55a – 110
+ +
0
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2
Example 18: Divide (x – 16 + 21x – 8x ) by (x – 1) and find the quotient and remainder.
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Solution: We have (x – 16 + 21x – 8x ) ÷ (x – 1)
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First, arrange the dividend and the divisor in descending order of degree, i.e., x – 8x + 21x – 16.
Then, proceed as follows:
x – 7x + 14
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x – 1 x – 8x + 21x – 16
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x – x 2 Multiplying (x – 1) by x 2
– +
2
– 7x + 21x – 16
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– 7x + 7x Multiplying (x – 1) by (–7x)
+ –
14x – 16
14x – 14 Multiplying (x – 1) by 14
– +

–2 (On subtracting)
2
Thus, quotient = x – 7x + 14 and remainder = –2.

205 Algebraic Expressions