10. Division of Polynomials
Solutions to Practice Sets 10.1 and 10.2, with every long division written out in full. Maharashtra State Board (Balbharati), Standard Eight Mathematics.
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Practice Set 10.1
Divide the first term of the dividend by the divisor, multiply, change the signs and subtract. Stop when the remainder is 0 or its degree is less than the degree of the divisor. (The small red signs are the changed signs used for subtraction.)
(1) (21m2) ÷ (7m)
| 3m | |
| 7m | 21m2 |
| 21m2 | |
| − | |
| 0 |
Quotient = 3m Remainder = 0
(2) (40a3) ÷ (−10a)
| −4a2 | |
| −10a | 40a3 |
| 40a3 | |
| − | |
| 0 |
Quotient = −4a2 Remainder = 0
(3) (−48p4) ÷ (−9p2)
| 163p2 | |
| −9p2 | −48p4 |
| −48p4 | |
| + | |
| 0 |
Quotient = 163p2 Remainder = 0
Two negatives give a positive, so the quotient is +163p². See the note at the end of this page about the textbook answer.
(4) (40m5) ÷ (30m3)
| 43m2 | |
| 30m3 | 40m5 |
| 40m5 | |
| − | |
| 0 |
Quotient = 43m2 Remainder = 0
(5) (5x3 − 3x2) ÷ (x2)
| 5x | − 3 | |
| x2 | 5x3 | − 3x2 |
| 5x3 | ||
| − | ||
| 0 | − 3x2 | |
| −3x2 | ||
| + | ||
| 0 |
Quotient = 5x − 3 Remainder = 0
(6) (8p3 − 4p2) ÷ (2p2)
| 4p | − 2 | |
| 2p2 | 8p3 | − 4p2 |
| 8p3 | ||
| − | ||
| 0 | − 4p2 | |
| −4p2 | ||
| + | ||
| 0 |
Quotient = 4p − 2 Remainder = 0
(7) (2y3 + 4y2 + 3) ÷ (2y2)
| y | + 2 | |||
| 2y2 | 2y3 | + 4y2 | + 0y | + 3 |
| 2y3 | ||||
| − | ||||
| 0 | + 4y2 | + 3 | ||
| 4y2 | ||||
| − | ||||
| 0 | + 3 |
Quotient = y + 2 Remainder = 3
The remainder 3 has degree 0, less than the degree of 2y² (which is 2), so the division stops.
(8) (21x4 − 14x2 + 7x) ÷ (7x3)
| 3x | ||||
| 7x3 | 21x4 | + 0x3 | − 14x2 | + 7x |
| 21x4 | ||||
| − | ||||
| 0 | − 14x2 | + 7x |
Quotient = 3x Remainder = −14x2 + 7x
The remainder −14x² + 7x has degree 2, less than the degree of 7x³ (which is 3), so the division stops.
(9) (6x5 − 4x4 + 8x3 + 2x2) ÷ (2x2)
| 3x3 | − 2x2 | + 4x | + 1 | |
| 2x2 | 6x5 | − 4x4 | + 8x3 | + 2x2 |
| 6x5 | ||||
| − | ||||
| 0 | − 4x4 | + 8x3 | + 2x2 | |
| −4x4 | ||||
| + | ||||
| 0 | + 8x3 | + 2x2 | ||
| 8x3 | ||||
| − | ||||
| 0 | + 2x2 | |||
| 2x2 | ||||
| − | ||||
| 0 |
Quotient = 3x3 − 2x2 + 4x + 1 Remainder = 0
(10) (25m4 − 15m3 + 10m + 8) ÷ (5m3)
| 5m | − 3 | ||||
| 5m3 | 25m4 | − 15m3 | + 0m2 | + 10m | + 8 |
| 25m4 | |||||
| − | |||||
| 0 | − 15m3 | + 10m | + 8 | ||
| −15m3 | |||||
| + | |||||
| 0 | + 10m | + 8 |
Quotient = 5m − 3 Remainder = 10m + 8
The remainder 10m + 8 has degree 1, less than 3, so the division stops.
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Practice Set 10.2
First write the dividend and the divisor in descending order of indices. Write any missing term with coefficient 0.
(1) (y² + 10y + 24) ÷ (y + 4)
| y | + 6 | ||
| y + 4 | y2 | + 10y | + 24 |
| y2 | + 4y | ||
| − | − | ||
| 0 | + 6y | + 24 | |
| 6y | + 24 | ||
| − | − | ||
| 0 |
Quotient = y + 6 Remainder = 0
(2) (p² + 7p − 5) ÷ (p + 3)
| p | + 4 | ||
| p + 3 | p2 | + 7p | − 5 |
| p2 | + 3p | ||
| − | − | ||
| 0 | + 4p | − 5 | |
| 4p | + 12 | ||
| − | − | ||
| 0 | − 17 |
Quotient = p + 4 Remainder = −17
(3) (3x + 2x² + 4x³) ÷ (x − 4)
| 4x2 | + 18x | + 75 | ||
| x − 4 | 4x3 | + 2x2 | + 3x | + 0 |
| 4x3 | − 16x2 | |||
| − | + | |||
| 0 | + 18x2 | + 3x | ||
| 18x2 | − 72x | |||
| − | + | |||
| 0 | + 75x | |||
| 75x | − 300 | |||
| − | + | |||
| 0 | + 300 |
Quotient = 4x2 + 18x + 75 Remainder = 300
In descending order the dividend is 4x³ + 2x² + 3x + 0.
(4) (2m³ + m² + m + 9) ÷ (2m − 1)
| m2 | + m | + 1 | ||
| 2m − 1 | 2m3 | + m2 | + m | + 9 |
| 2m3 | − m2 | |||
| − | + | |||
| 0 | + 2m2 | + m | + 9 | |
| 2m2 | − m | |||
| − | + | |||
| 0 | + 2m | + 9 | ||
| 2m | − 1 | |||
| − | + | |||
| 0 | + 10 |
Quotient = m2 + m + 1 Remainder = 10
(5) (3x − 3x² − 12 + x⁴ + x³) ÷ (2 + x²)
| x2 | + x | − 5 | |||
| x2 + 2 | x4 | + x3 | − 3x2 | + 3x | − 12 |
| x4 | + 2x2 | ||||
| − | − | ||||
| 0 | + x3 | − 5x2 | + 3x | − 12 | |
| x3 | + 2x | ||||
| − | − | ||||
| 0 | − 5x2 | + x | − 12 | ||
| −5x2 | − 10 | ||||
| + | + | ||||
| 0 | + x | − 2 |
Quotient = x2 + x − 5 Remainder = x − 2
In descending order: dividend x⁴ + x³ − 3x² + 3x − 12 and divisor x² + 2. The remainder x − 2 has degree 1, less than 2.
(6★) (a⁴ − a³ + a² − a + 1) ÷ (a³ − 2)
| a | − 1 | ||||
| a3 − 2 | a4 | − a3 | + a2 | − a | + 1 |
| a4 | − 2a | ||||
| − | + | ||||
| 0 | − a3 | + a2 | + a | + 1 | |
| −a3 | + 2 | ||||
| + | − | ||||
| 0 | + a2 | + a | − 1 |
Quotient = a − 1 Remainder = a2 + a − 1
The remainder a² + a − 1 has degree 2, less than the degree 3 of a³ − 2, so the division stops.
(7★) (4x⁴ − 5x³ − 7x + 1) ÷ (4x − 1)
| x3 | − x2 | − 14x | − 2916 | ||
| 4x − 1 | 4x4 | − 5x3 | + 0x2 | − 7x | + 1 |
| 4x4 | − x3 | ||||
| − | + | ||||
| 0 | − 4x3 | − 7x | + 1 | ||
| −4x3 | + x2 | ||||
| + | − | ||||
| 0 | − x2 | − 7x | + 1 | ||
| −x2 | + 14x | ||||
| + | − | ||||
| 0 | − 294x | + 1 | |||
| −294x | + 2916 | ||||
| + | − | ||||
| 0 | − 1316 |
Quotient = x3 − x2 − 14x − 2916 Remainder = −1316
The x² term is missing, so it is written as 0x². Check: (4x − 1)(quotient) + remainder gives back the dividend.
Note on the textbook answer key: for Practice Set 10.1 (3), the key prints the quotient as −163p². Since (−48p⁴) ÷ (−9p²) is a negative divided by a negative, the correct quotient is +163p², with remainder 0.