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Forming Equation from Solution

Divide. Write...

Question

Divide. Write the quotient and the remainder.
(1) 21m² ÷ 7m
(2) 40a³ ÷ (−10a)
(3) (−48p⁴) ÷ (−9p²)
(4) 40m⁵ ÷ 30m³
(5) (5x³ − 3x²) ÷ x²
(6) (8p³ − 4p²) ÷ 2p²
(7) (2y³ + 4y² + 3) ÷ 2y²
(8) (21x⁴ − 14x² + 7x) ÷ 7x³
(9) (6x⁵ − 4x⁴ + 8x³ + 2x²) ÷ 2x²
(10) (25m⁴ − 15m³ + 10m + 8) ÷ 5m³

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Solution

(1) 21m² ÷ 7m

$= \frac{21 m^{2}}{7 m} = \frac{3 \times 7 \times m \times m}{7 \times m} = 3 m$
So, quotient = 3m and remainder = 0

(2) 40a³ ÷ (−10a)
$= \frac{40 a^{3}}{(- 10 a)} = \frac{(- 4) \times (- 10 a) \times a^{2}}{(- 10 a)} = - 4 a^{2}$
So, quotient = $- 4 a^{2}$ and remainder = 0

(3) (−48p⁴) ÷ (−9p²)
$= \frac{- 48 p^{4}}{- 9 p^{2}} = \frac{- 48 p^{2} \times p^{2}}{- 9 p^{2}} = \frac{16 p^{2}}{3}$
So, quotient = $\frac{16 p^{2}}{3}$ and remainder = 0.

(4) 40m⁵ ÷ 30m³
$= \frac{40 m^{5}}{30 m^{3}} = \frac{4 \times 10 m^{3} \times m^{2}}{3 \times 10 m^{3}} = \frac{4 m^{2}}{3}$
So, quotient = $\frac{4 m^{2}}{3}$ and remainder = 0.

(5) (5x³ − 3x²) ÷ x²
$= \frac{5 x^{3} - 3 x^{2}}{x^{2}} = \frac{x^{2} (5 x - 3)}{x^{2}} = 5 x - 3$
So, quotient = $5 x - 3$ and remainder = 0

(6) (8p³ − 4p²) ÷ 2p²
$= \frac{8 p^{3} - 4 p^{2}}{2 p^{2}} = \frac{2 p^{2} (4 p - 2)}{2 p^{2}} = 4 p - 2$
So, quotient = $4 p - 2$ and remainder = 0

(7) (2y³ + 4y² + 3) ÷ 2y²
$= \frac{2 y^{3} + 4 y^{2} + 3}{2 y^{2}} = \frac{2 y^{2} (y + 2) + 3}{2 y^{2}} = \frac{2 y^{2} (y + 2)}{2 y^{2}} + \frac{3}{2 y^{2}} = (y + 2) + \frac{3}{2 y^{2}}$
So, quotient = $(y + 2)$ and remainder = 3

(8) (21x⁴ − 14x² + 7x) ÷ 7x³
$= \frac{21 x^{4} - 14 x^{2} + 7 x}{7 x^{3}} = \frac{7 x^{3} \times 3 x - 14 x^{2} + 7 x}{7 x^{3}} = \frac{7 x^{3} \times 3 x}{7 x^{3}} + \frac{- 14 x^{2} + 7 x}{7 x^{3}} = 3 x + \frac{- 14 x^{2} + 7 x}{7 x^{3}}$
So, quotient = 3x and remainder = $- 14 x^{2} + 7 x$

(9) (6x⁵ − 4x⁴ + 8x³ + 2x²) ÷ 2x²
$= \frac{6 x^{5} - 4 x^{4} + 8 x^{3} + 2 x^{2}}{2 x^{2}} = \frac{2 x^{2} (3 x^{3} - 2 x^{2} + 4 x + 1)}{2 x^{2}} = 3 x^{3} - 2 x^{2} + 4 x + 1$
So, quotient = $3 x^{3} - 2 x^{2} + 4 x + 1$ and remainder = 0

(10) (25m⁴ − 15m³ + 10m + 8) ÷ 5m³
$= \frac{25 m^{4} - 15 m^{3} + 10 m + 8}{5 m^{3}} = \frac{5 m^{3} (5 m - 3) + (10 m + 8)}{5 m^{3}} = \frac{5 m^{3} (5 m - 3)}{5 m^{3}} + \frac{(10 m + 8)}{5 m^{3}} = (5 m - 3) + \frac{(10 m + 8)}{5 m^{3}}$
So, quotient = $5 m - 3$ and remainder = $10 m + 8$

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