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

Answer 1

(1) 21m² ÷ 7m

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

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

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

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

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

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

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

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

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

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