Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: Class 9
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: Class 9
Question 1.
Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: Class 9
i. $\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}$
ii. $\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}$
iii. $\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}$
iv. $\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}$
v. $\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}$
vi. $\frac{p^4-16}{p^2-4 p+4}$
Solution:
i. $\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}$
3p² - 3pq - 18q² = 3(p² - pq - 6q²)
p² - pq - 6q²:
a + b = - 1 and ab = - 6
Since -3 + 2 = -1 and (-3) × 2 = -6
∴ 3p² - 3pq - 18q² = 3(p - 3q)(p + 2q)
p² + 3pq - 10q²:
a + b = 3 and ab = -10
Since 5 + (-2) = 3 and 5 × (-2) = -10
∴ p² + 3pq - 10q² = (p + 5q)(p - 2q)
$\therefore \quad \frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}=\frac{3(p-3 q)(p+2 q)}{(p+5 q)(p-2 q)}$
ii. $\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}$
n³ - 3n²m + 3nm² - m³ = (n - m)³
...... [∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
5m² - 10mn + 5n² = 5(m² - 2mn + n²)
= 5(m - n)²
..... [∵ (a- b)² = a² - 2ab + b²]
$\begin{aligned} \therefore \quad & \frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2} \\ & =\frac{(n-m)^3}{5(m-n)^2} \\ & =\frac{(n-m)^3}{5(n-m)^2} \quad \ldots\left[\because(m-n)^2=(n-m)^2\right] \\ & =\frac{n-m}{5}\end{aligned}$
... [Cancelling the common factor (n - m)²]
iii. $\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}$
w³ - v³ + x³ + 3wvx
= w³ + (-v)³ + x³ - 3(w)(-v)(x)
Here, a = w, b = -v, c = x
∴ w³ - v³ + x³ + 3wvx
= [w + (-v) + x][w² + (-v)² + x² - (w)(-v) - (-v)(x) - (w)(x)]
........... [∵ a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)]
= (w - v + x)(w² + v² + x² + wv + vx - wx)
w² + v² + x² - 2wv - 2vx + 2wx
= w² + (-v)² + x² + 2(w)(-v) + 2(-v)(x) + 2(w)(x)
Here, a = w, b = -v, c = x
∴ w² + v² + x² - 2wv - 2vx + 2wx
= (w + (-v) + x)²
.....[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= (w - v + x)²
= (w - v + x)(w - v + x)
$\begin{aligned} \therefore \quad & \frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x} \\ & =\frac{(w-v+x)\left(w^2+v^2+x^2+w v+v x-w x\right)}{(w-v+x)(w-v+x)} \\ & =\frac{w^2+v^2+x^2+w v+v x-w x}{w-v+x}\end{aligned}$
.......[Cancelling the common factor (w - v + x)]
iv. $\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}$
4y² - 20yz + 25z² = 25z² - 20yz + 4y²
= (5z)² - 2(5z)(2y) + (2y)²
Here, a = 5z, b = 2y
∴ 4y² - 20yz + 25z² = (5z - 2y)²
...[∵ (a - b)² = a² - 2ab + b²]
= (5z - 2y)(5z - 2y)
∴ 25z² - 4y² = (5z)² - (2y)²
Here, a = 5z, b = 2y
∴ 25z² - 4y² = (5z + 2y)(5z - 2y)
...[∵ a² - b² = (a + b)(a - b)]
$\begin{aligned} \therefore \quad \frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)} & =\frac{(5 z-2 y)(5 z-2 y)}{(5 z+2 y)(5 z-2 y)} \\ & =\frac{5 z-2 y}{5 z+2 y}\end{aligned}$
....[Cancelling the common factor (5z - 2y)]
v. $\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}$
x² + x - 6:
a + b = 1 and ab = - 6
Since 3 + (-2) = 1 and 3 × (-2) = -6
∴ x² + x - 6 = (x + 3) (x - 2)
x² - 7x + 12:
a + b = -7 and ab = 12
Since (-3) + (- 4) = -7 and (-3) × (-4) = 12
∴ x² - 7x + 12 = (x - 3)(x - 4)
x² - 6x + 8:
a + b = - 6 and ab = 8
Since (- 2) + (- 4) = - 6 and (-2) × (- 4) = 8
∴ x² - 6x + 8 = (x - 2)(x - 4)
x² - 9 = (x + 3)( x - 3)
...[∵ a² - b² = (a + b)(a - b)]
$\begin{aligned} \therefore \quad & \frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)} \\ & =\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x+3)(x-3)}\end{aligned}$
= 1 ...[Cancelling common factors]
vi. $\frac{p^4-16}{p^2-4 p+4}$
p4 - 16 = (p²)² - 4²
= (p² + 4)(p² - 4)
....[∵ a² - b² = (a + b)(a - b)]
= (p² + 4)(p² - 2²)
= (p² + 4)(p + 2)(p - 2)
...[∵ a² - b² = (a + b)(a - b)]
p² - 4p + 4 = p² - 2(p)(2) + 2²
= (p - 2)²
...[∵ (a - b)² = a² - 2ab + b²]
= (p - 2)(p - 2)
$\begin{aligned} \therefore \quad \frac{p^4-16}{p^2-4 p+4} & =\frac{\left(p^2+4\right)(p+2)(p-2)}{(p-2)(p-2)} \\ & =\frac{\left(p^2+4\right)(p+2)}{(p-2)}\end{aligned}$
...[Cancelling common factors (p - 2)]