Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Question 1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero.
(i) $\dfrac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}$
Solution:
Given expression is
$$ \frac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}. $$Step 1: Factorise the numerator.
Take $3$ common from the numerator.
$$ \begin{aligned} 3p^2-3pq-18q^2 &=3(p^2-pq-6q^2). \end{aligned} $$Now factorise the quadratic expression.
We need two numbers whose product is $-6$ and whose sum is $-1$.
The required numbers are $-3$ and $2$.
$$ \begin{aligned} p^2-pq-6q^2 &=p^2-3pq+2pq-6q^2\\ &=p(p-3q)+2q(p-3q)\\ &=(p-3q)(p+2q). \end{aligned} $$Hence,
$$ 3p^2-3pq-18q^2 = 3(p-3q)(p+2q). $$Step 2: Factorise the denominator.
$$ p^2+3pq-10q^2. $$We need two numbers whose product is $-10$ and whose sum is $3$.
The required numbers are $5$ and $-2$.
$$ \begin{aligned} p^2+3pq-10q^2 &=p^2+5pq-2pq-10q^2\\ &=p(p+5q)-2q(p+5q)\\ &=(p-2q)(p+5q). \end{aligned} $$Step 3: Write the factorised form.
$$ \begin{aligned} \frac{3p^2-3pq-18q^2}{p^2+3pq-10q^2} = \frac{3(p-3q)(p+2q)} {(p-2q)(p+5q)}. \end{aligned} $$There is no common factor in the numerator and denominator.
Answer:
$$ \boxed{ \frac{3(p-3q)(p+2q)} {(p-2q)(p+5q)} } $$(ii) $\dfrac{n^3-3n^2m+3nm^2-m^3}{5m^2-10mn+5n^2}$
Solution:
Given expression is
$$ \frac{n^3-3n^2m+3nm^2-m^3} {5m^2-10mn+5n^2}. $$Step 1: Factorise the numerator.
The numerator is the expansion of
$$ (a-b)^3=a^3-3a^2b+3ab^2-b^3. $$Taking
$$ a=n,\qquad b=m, $$we get
$$ \begin{aligned} n^3-3n^2m+3nm^2-m^3 &=(n-m)^3. \end{aligned} $$Step 2: Factorise the denominator.
Take $5$ common.
$$ \begin{aligned} 5m^2-10mn+5n^2 &=5(m^2-2mn+n^2). \end{aligned} $$Using the identity
$$ a^2-2ab+b^2=(a-b)^2, $$we get
$$ \begin{aligned} 5(m^2-2mn+n^2) &=5(m-n)^2. \end{aligned} $$Since
$$ (m-n)^2=(n-m)^2, $$the denominator becomes
$$ 5(n-m)^2. $$Step 3: Cancel the common factor.
$$ \begin{aligned} \frac{(n-m)^3} {5(n-m)^2} &=\frac{n-m}{5}. \end{aligned} $$Answer:
$$ \boxed{\frac{n-m}{5}} $$(iii) $\dfrac{w^3-v^3+x^3+3vwx}{w^2+v^2+x^2-2wv-2vx+2wx}$
Solution:
The given expression is
$$ \frac{w^3-v^3+x^3+3vwx} {w^2+v^2+x^2-2wv-2vx+2wx}. $$Step 1: Factorise the numerator.
The numerator is of the form
$$ a^3+b^3+c^3-3abc =(a+b+c)(a^2+b^2+c^2-ab-bc-ca). $$Take
$$ a=w,\qquad b=x,\qquad c=-v. $$Then
$$ \begin{aligned} a^3+b^3+c^3-3abc &=w^3+x^3-v^3+3vwx. \end{aligned} $$Therefore,
$$ \begin{aligned} w^3-v^3+x^3+3vwx &=(w+x-v)\\ &\quad\times(w^2+x^2+v^2-wx+xv+vw). \end{aligned} $$Step 2: Factorise the denominator.
Using the identity
$$ (a+b-c)^2 = a^2+b^2+c^2 +2ab-2ac-2bc, $$we get
$$ \begin{aligned} (w+x-v)^2 &=w^2+x^2+v^2\\ &\quad+2wx-2wv-2vx. \end{aligned} $$Hence,
$$ w^2+v^2+x^2-2wv-2vx+2wx =(w+x-v)^2. $$Step 3: Write the factorised form.
$$ \begin{aligned} \frac{w^3-v^3+x^3+3vwx} {w^2+v^2+x^2-2wv-2vx+2wx} &= \frac{(w+x-v)(w^2+x^2+v^2-wx+xv+vw)} {(w+x-v)^2}. \end{aligned} $$Cancel the common factor.
$$ \begin{aligned} &= \frac{w^2+x^2+v^2-wx+xv+vw} {w+x-v}. \end{aligned} $$Answer:
$$ \boxed{ \frac{w^2+x^2+v^2-wx+xv+vw} {w+x-v} } $$(iv) $\dfrac{4y^2-20yz+25z^2}{25z^2-4y^2}$
Solution:
The given expression is
$$ \frac{4y^2-20yz+25z^2} {25z^2-4y^2}. $$Step 1: Factorise the numerator.
Using the identity
$$ a^2-2ab+b^2=(a-b)^2, $$where
$$ a=2y,\qquad b=5z, $$we get
$$ \begin{aligned} 4y^2-20yz+25z^2 &=(2y-5z)^2. \end{aligned} $$Step 2: Factorise the denominator.
Using the identity
$$ a^2-b^2=(a-b)(a+b), $$we get
$$ \begin{aligned} 25z^2-4y^2 &=(5z)^2-(2y)^2\\ &=(5z-2y)(5z+2y). \end{aligned} $$Since
$$ 2y-5z=-(5z-2y), $$the numerator becomes
$$ (2y-5z)^2=(5z-2y)^2. $$Therefore,
$$ \begin{aligned} \frac{(5z-2y)^2} {(5z-2y)(5z+2y)} &= \frac{5z-2y}{5z+2y}. \end{aligned} $$Answer:
$$ \boxed{ \frac{5z-2y}{5z+2y} } $$(v) $\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$
Solution:
The given expression is
$$ \frac{(x^2+x-6)(x^2-7x+12)} {(x^2-6x+8)(x^2-9)}. $$Step 1: Factorise each quadratic expression.
Factorising the first quadratic,
$$ \begin{aligned} x^2+x-6 &=(x+3)(x-2). \end{aligned} $$Factorising the second quadratic,
$$ \begin{aligned} x^2-7x+12 &=(x-3)(x-4). \end{aligned} $$Factorising the third quadratic,
$$ \begin{aligned} x^2-6x+8 &=(x-2)(x-4). \end{aligned} $$Using the identity $a^2-b^2=(a-b)(a+b)$,
$$ \begin{aligned} x^2-9 &=x^2-3^2\\ &=(x-3)(x+3). \end{aligned} $$Step 2: Substitute the factorised forms.
$$ \begin{aligned} &\frac{(x+3)(x-2)(x-3)(x-4)} {(x-2)(x-4)(x-3)(x+3)}. \end{aligned} $$Step 3: Cancel the common factors.
The common factors are
$$ (x+3),\quad (x-2),\quad (x-3),\quad (x-4). $$After cancelling all the common factors, we get
$$ 1. $$Answer:
$$ \boxed{1} $$(vi) $\dfrac{p^4-16}{p^2-4p+4}$
Solution:
The given expression is
$$ \frac{p^4-16}{p^2-4p+4}. $$Step 1: Factorise the numerator.
Using the identity
$$ a^2-b^2=(a-b)(a+b), $$we get
$$ \begin{aligned} p^4-16 &=(p^2)^2-4^2\\ &=(p^2-4)(p^2+4). \end{aligned} $$Now factorise
$$ p^2-4. $$ $$ \begin{aligned} p^2-4 &=(p-2)(p+2). \end{aligned} $$Therefore,
$$ \begin{aligned} p^4-16 &=(p-2)(p+2)(p^2+4). \end{aligned} $$Step 2: Factorise the denominator.
Using the identity
$$ a^2-2ab+b^2=(a-b)^2, $$we obtain
$$ \begin{aligned} p^2-4p+4 &=(p-2)^2. \end{aligned} $$Step 3: Write the factorised form.
$$ \begin{aligned} \frac{p^4-16}{p^2-4p+4} &= \frac{(p-2)(p+2)(p^2+4)} {(p-2)^2}. \end{aligned} $$Cancel one common factor $(p-2)$.
$$ \begin{aligned} &= \frac{(p+2)(p^2+4)} {p-2}. \end{aligned} $$Answer:
$$ \boxed{ \frac{(p+2)(p^2+4)} {p-2} } $$