**Intext (NCERT Pg 87)** **Q1** Try to simplify the following rational expression: 36s2−12st+t2t2+2ts−48s2=(6s−t)2(⋯+… )(⋯+… )\frac{36s^2 - 12st + t^2}{t^2 + 2ts - 48s^2} = \frac{(6s - t)^2}{(\dots + \dots)(\dots + \dots)} (Hint: Factorize t2+2ts−48s2t^2 + 2ts - 48s^2 and simplify the rational expressions assuming that t2+2ts−48s2≠0t^2 + 2ts - 48s^2 \neq 0). --- **Exercise 4.5 (NCERT Pg 87)** **Q1** Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero. (i) 3p2−3pq−18q2p2+3pq−10q2\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} (ii) n3−3n2m+3nm2−m35m2−10mn+5n2\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2} (iii) w3−v3+x3+3wvxw2+v2+x2−2wv−2vx+2wx\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} (iv) 4y2−20yz+25z2(25z2−4y2)\dfrac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)} (v) (x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} (vi) p4−16p2−4q+4\dfrac{p^4 - 16}{p^2 - 4q + 4}

Answer: t−6st+8s\frac{t - 6s}{t + 8s}

Step-by-step solution

Step 1: Factor the numerator using algebraic identity

We recognize the numerator as a perfect square trinomial matching the algebraic identity a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2. Here, a=6sa = 6s and b=tb = t, giving (6s−t)2(6s - t)^2. Since (a−b)2=(b−a)2(a - b)^2 = (b - a)^2, this can also be written as (t−6s)2(t - 6s)^2.

Step 2: Factor the denominator by splitting the middle term

To factor the quadratic trinomial in the denominator, we split the middle term 2ts2ts into 8ts−6ts8ts - 6ts because 8×(−6)=−488 \times (-6) = -48. Factoring by grouping gives t(t+8s)−6s(t+8s)=(t+8s)(t−6s)t(t + 8s) - 6s(t + 8s) = (t + 8s)(t - 6s).

Step 3: Cancel common factors and simplify

We rewrite the rational expression with both numerator and denominator in factored form. We then divide both the numerator and the denominator by the non-zero common factor (t−6s)(t - 6s) to obtain the simplified expression t−6st+8s\frac{t - 6s}{t + 8s}.

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