Exploring Algebraic Identities | Exercise 4.5

Question 1

Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(i) $\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

(ii) $\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

(iii) $\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$

(iv) $\dfrac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}$

(v) $\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$

(vi) $\dfrac{p^4 - 16}{p^2 - 4p + 4}$

Solution

Understand the Question

To simplify rational algebraic expressions, factorise both the numerator and the denominator completely using standard algebraic identities and polynomial factorisation techniques.
Cancel out any non-zero common factors between the numerator and denominator to reduce the expression to its simplest form.

(i) Simplify $\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$

Step 1 · Factorise Numerator and Denominator

Factorising the numerator by taking out the common factor $3$ and splitting the middle term: $3p^2 - 3pq - 18q^2 = 3(p^2 - pq - 6q^2)$ $= 3(p-3q)(p+2q)$

Factorising the denominator: $p^2 + 3pq - 10q^2 = (p+5q)(p-2q)$

Writing the expression with factored terms: $\dfrac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \dfrac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}$

Since there are no common factors, the expression is already in its simplest form.

Answer

(i) $\dfrac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}$

(ii) Simplify $\dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}$

Step 1 · Factorise and Cancel Common Factors

Factorising the numerator using the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ : $n^3 - 3n^2m + 3nm^2 - m^3 = (n-m)^3$

Factorising the denominator:

\begin{aligned} 5m^2 - 10mn + 5n^2 &= 5(m^2 - 2mn + n^2) \\ &= 5(m-n)^2 \\ &= 5(n-m)^2 \end{aligned}

Substituting into the rational expression and cancelling $(n-m)^2$ :

\begin{aligned} \dfrac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2} &= \dfrac{(n-m)^3}{5(n-m)^2} \\[0.6em] &= \dfrac{n-m}{5} \end{aligned}

Answer

(ii) $\dfrac{n-m}{5}$

(iii) Simplify $\dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}$

Step 1 · Factorise and Cancel Common Factors

Using the identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ with $a=w$ , $b=-v$ , and $c=x$ :

\begin{aligned} w^3 - v^3 + x^3 + 3wvx &= w^3 + (-v)^3 + x^3 - 3w(-v)x \\[0.6em] &= (w-v+x)(w^2+(-v)^2+x^2 - w(-v) - (-v)x - xw) \\[0.6em] &= (w-v+x)(w^2+v^2+x^2+wv+vx-wx) \end{aligned}

Factorising the denominator using the expansion of $(a+b+c)^2$ : $w^2 + v^2 + x^2 - 2wv - 2vx + 2wx = (w-v+x)^2$

Substituting and cancelling the common factor $(w-v+x)$ :

\begin{aligned} \dfrac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} &= \dfrac{(w-v+x)(w^2+v^2+x^2+wv+vx-wx)}{(w-v+x)^2} \\[0.6em] &= \dfrac{w^2+v^2+x^2+wv+vx-wx}{w-v+x} \end{aligned}

Answer

(iii) $\dfrac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}$

(iv) Simplify $\dfrac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}$

Step 1 · Factorise and Cancel Common Factors

Factorising the numerator as a perfect square: $4y^2 - 20yz + 25z^2 = (2y-5z)^2 = (5z-2y)^2$

Factorising the denominator using difference of squares:

\begin{aligned} 25z^2 - 4y^2 &= (5z)^2 - (2y)^2 \\[0.6em] &= (5z-2y)(5z+2y) \end{aligned}

Substituting and cancelling $(5z-2y)$ :

\begin{aligned} \dfrac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2} &= \dfrac{(5z-2y)^2}{(5z-2y)(5z+2y)} \\[0.6em] &= \dfrac{5z-2y}{5z+2y} \end{aligned}

Answer

(iv) $\dfrac{5z-2y}{5z+2y}$

(v) Simplify $\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}$

Step 1 · Factorise Each Quadratic Polynomial

Factorising each expression individually: $x^2 + x - 6 = (x+3)(x-2)$ $x^2 - 7x + 12 = (x-3)(x-4)$ $x^2 - 6x + 8 = (x-2)(x-4)$ $x^2 - 9 = (x-3)(x+3)$

Step 2 · Substitute and Cancel Common Factors

Substituting the factored terms into the expression: $\dfrac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} = \dfrac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}$

Cancelling all common factors: $= 1$

Answer

(v) $1$

(vi) Simplify $\dfrac{p^4 - 16}{p^2 - 4p + 4}$

Step 1 · Factorise and Cancel Common Factors

Factorising the numerator using difference of squares repeatedly:

\begin{aligned} p^4 - 16 &= (p^2)^2 - 4^2 \\[0.6em] &= (p^2-4)(p^2+4) \\[0.6em] &= (p-2)(p+2)(p^2+4) \end{aligned}

Factorising the denominator: $p^2 - 4p + 4 = (p-2)^2$

Substituting and cancelling $(p-2)$ :

\begin{aligned} \dfrac{p^4 - 16}{p^2 - 4p + 4} &= \dfrac{(p-2)(p+2)(p^2+4)}{(p-2)^2} \\[0.6em] &= \dfrac{(p+2)(p^2+4)}{p-2} \end{aligned}

Answer

(vi) $\dfrac{(p+2)(p^2+4)}{p-2}$

Common Mistakes

Sign in Squared Differences: Remember that $(a - b)^2 = (b - a)^2$ , but $(a - b) = -(b - a)$ . In part (iv), $(2y - 5z)^2 = (5z - 2y)^2$ , but without squaring, reversing the terms introduces a negative sign.
Incomplete Factorisation: In part (vi), do not stop at $(p^2 - 4)(p^2 + 4)$ ; remember to further factorise $(p^2 - 4)$ into $(p - 2)(p + 2)$ so common factors can be cancelled.
Direct Term Cancellation: Never cancel terms directly across addition or subtraction signs before completely factorising both expressions into products.