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) 3p2−3pq−18q2p2+3pq−10q2\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}

(ii) n3−3n2m+3nm2−m35m2−10mn+5n2\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}

(iii) w3−v3+x3+3wvxw2+v2+x2−2wv−2vx+2wx\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}

(iv) 4y2−20yz+25z2(25z2−4y2)\frac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}

(v) (x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}

(vi) p4−16p2−4p+4\frac{p^4 - 16}{p^2 - 4p + 4}

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Solution

We factor the numerator and denominator to find common terms and cancel them.

Step 1 — Simplify Expression (i)

Let's factor the numerator first. We take out the common factor 3. 3p2−3pq−18q2=3(p2−pq−6q2)3p^2 - 3pq - 18q^2 = 3(p^2 - pq - 6q^2) We factor the quadratic expression. =3(p−3q)(p+2q)= 3(p-3q)(p+2q) Now, let's factor the denominator. We look for two numbers that multiply to -10 and add to 3. p2+3pq−10q2=(p+5q)(p−2q)p^2 + 3pq - 10q^2 = (p+5q)(p-2q) We write the rational expression with factored terms. 3p2−3pq−18q2p2+3pq−10q2=3(p−3q)(p+2q)(p+5q)(p−2q)\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2} = \frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)} We check for any common factors. No common factors are present. The expression is already in its simplest form.

3(p−3q)(p+2q)(p+5q)(p−2q)\boxed{\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}}

Step 2 — Simplify Expression (ii)

Let's factor the numerator. This is a perfect cube expansion. n3−3n2m+3nm2−m3=(n−m)3n^3 - 3n^2m + 3nm^2 - m^3 = (n-m)^3 Now, let's factor the denominator. We take out the common factor 5. 5m2−10mn+5n2=5(m2−2mn+n2)5m^2 - 10mn + 5n^2 = 5(m^2 - 2mn + n^2) This is a perfect square. =5(m−n)2= 5(m-n)^2 We know that (m−n)2(m-n)^2 is the same as (n−m)2(n-m)^2. =5(n−m)2= 5(n-m)^2 We write the rational expression with factored terms. n3−3n2m+3nm2−m35m2−10mn+5n2=(n−m)35(n−m)2\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2} = \frac{(n-m)^3}{5(n-m)^2} We cancel the common factor (n−m)2(n-m)^2. =n−m5= \frac{n-m}{5}

n−m5\boxed{\frac{n-m}{5}}

Step 3 — Simplify Expression (iii)

Let's factor the numerator. We use the identity a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca). Here, a=wa=w, b=−vb=-v, and c=xc=x. w3−v3+x3+3wvx=w3+(−v)3+x3−3w(−v)xw^3 - v^3 + x^3 + 3wvx = w^3 + (-v)^3 + x^3 - 3w(-v)x =(w−v+x)(w2+(−v)2+x2−w(−v)−(−v)x−xw)= (w-v+x)(w^2+(-v)^2+x^2 - w(-v) - (-v)x - xw) =(w−v+x)(w2+v2+x2+wv+vx−wx)= (w-v+x)(w^2+v^2+x^2+wv+vx-wx) Now, let's factor the denominator. This is a perfect square expansion. w2+v2+x2−2wv−2vx+2wx=(w−v+x)2w^2 + v^2 + x^2 - 2wv - 2vx + 2wx = (w-v+x)^2 We write the rational expression with factored terms. w3−v3+x3+3wvxw2+v2+x2−2wv−2vx+2wx=(w−v+x)(w2+v2+x2+wv+vx−wx)(w−v+x)2\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx} = \frac{(w-v+x)(w^2+v^2+x^2+wv+vx-wx)}{(w-v+x)^2} We cancel the common factor (w−v+x)(w-v+x). =w2+v2+x2+wv+vx−wxw−v+x= \frac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}

w2+v2+x2+wv+vx−wxw−v+x\boxed{\frac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}}

Step 4 — Simplify Expression (iv)

Let's factor the numerator. This is a perfect square trinomial. 4y2−20yz+25z2=(2y−5z)24y^2 - 20yz + 25z^2 = (2y-5z)^2 Now, let's factor the denominator. This is a difference of squares. 25z2−4y2=(5z)2−(2y)225z^2 - 4y^2 = (5z)^2 - (2y)^2 =(5z−2y)(5z+2y)= (5z-2y)(5z+2y) We know that (2y−5z)2(2y-5z)^2 is the same as (5z−2y)2(5z-2y)^2. 4y2−20yz+25z225z2−4y2=(5z−2y)2(5z−2y)(5z+2y)\frac{4y^2 - 20yz + 25z^2}{25z^2 - 4y^2} = \frac{(5z-2y)^2}{(5z-2y)(5z+2y)} We cancel the common factor (5z−2y)(5z-2y). =5z−2y5z+2y= \frac{5z-2y}{5z+2y}

5z−2y5z+2y\boxed{\frac{5z-2y}{5z+2y}}

Step 5 — Simplify Expression (v)

Let's factor each quadratic expression. For x2+x−6x^2 + x - 6: x2+x−6=(x+3)(x−2)x^2 + x - 6 = (x+3)(x-2) For x2−7x+12x^2 - 7x + 12: x2−7x+12=(x−3)(x−4)x^2 - 7x + 12 = (x-3)(x-4) For x2−6x+8x^2 - 6x + 8: x2−6x+8=(x−2)(x−4)x^2 - 6x + 8 = (x-2)(x-4) For x2−9x^2 - 9: x2−9=(x−3)(x+3)x^2 - 9 = (x-3)(x+3) We substitute these factored terms into the expression. (x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)=(x+3)(x−2)(x−3)(x−4)(x−2)(x−4)(x−3)(x+3)\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)} = \frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)} We cancel all the common factors. =1= 1

1\boxed{1}

Step 6 — Simplify Expression (vi)

Let's factor the numerator. This is a difference of squares. p4−16=(p2)2−42p^4 - 16 = (p^2)^2 - 4^2 =(p2−4)(p2+4)= (p^2-4)(p^2+4) We factor the term (p2−4)(p^2-4) further. =(p−2)(p+2)(p2+4)= (p-2)(p+2)(p^2+4) Now, let's factor the denominator. This is a perfect square trinomial. p2−4p+4=(p−2)2p^2 - 4p + 4 = (p-2)^2 We write the rational expression with factored terms. p4−16p2−4p+4=(p−2)(p+2)(p2+4)(p−2)2\frac{p^4 - 16}{p^2 - 4p + 4} = \frac{(p-2)(p+2)(p^2+4)}{(p-2)^2} We cancel the common factor (p−2)(p-2). =(p+2)(p2+4)p−2= \frac{(p+2)(p^2+4)}{p-2}

(p+2)(p2+4)p−2\boxed{\frac{(p+2)(p^2+4)}{p-2}}

Answer

(i) 3(p−3q)(p+2q)(p+5q)(p−2q)\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)} (ii) n−m5\frac{n-m}{5} (iii) w2+v2+x2+wv+vx−wxw−v+x\frac{w^2+v^2+x^2+wv+vx-wx}{w-v+x} (iv) 5z−2y5z+2y\frac{5z-2y}{5z+2y} (v) 11 (vi) (p+2)(p2+4)p−2\frac{(p+2)(p^2+4)}{p-2}

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