Ch 4 Exploring Algebraic Identities NCERT Solutions Class 9| Ganita Manjari

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Master the fundamental concepts of polynomials with our comprehensive guide on Ch 4 Exploring Algebraic Identities NCERT Solutions Class 9.

Ganita Manjari’s mathematics book for class 9 has questions that require the apt use of formulas and concepts.

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Exercise Set 4.1

Exercise 4.1 – Question 1

Identity used in all the questions here is –

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

(i) (7x+4y)2(7x+4y)^2

Here, a=7x, b=4ya=7x,\ b=4y
=(7x)2+2(7x)(4y)+(4y)2=(7x)^2 + 2(7x)(4y) + (4y)^2

=49x2+56xy+16y2 Ans=49x^2 + 56xy + 16y^2 \ \text{Ans}

(ii) (7x5+3y2)2\left(\frac{7x}{5}+\frac{3y}{2}\right)^2

Here,

a=7x5, b=3y2a=\frac{7x}{5},\ b=\frac{3y}{2}=(7x5)2+2(7x5)(3y2)+(3y2)2=\left(\frac{7x}{5}\right)^2 + 2\left(\frac{7x}{5}\right)\left(\frac{3y}{2}\right) + \left(\frac{3y}{2}\right)^2
=49x225+21xy5+9y24 Ans=\frac{49x^2}{25} + \frac{21xy}{5} + \frac{9y^2}{4} \ \text{Ans}

(iii) (2.5p+1.5q)2(2.5p+1.5q)^2

Here, a=2.5p, b=1.5qa=2.5p,\ b=1.5q

=(2.5p)2+2(2.5p)(1.5q)+(1.5q)2=(2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2
=6.25p2+7.5pq+2.25q2 Ans=6.25p^2 + 7.5pq + 2.25q^2 \ \text{Ans}

(iv) (3s4+8t)2\left(\frac{3s}{4}+8t\right)^2

Here, a=3s4, b=8ta=\frac{3s}{4},\ b=8t=(3s4)2+2(3s4)(8t)+(8t)2=\left(\frac{3s}{4}\right)^2 + 2\left(\frac{3s}{4}\right)(8t) + (8t)^2
=9s216+12st+64t2 Ans=\frac{9s^2}{16} + 12st + 64t^2 \ \text{Ans}

(v) (x+12y)2\left(x+\frac{1}{2y}\right)^2

Here, a=x, b=12ya=x,\ b=\frac{1}{2y}

=x2+2x(12y)+(12y)2=x^2 + 2x\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2
=x2+xy+14y2 Ans=x^2 + \frac{x}{y} + \frac{1}{4y^2} \ \text{Ans}

(vi) (1x+1y)2\left(\frac{1}{x}+\frac{1}{y}\right)^2

Here, a=1x, b=1ya=\frac{1}{x},\ b=\frac{1}{y}
=(1x)2+2(1x)(1y)+(1y)2=\left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2
=1x2+2xy+1y2 Ans=\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2} \ \text{Ans}

Exercise Set 4.2

  1. Factor completely:
    (i) 9x2 + 24xy + 16y2
    (ii) 4s2 + 20st + 25t2
    (iii) 49x2 + 28xy + 4y2
    (iv) 64p2+32/3pq+4/9q2
    (v) 3a2 + 4ab +4/3 b2
    (vi) 9/5s2+6sv+5v2

Solutions:

Direct Answers

(i) (3x+4y)2(3x + 4y)^2
(ii) (2s+5t)2(2s + 5t)^2
(iii) (7x+2y)2(7x + 2y)^2
(iv) (8p+23q)2\left(8p + \frac{2}{3}q\right)^2
(v) 13(3a+2b)2\frac{1}{3}(3a + 2b)^2
(vi) 15(3s+5v)2\frac{1}{5}(3s + 5v)^2

(i) 9x2+24xy+16y29x^2 + 24xy + 16y^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,9x2+24xy+16y2=(3x)2+2×3x×4y+(4y)29x^2 + 24xy + 16y^2 = (3x)^2 + 2 \times 3x \times 4y + (4y)^2

So,a=3x,b=4ya = 3x,\quad b = 4y

Hence,
9x2+24xy+16y2=(3x+4y)2 Ans9x^2 + 24xy + 16y^2 = (3x + 4y)^2 \ \text{Ans}

(ii) 4s2+20st+25t24s^2 + 20st + 25t^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,4s2+20st+25t2=(2s)2+2×2s×5t+(5t)24s^2 + 20st + 25t^2 = (2s)^2 + 2 \times 2s \times 5t + (5t)^2

So,a=2s,b=5ta = 2s,\quad b = 5t

Hence,4s2+20st+25t2=(2s+5t)2 Ans4s^2 + 20st + 25t^2 = (2s + 5t)^2 \ \text{Ans}

(iii) 49x2+28xy+4y249x^2 + 28xy + 4y^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,49x2+28xy+4y2=(7x)2+2×7x×2y+(2y)249x^2 + 28xy + 4y^2 = (7x)^2 + 2 \times 7x \times 2y + (2y)^2

So,a=7x,b=2ya = 7x,\quad b = 2y

Hence,49x2+28xy+4y2=(7x+2y)2 Ans49x^2 + 28xy + 4y^2 = (7x + 2y)^2 \ \text{Ans}

(iv) 64p2+323pq+49q264p^2 + \frac{32}{3}pq + \frac{4}{9}q^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,64p2+323pq+49q2=(8p)2+2×8p×23q+(23q)264p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 = (8p)^2 + 2 \times 8p \times \frac{2}{3}q + \left(\frac{2}{3}q\right)^2

So,a=8p,b=23qa = 8p,\quad b = \frac{2}{3}q

Hence,64p2+323pq+49q2=(8p+23q)2 Ans64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 = \left(8p + \frac{2}{3}q\right)^2 \ \text{Ans}

(v) 3a2+4ab+43b23a^2 + 4ab + \frac{4}{3}b^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,3a2+4ab+43b2=13(9a2+12ab+4b2)3a^2 + 4ab + \frac{4}{3}b^2 = \frac{1}{3}(9a^2 + 12ab + 4b^2)=13[(3a)2+2×3a×2b+(2b)2]= \frac{1}{3}\left[(3a)^2 + 2 \times 3a \times 2b + (2b)^2\right]

So,a=3a,b=2ba = 3a,\quad b = 2b

Hence,3a2+4ab+43b2=13(3a+2b)2 Ans3a^2 + 4ab + \frac{4}{3}b^2 = \frac{1}{3}(3a + 2b)^2 \ \text{Ans}

(vi) 95s2+6sv+5v2\frac{9}{5}s^2 + 6sv + 5v^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,95s2+6sv+5v2=15(9s2+30sv+25v2)\frac{9}{5}s^2 + 6sv + 5v^2 = \frac{1}{5}(9s^2 + 30sv + 25v^2)=15[(3s)2+2×3s×5v+(5v)2]= \frac{1}{5}\left[(3s)^2 + 2 \times 3s \times 5v + (5v)^2\right]

So,a=3s,b=5va = 3s,\quad b = 5v

Hence,95s2+6sv+5v2=15(3s+5v)2 Ans\frac{9}{5}s^2 + 6sv + 5v^2 = \frac{1}{5}(3s + 5v)^2 \ \text{Ans}

2. Find the values of the following using the identity
(a – b)2 = a2 – 2ab + b2.
(i) (79)2 (ii) (193)2 (iii) (299)2

Final Answers

(i) 6241
(ii) 37249
(iii) 89401

(i) 79279^2

Now, using identity,

(a−b)2=a2−2ab+b2(a-b)^2 = a^2 – 2ab + b^2

We have,
79=80−179 = 80 – 1792=(80−1)279^2 = (80 – 1)^2=(80)2−2×80×1+(1)2= (80)^2 – 2 \times 80 \times 1 + (1)^2
We have,79=80−179 = 80 – 1792=(80−1)279^2 = (80 – 1)^2=(80)2−2×80×1+(1)2= (80)^2 – 2 \times 80 \times 1 + (1)^2

So,a=80,b=1a = 80,\quad b = 1=6400−160+1=6241= 6400 – 160 + 1 = 6241

Hence,792=6241 Ans79^2 = 6241 \ \text{Ans}

(ii) 1932193^2

Now, using identity

(a−b)2=a2−2ab+b2(a-b)^2 = a^2 – 2ab + b^2

1932=(200−7)2193^2 = (200 – 7)^2=(200)2−2×200×7+(7)2= (200)^2 – 2 \times 200 \times 7 + (7)^2

So,a=200,b=7a = 200,\quad b = 7=40000−2800+49=37249= 40000 – 2800 + 49 = 37249

Hence,1932=37249 Ans193^2 = 37249 \ \text{Ans}

(iii) 2992299^2

Now, using identity

(a−b)2=a2−2ab+b2(a-b)^2 = a^2 – 2ab + b^2

We have,299=300−1299 = 300 – 12992=(300−1)2299^2 = (300 – 1)^2=(300)2−2×300×1+(1)2= (300)^2 – 2 \times 300 \times 1 + (1)^2

So,a=300,b=1a = 300,\quad b = 1=90000−600+1=89401= 90000 – 600 + 1 = 89401

Hence,2992=89401 Ans299^2 = 89401 \ \text{Ans}

Exercise Set 4.3

Important identities:

  1. (a + b)2 = a2 + 2ab + b2
  2. (a – b)2 = a2 – 2ab + b2
  3. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
  1. Find the following squares using one of the above identities.
    Determine which of these identities will make these calculations
    easier.
    (i) 1172 (ii) 782 (iii) 1982
    (iv) 2142 (v) 11042 (vi) 11202

(i) 1172117^2

Now, using identity

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

We have,

117=100+17117 = 100 + 17

So,

a=100, b=17a = 100,\ b = 17

1172=(100+17)2117^2 = (100 + 17)^2

=(100)2+2×100×17+(17)2= (100)^2 + 2 \times 100 \times 17 + (17)^2 ( Using , (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 )

=10000+3400+289= 10000 + 3400 + 289

=13689= 13689

Hence,

1172=13689117^2 = 13689 Ans

(ii) 78278^2

Now, using identity

(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,

78=80−278 = 80 – 2

So,

a=80, b=2a = 80,\ b = 2

782=(80−2)278^2 = (80 – 2)^2

=(80)2−2×80×2+(2)2= (80)^2 – 2 \times 80 \times 2 + (2)^2 ( Using , (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 )

=6400−320+4= 6400 – 320 + 4

=6084= 6084

Hence,

782=608478^2 = 6084 Ans

(iii) 1982198^2

Now, using identity

(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,

198=200−2198 = 200 – 2

So,

a=200, b=2a = 200,\ b = 2

1982=(200−2)2198^2 = (200 – 2)^2

=(200)2−2×200×2+(2)2= (200)^2 – 2 \times 200 \times 2 + (2)^2 (Using (a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2 )

=40000−800+4= 40000 – 800 + 4

=39204= 39204

Hence,

1982=39204198^2 = 39204 Ans

(iv) 2142214^2

Now, using identity

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

We have,

214=200+14214 = 200 + 14

So,

a=200, b=14a = 200,\ b = 14

2142=(200+14)2214^2 = (200 + 14)^2

=(200)2+2×200×14+(14)2= (200)^2 + 2 \times 200 \times 14 + (14)^2 (Using (a−b)2=a2−2ab+b2 )

=40000+5600+196= 40000 + 5600 + 196

=45796= 45796

Hence,

2142=45796214^2 = 45796 Ans

(v) 110421104^2

Now, using identity

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

We have,

1104=1100+41104 = 1100 + 4

So,

a=1100, b=4a = 1100,\ b = 4

11042=(1100+4)21104^2 = (1100 + 4)^2

=(1100)2+2×1100×4+(4)2= (1100)^2 + 2 \times 1100 \times 4 + (4)^2 ((a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2)

=1210000+8800+16= 1210000 + 8800 + 16

=1218816= 1218816

Hence,

11042=12188161104^2 = 1218816 Ans

(vi) 112021120^2

Now, using identity

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

We have,

1120=1100+201120 = 1100 + 20

So,

a=1100, b=20a = 1100,\ b = 20

11202=(1100+20)21120^2 = (1100 + 20)^2

=(1100)2+2×1100×20+(20)2= (1100)^2 + 2 \times 1100 \times 20 + (20)^2

=1210000+44000+400= 1210000 + 44000 + 400

=1254400= 1254400

Hence,

11202=12544001120^2 = 1254400 Ans

Q2. Factor using suitable identities

(i) 16y2−24y+916y^2 – 24y + 9

Now, using identity

(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,

16y2=(4y)216y^2 = (4y)^2

9=(3)29 = (3)^2

=(4y)2−2×4y×3+(3)2= (4y)^2 – 2 \times 4y \times 3 + (3)^2

So,
a=4y, b=3a = 4y,\ b = 3

Hence,

16y2−24y+9=(4y−3)216y^2 – 24y + 9 = (4y – 3)^2 Ans

(ii) 94s2+6st+4t2\frac{9}{4}s^2 + 6st + 4t^2

Now, using identity

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

We have,

94s2=(32s)2\frac{9}{4}s^2 = \left(\frac{3}{2}s\right)^2

4t2=(2t)24t^2 = (2t)^2

=(32s)2+2×32s×2t+(2t)2= \left(\frac{3}{2}s\right)^2 + 2 \times \frac{3}{2}s \times 2t + (2t)^2

So,

a=32s, b=2ta = \frac{3}{2}s,\ b = 2t

Hence,

94s2+6st+4t2=(32s+2t)2\frac{9}{4}s^2 + 6st + 4t^2 = \left(\frac{3}{2}s + 2t\right)^2 Ans

(iii) m2+mk+k2+3nk+2mn+9n2m^2 + mk + k^2 + 3nk + 2mn + 9n^2

Now, using identity

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

We have,

a=m, b=k, c=3na = m,\ b = k,\ c = 3n

Hence,

=(m+k+3n)2= (m + k + 3n)^2 Ans

(iv) p216−2+16p2\frac{p^2}{16} – 2 + \frac{16}{p^2}

Now, using identity

(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,

p216=(p4)2\frac{p^2}{16} = \left(\frac{p}{4}\right)^2

16p2=(4p)2\frac{16}{p^2} = \left(\frac{4}{p}\right)^2

=(p4)2−2×p4×4p+(4p)2= \left(\frac{p}{4}\right)^2 – 2 \times \frac{p}{4} \times \frac{4}{p} + \left(\frac{4}{p}\right)^2

So,
a=p4, b=4pa = \frac{p}{4},\ b = \frac{4}{p}

Hence,

p216−2+16p2=(p4−4p)2\frac{p^2}{16} – 2 + \frac{16}{p^2} = \left(\frac{p}{4} – \frac{4}{p}\right)^2

(v) 9a2+4b2+c2−12ab+6ac−4bc9a^2 + 4b^2 + c^2 – 12ab + 6ac – 4bc

Now, using identity

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

We have,

=(3a)2+(2b)2+c2−2×3a×2b+2×3a×c−2×2b×c= (3a)^2 + (2b)^2 + c^2 – 2 \times 3a \times 2b + 2 \times 3a \times c – 2 \times 2b \times c

So,
a=3a, b=−2b, c=ca = 3a,\ b = -2b,\ c = c

Hence,

=(3a−2b+c)2= (3a – 2b + c)^2 Ans

Q3. Expand the following using the identity


(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(i) (p+3q+7r)2(p + 3q + 7r)^2

(ii) (3x−2y+4z)2

(i) (p+3q+7r)2(p + 3q + 7r)^2

Now, using identity

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

We have,

=p2+9q2+49r2+6pq+42qr+14pr= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr Ans

(ii) (3x−2y+4z)2(3x – 2y + 4z)^2

Now, using identity

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca(a+b+c)

We have,

=9x2+4y2+16z2−12xy−16yz+24xz= 9x^2 + 4y^2 + 16z^2 – 12xy – 16yz + 24xz Ans

Q4. Is this an identity?

(a+b−c)2+(a−b+c)2(a + b – c)^2 + (a – b + c)^2

Expand:

=a2+b2+c2+2ab−2ac−2bc= a^2 + b^2 + c^2 + 2ab – 2ac – 2bc

+a2+b2+c2−2ab+2ac−2bc+ a^2 + b^2 + c^2 – 2ab + 2ac – 2bc

=2a2+2b2+2c2−4bc= 2a^2 + 2b^2 + 2c^2 – 4bc

Right side:

2a2+2b2+2c22a^2 + 2b^2 + 2c^2

Not equal

Hence, not an identity

Ans: Not an identity

Exercise Set 4.4

  1. Fill in the blanks to complete the following identities:
    (i) s2 – 11s + 24 = (________) (________)
    (ii) (________) (x + 1) = (3x2 – 4x –7)
    (iii) 10x2 – 11x – 6 = (2x ________) ( + 2)
    (iv) 6x2 + 7x + 2 = (__________    ) (_________)

Solution:

(i) s2−11s+24s^2 – 11s + 24

Now, using identity(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

We have,

a+b=−11,ab=24a + b = -11,\quad ab = 24

Numbers are: −3-3 and −8-8

Hence,s2−11s+24=(s−3)(s−8)s^2 – 11s + 24 = (s – 3)(s – 8)

Ans: (s−3)(s−8)(s – 3)(s – 8)

(ii) (___)(x+1)=3x2−4x−7(\_\_\_)(x + 1) = 3x^2 – 4x – 7(___)

Now, using identity(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

We factor RHS:3x2−4x−73x^2 – 4x – 7

Split middle term:=3x2−7x+3x−7= 3x^2 – 7x + 3x – 7=(3x2−7x)+(3x−7)= (3x^2 – 7x) + (3x – 7)=x(3x−7)+1(3x−7)= x(3x – 7) + 1(3x – 7)=(x+1)(3x−7)= (x + 1)(3x – 7)

Hence,(___)(x+1)=(3x−7)(x+1)(\_\_\_)(x + 1) = (3x – 7)(x + 1)

Ans: 3x−73x – 7

(iii) 10x2−11x−6=(2x−__)(__+2)10x^2 – 11x – 6 = (2x – \_\_)(\_\_ + 2)

Now, using identity(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

We factor:10x2−11x−610x^2 – 11x – 6

Split middle term:=10x2−15x+4x−6= 10x^2 – 15x + 4x – 6=(10x2−15x)+(4x−6)= (10x^2 – 15x) + (4x – 6)=5x(2x−3)+2(2x−3)= 5x(2x – 3) + 2(2x – 3)=(5x+2)(2x−3)= (5x + 2)(2x – 3)

Hence,=(2x−3)(5x+2)= (2x – 3)(5x + 2)

Ans: 3, 5x3,\ 5x

(iv) 6x2+7x+2=(__)(__)6x^2 + 7x + 2 = (\_\_)(\_\_)

Now, using identity(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

We factor:6x2+7x+26x^2 + 7x + 2

Split middle term:=6x2+3x+4x+2= 6x^2 + 3x + 4x + 2=(6x2+3x)+(4x+2)= (6x^2 + 3x) + (4x + 2)=3x(2x+1)+2(2x+1)= 3x(2x + 1) + 2(2x + 1)=(3x+2)(2x+1)= (3x + 2)(2x + 1)

Ans: (3x+2)(2x+1)(3x + 2)(2x + 1)

2. Select and use the identity that will help you to find the following
products without multiplying directly:

(i) (41)2 (ii) (27)2
(iii) (23 × 17) (iv) (135)2
(v) (97)2 (vi) (18 × 29)
(vii) (34 × 43) (viii) (205)2

Solutions :

(i) 41241^2

Now, using identity(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

We have,

41=40+141 = 40 + 1412=(40+1)2=402+2×40×1+1241^2 = (40 + 1)^2 = 40^2 + 2 \times 40 \times 1 + 1^2=1600+80+1=1681= 1600 + 80 + 1 = 1681

Ans: 1681

(ii) 27227^2

Now, using identity(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,

27=30−327 = 30 – 3272=(30−3)2=900−180+9=72927^2 = (30 – 3)^2 = 900 – 180 + 9 = 729

Ans: 729

(iii) 23×1723 \times 17

Now, using identity(a+b)(a−b)=a2−b2(a + b)(a – b) = a^2 – b^2

We have,

23=20+3,17=20−323 = 20 + 3,\quad 17 = 20 – 3
23×17=(20+3)(20−3)=202−32=400−9=39123 \times 17 = (20 + 3)(20 – 3) = 20^2 – 3^2 = 400 – 9 = 391

Ans: 391

(iv) 1352135^2

Now, using identity(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

We have,

135=100+35135 = 100 + 351352=1002+2×100×35+352135^2 = 100^2 + 2 \times 100 \times 35 + 35^2=10000+7000+1225=18225= 10000 + 7000 + 1225 = 18225

Ans: 18225

(i) 9a2+b2+4c2−6ab+12ac−4bc9a^2 + b^2 + 4c^2 – 6ab + 12ac – 4bc

Now, using identity(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

We write:=(3a)2+b2+(2c)2−2(3a)(b)+2(3a)(2c)−2(b)(2c)= (3a)^2 + b^2 + (2c)^2 – 2(3a)(b) + 2(3a)(2c) – 2(b)(2c)

So,

a=3a, b=−b, c=2ca = 3a,\ b = -b,\ c = 2c

Hence,=(3a−b+2c)2= (3a – b + 2c)^2

Ans: (3a−b+2c)2(3a – b + 2c)^2

(ii) 16s2+25t2−40st16s^2 + 25t^2 – 40st

Now, using identity(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,16s2=(4s)2,25t2=(5t)216s^2 = (4s)^2,\quad 25t^2 = (5t)^2=(4s)2−2(4s)(5t)+(5t)2= (4s)^2 – 2(4s)(5t) + (5t)^2

Hence,=(4s−5t)2= (4s – 5t)^2

Ans: (4s−5t)2(4s – 5t)^2

(iii) r2−r−42r^2 – r – 42

Now, using identity(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

We have,

a+b=−1,ab=−42a + b = -1,\quad ab = -42

Numbers are: 66 and −7-7

Hence,r2−r−42=(r+6)(r−7)r^2 – r – 42 = (r + 6)(r – 7)

Ans: (r+6)(r−7)(r + 6)(r – 7)

Exercise Set 4.5

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

(i)3p2−3pq−18q2p2+3pq−10q2(i) \frac{3p^2 – 3pq – 18q^2}{p^2 + 3pq – 10q^2}
(ii)n3−3n2m+3nm2−m35m2−10mn+5n2(ii) \frac{n^3 – 3n^2m + 3nm^2 – m^3}{5m^2 – 10mn + 5n^2}
(iii)w3−v3+x3+3wvxw2+v2+x2−2wv−2vx+2wx(iii) \frac{w^3 – v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 – 2wv – 2vx + 2wx}
(iv)4y2−20yz+25z2(25z2−4y2)(iv) \frac{4y^2 – 20yz + 25z^2}{(25z^2 – 4y^2)}
(v)(x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)(v) \frac{(x^2 + x – 6)(x^2 – 7x + 12)}{(x^2 – 6x + 8)(x^2 – 9)}
(vi)p4−16p2−4p+4(vi) \frac{p^4 – 16}{p^2 – 4p + 4}

Solution:

(i) 3p2−3pq−18q2p2+pq−10q2\dfrac{3p^2 – 3pq – 18q^2}{p^2 + pq – 10q^2}

Now, using identity(x+a)(x+b)=x2+(a+b)x+ab(x + a)(x + b) = x^2 + (a + b)x + ab

We factor numerator:3p2−3pq−18q2=3(p2−pq−6q2)3p^2 – 3pq – 18q^2 = 3(p^2 – pq – 6q^2)=3(p2−3pq+2pq−6q2)= 3(p^2 – 3pq + 2pq – 6q^2)=3[(p2−3pq)+(2pq−6q2)]= 3[(p^2 – 3pq) + (2pq – 6q^2)]=3[p(p−3q)+2q(p−3q)]= 3[p(p – 3q) + 2q(p – 3q)]=3(p−3q)(p+2q)= 3(p – 3q)(p + 2q)

Denominator:p2+pq−10q2=(p+5q)(p−2q)p^2 + pq – 10q^2 = (p + 5q)(p – 2q)

Hence,3(p−3q)(p+2q)(p+5q)(p−2q)\dfrac{3(p – 3q)(p + 2q)}{(p + 5q)(p – 2q)}

Ans: 3(p−3q)(p+2q)(p+5q)(p−2q)\dfrac{3(p – 3q)(p + 2q)}{(p + 5q)(p – 2q)}

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

Now, using identity(a−b)3=a3−3a2b+3ab2−b3(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3

We factor numerator:=(n−m)3= (n – m)^3

Denominator:5m2−10mn+5n2=5(m−n)25m^2 – 10mn + 5n^2 = 5(m – n)^2=5(n−m)2= 5(n – m)^2

Hence,(n−m)35(n−m)2=n−m5\dfrac{(n – m)^3}{5(n – m)^2} = \dfrac{n – m}{5}

Ans: n−m5\dfrac{n – m}{5}

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

Now, using identitya3+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)

We rewrite numerator:=w3+v3+x3−3wvx= w^3 + v^3 + x^3 – 3wvx=(w+v+x)(w2+v2+x2−wv−vx−wx)= (w + v + x)(w^2 + v^2 + x^2 – wv – vx – wx)


Hence,(w+v+x)(w2+v2+x2−wv−vx−wx)w2+v2+x2−wv−vx−wx\dfrac{(w + v + x)(w^2 + v^2 + x^2 – wv – vx – wx)}{w^2 + v^2 + x^2 – wv – vx – wx}

=w+v+x= w + v + x

Ans: w+v+xw + v + x

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

Now, using identity(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

Numerator:=(2y−5z)2= (2y – 5z)^2

Denominator:2z2−4y2=2(z2−2y2)2z^2 – 4y^2 = 2(z^2 – 2y^2)=2(z−2y)(z+2y)= 2(z – \sqrt{2}y)(z + \sqrt{2}y)

Ans: (2y−5z)22(z2−2y2)\dfrac{(2y – 5z)^2}{2(z^2 – 2y^2)}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)}

Now, using identity

(x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b) = x^2 + (a+b)x + ab

We factor:x2+x−6=(x+3)(x−2)x^2 + x – 6 = (x + 3)(x – 2)x2−7x+12=(x−3)(x−4)x^2 – 7x + 12 = (x – 3)(x – 4)x2−6x+8=(x−2)(x−4)x^2 – 6x + 8 = (x – 2)(x – 4)x2−9=(x−3)(x+3)x^2 – 9 = (x – 3)(x + 3)

Hence,(x+3)(x−2)(x−3)(x−4)(x−2)(x−4)(x−3)(x+3)\dfrac{(x + 3)(x – 2)(x – 3)(x – 4)}{(x – 2)(x – 4)(x – 3)(x + 3)}

Cancel common terms:=1= 1

Ans: 11

(vi) p4−16p2−4p+4\dfrac{p^4 – 16}{p^2 – 4p + 4}

Now, using identitya2−b2=(a+b)(a−b)a^2 – b^2 = (a + b)(a – b)p4−16=(p2)2−42=(p2−4)(p2+4)p^4 – 16 = (p^2)^2 – 4^2 = (p^2 – 4)(p^2 + 4)=(p−2)(p+2)(p2+4)= (p – 2)(p + 2)(p^2 + 4)

Denominator:p2−4p+4=(p−2)2p^2 – 4p + 4 = (p – 2)^2

Hence,(p−2)(p+2)(p2+4)(p−2)2\dfrac{(p – 2)(p + 2)(p^2 + 4)}{(p – 2)^2}=(p+2)(p2+4)p−2= \dfrac{(p + 2)(p^2 + 4)}{p – 2}

Ans: (p+2)(p2+4)p−2\dfrac{(p + 2)(p^2 + 4)}{p – 2}

End-of-Chapter Exercises

1. Use suitable identities to find the following products:

(i) (−3x+4)2(ii) (2s+7)(2s−7)(iii) (p2+12)(p2−12)(iv) (2n+7)(2n−7)(v) (s−2t)(s2+2st+4t2)(vi) (12r−4r)2(vii) (−3m+4k−l)2(viii) (x−13y)3(ix) (72k−23m)3 \begin{matrix} \text{(i) } (-3x + 4)^2 & \text{(ii) } (2s + 7)(2s – 7) \\ \\ \text{(iii) } (p^2 + \frac{1}{2})(p^2 – \frac{1}{2}) & \text{(iv) } (2n + 7)(2n – 7) \\ \\ \text{(v) } (s – 2t)(s^2 + 2st + 4t^2) & \text{(vi) } (\frac{1}{2r} – 4r)^2 \\ \\ \text{(vii) } (-3m + 4k – l)^2 & \text{(viii) } (x – \frac{1}{3}y)^3 \\ \\ \text{(ix) } (\frac{7}{2}k – \frac{2}{3}m)^3 & \end{matrix}

(i) (−3x+4)2(-3x + 4)^2

Now, using identity

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

We have,

a=−3x, b=4a = -3x,\ b = 4(−3x+4)2=(−3x)2+2(−3x)(4)+42(-3x + 4)^2 = (-3x)^2 + 2(-3x)(4) + 4^2=9x2−24x+16= 9x^2 – 24x + 16

Hence,(−3x+4)2=9x2−24x+16 Ans(-3x + 4)^2 = 9x^2 – 24x + 16 \ \text{Ans}

(ii) (2s+7)(2s−7)(2s + 7)(2s – 7)

Now, using identity(a+b)(a−b)=a2−b2(a + b)(a – b) = a^2 – b^2

We have,

a=2s, b=7a = 2s,\ b = 7=(2s)2−72=4s2−49= (2s)^2 – 7^2 = 4s^2 – 49

Hence,(2s+7)(2s−7)=4s2−49 Ans(2s + 7)(2s – 7) = 4s^2 – 49 \ \text{Ans}

(iii) (p2+12)(p2−12)\left(p^2 + \frac{1}{2}\right)\left(p^2 – \frac{1}{2}\right)

Now, using identity(a+b)(a−b)=a2−b2(a + b)(a – b) = a^2 – b^2

We have,

a=p2, b=12a = p^2,\ b = \frac{1}{2}=(p2)2−(12)2= (p^2)^2 – \left(\frac{1}{2}\right)^2=p4−14= p^4 – \frac{1}{4}

Hence,=p4−14 Ans= p^4 – \frac{1}{4} \ \text{Ans}

(iv) (2n+7)(2n−7)(2n + 7)(2n – 7)

Now, using identity(a+b)(a−b)=a2−b2(a + b)(a – b) = a^2 – b^2

We have,

a=2n, b=7a = 2n,\ b = 7=(2n)2−72=4n2−49= (2n)^2 – 7^2 = 4n^2 – 49

Hence,=4n2−49 Ans= 4n^2 – 49 \ \text{Ans}

(v) (s−2t)(s2+2st+4t2)(s – 2t)(s^2 + 2st + 4t^2)

Now, using identity(a−b)(a2+ab+b2)=a3−b3(a – b)(a^2 + ab + b^2) = a^3 – b^3

We have,

a=s, b=2ta = s,\ b = 2t=s3−(2t)3=s3−8t3= s^3 – (2t)^3 = s^3 – 8t^3

Hence,=s3−8t3 Ans= s^3 – 8t^3 \ \text{Ans}

(vi) (12r−4r)2\left(\frac{1}{2r} – 4r\right)^2

Now, using identity(a−b)2=a2−2ab+b2(a – b)^2 = a^2 – 2ab + b^2

We have,

a=12r, b=4ra = \frac{1}{2r},\ b = 4r=(12r)2−2(12r)(4r)+(4r)2= \left(\frac{1}{2r}\right)^2 – 2\left(\frac{1}{2r}\right)(4r) + (4r)^2=14r2−4+16r2= \frac{1}{4r^2} – 4 + 16r^2

Hence,=16r2−4+14r2 Ans= 16r^2 – 4 + \frac{1}{4r^2} \ \text{Ans}

(vii) (−3m+4k−l)2(-3m + 4k – l)^2

Now, using identity(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

We have,

a=−3m, b=4k, c=−la = -3m,\ b = 4k,\ c = -l
=(−3m)2+(4k)2+(−l)2+2(−3m)(4k)+2(4k)(−l)+2(−3m)(−l)= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-3m)(-l)=9m2+16k2+l2−24mk−8kl+6ml= 9m^2 + 16k^2 + l^2 – 24mk – 8kl + 6ml

Hence,=9m2+16k2+l2−24mk−8kl+6ml Ans= 9m^2 + 16k^2 + l^2 – 24mk – 8kl + 6ml \ \text{Ans}

(viii) (x−13y)3\left(x – \frac{1}{3}y\right)^3

Now, using identity(a−b)3=a3−3a2b+3ab2−b3(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3

We have,

a=x, b=13ya = x,\ b = \frac{1}{3}y=x3−3x2(13y)+3x(13y)2−(13y)3= x^3 – 3x^2\left(\frac{1}{3}y\right) + 3x\left(\frac{1}{3}y\right)^2 – \left(\frac{1}{3}y\right)^3=x3−x2y+13xy2−127y3= x^3 – x^2y + \frac{1}{3}xy^2 – \frac{1}{27}y^3

Hence,=x3−x2y+13xy2−127y3 Ans= x^3 – x^2y + \frac{1}{3}xy^2 – \frac{1}{27}y^3 \ \text{Ans}

(ix) (72k−23m)3\left(\frac{7}{2}k – \frac{2}{3}m\right)^3

Now, using identity(a−b)3=a3−3a2b+3ab2−b3(a – b)^3 = a^3 – 3a^2b + 3ab^2 – b^3

We have,

a=72k, b=23ma = \frac{7}{2}k,\ b = \frac{2}{3}m=(72k)3−3(72k)2(23m)+3(72k)(23m)2−(23m)3= \left(\frac{7}{2}k\right)^3 – 3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2 – \left(\frac{2}{3}m\right)^3

=3438k3−492k2m+143km2−827m3= \frac{343}{8}k^3 – \frac{49}{2}k^2m + \frac{14}{3}km^2 – \frac{8}{27}m^3

Hence,=3438k3−492k2m+143km2−827m3 Ans= \frac{343}{8}k^3 – \frac{49}{2}k^2m + \frac{14}{3}km^2 – \frac{8}{27}m^3 \ \text{Ans}

  1. Find the values using suitable identities:
    (i) 17 × 21 (ii) 104 × 96
    (iii) 24 × 16 (iv) 1473
    (v) 1993 (vi) 1273
    (vii) (–107)3 (viii) (–299)3

Solution:

(i) 17×2117 \times 21

Using identity

(a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2

We have,

17×21=(19−2)(19+2)17 \times 21 = (19 – 2)(19 + 2)
=192−22= 19^2 – 2^2=361−4= 361 – 4
=357= 357

357 Ans

(ii) 104×96104 \times 96

Using identity

(a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2

We have,

104×96=(100+4)(100−4)104 \times 96 = (100 + 4)(100 – 4)=1002−42= 100^2 – 4^2=10000−16= 10000 – 16=9984= 9984

Ans: 9984

(iii) 24×1624 \times 16

Using identity(a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2

We have,

24×16=(20+4)(20−4)24 \times 16 = (20 + 4)(20 – 4)=202−42= 20^2 – 4^2=400−16= 400 – 16=384= 384

Ans: 384

(iv) 1473147^3

Using identity

(a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3

We have,

147=150−3147 = 150 – 3(150−3)3(150 – 3)^3=1503−3⋅1502⋅3+3⋅150⋅32−33= 150^3 – 3 \cdot 150^2 \cdot 3 + 3 \cdot 150 \cdot 3^2 – 3^3=3375000−202500+4050−27= 3375000 – 202500 + 4050 – 27=3176523= 3176523

Ans: 3176523

(v) 1993199^3

Using identity(a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3

We have,

199=200−1199 = 200 – 1(200−1)3(200 – 1)^3=2003−3⋅2002+3⋅200−1= 200^3 – 3 \cdot 200^2 + 3 \cdot 200 – 1=8000000−120000+600−1= 8000000 – 120000 + 600 – 1=7880599= 7880599

Ans: 7880599

(vi) 1273127^3

Using identity

(a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3

We have,

127=100+27127 = 100 + 27(100+27)3(100 + 27)^3=1003+3⋅1002⋅27+3⋅100⋅272+273= 100^3 + 3 \cdot 100^2 \cdot 27 + 3 \cdot 100 \cdot 27^2 + 27^3=1000000+810000+218700+19683= 1000000 + 810000 + 218700 + 19683=2048383= 2048383

Ans: 2048383

(vii) (−107)3(-107)^3

Using identity(−a)3=−a3(-a)^3 = -a^3

We have,(−107)3=−(1073)(-107)^3 = -(107^3)107=100+7107 = 100 + 7=−(100+7)3= -(100 + 7)^3=−(1000000+210000+14700+343)= -(1000000 + 210000 + 14700 + 343)=−1225043= -1225043

Ans: −1225043-1225043

(viii) (−299)3(-299)^3

Using identity(a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3

We have,299=300−1299 = 300 – 1(−299)3=−(300−1)3(-299)^3 = -(300 – 1)^3=−(3003−3⋅3002+3⋅300−1)= -(300^3 – 3 \cdot 300^2 + 3 \cdot 300 – 1)=−(27000000−270000+900−1)= -(27000000 – 270000 + 900 – 1)=−26730899= -26730899

Ans: −26730899-26730899

Q3. Factor the following algebraic expressions:

Solutions:

(i) 4y2+1+116y24y^2 + 1 + \frac{1}{16y^2}

Using identity

a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2

We have,4y2=(2y)24y^2 = (2y)^2116y2=(14y)2\frac{1}{16y^2} = \left(\frac{1}{4y}\right)^21=2(2y)(14y)1 = 2(2y)\left(\frac{1}{4y}\right)

Hence,4y2+1+116y2=(2y+14y)24y^2 + 1 + \frac{1}{16y^2} = (2y + \frac{1}{4y})^2

Ans: (2y+14y)2(2y + \frac{1}{4y})^2

(ii) 9m2−125n29m^2 – \frac{1}{25n^2}

Using identitya2−b2=(a−b)(a+b)a^2 – b^2 = (a-b)(a+b)9m2=(3m)29m^2 = (3m)^2125n2=(15n)2\frac{1}{25n^2} = \left(\frac{1}{5n}\right)^2

Hence,=(3m−15n)(3m+15n)= (3m – \frac{1}{5n})(3m + \frac{1}{5n})

Ans: (3m−15n)(3m+15n)(3m – \frac{1}{5n})(3m + \frac{1}{5n})

(iii) 27b3−164b327b^3 – \frac{1}{64b^3}

Using identitya3−b3=(a−b)(a2+ab+b2)a^3 – b^3 = (a-b)(a^2 + ab + b^2)27b3=(3b)327b^3 = (3b)^3164b3=(14b)3\frac{1}{64b^3} = \left(\frac{1}{4b}\right)^3

Hence,=(3b−14b)(9b2+34+116b2)= (3b – \frac{1}{4b})(9b^2 + \frac{3}{4} + \frac{1}{16b^2})

(iv) x2+5x6+16x^2 + \frac{5x}{6} + \frac{1}{6}

Using identityx2+(a+b)x+ab=(x+a)(x+b)x^2 + (a+b)x + ab = (x+a)(x+b)

We have,a+b=56,ab=16a+b = \frac{5}{6}, \quad ab = \frac{1}{6}

Numbers:12,13\frac{1}{2}, \frac{1}{3}

Hence,=(x+12)(x+13)= (x + \frac{1}{2})(x + \frac{1}{3})

(v) 27u3−1125−27u25+9u2527u^3 – \frac{1}{125} – \frac{27u^2}{5} + \frac{9u}{25}

Using identity,

(a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^327u3=(3u)327u^3 = (3u)^31125=(15)3\frac{1}{125} = \left(\frac{1}{5}\right)^3

Hence,=(3u−15)3= (3u – \frac{1}{5})^3

(vi) 64y3+1125z364y^3 + \frac{1}{125}z^3

Using identitya3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 – ab + b^2)64y3=(4y)364y^3 = (4y)^31125z3=(z5)3\frac{1}{125}z^3 = \left(\frac{z}{5}\right)^3

Hence,=(4y+z5)(16y2−4yz5+z225)= (4y + \frac{z}{5})(16y^2 – \frac{4yz}{5} + \frac{z^2}{25})

(vii) p3+27q3+r3−9pqrp^3 + 27q^3 + r^3 – 9pqr

Using identitya3+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)a=p, b=3q, c=ra=p,\ b=3q,\ c=r

Hence,=(p+3q+r)(p2+9q2+r2−3pq−3qr−pr)= (p+3q+r)(p^2+9q^2+r^2-3pq-3qr-pr)

(viii) 9m2−12m+49m^2 – 12m + 4

Using identitya2−2ab+b2=(a−b)2a^2 – 2ab + b^2 = (a-b)^29m2=(3m)2,4=229m^2 = (3m)^2,\quad 4 = 2^2

Hence,=(3m−2)2= (3m – 2)^2

(ix)

9x3−83y3+z33+6xyz9x^3 – \frac{8}{3}y^3 + \frac{z^3}{3} + 6xyz

Using 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)

We match:9x3=(3x)39x^3 = (3x)^3−83y3=(−23y)3-\frac{8}{3}y^3 = \left(-\frac{2}{3}y\right)^3z33=(z3)3\frac{z^3}{3} = \left(\frac{z}{3}\right)^3

So,a=3x,b=−23y,c=z3a = 3x,\quad b = -\frac{2}{3}y,\quad c = \frac{z}{3}

Check middle term:−3abc=−3⋅3x⋅(−23y)⋅z3=6xyz-3abc = -3 \cdot 3x \cdot \left(-\frac{2}{3}y\right) \cdot \frac{z}{3} = 6xyz

Hence,=(3x−23y+z3)3= (3x – \frac{2}{3}y + \frac{z}{3})^3

Ans: (3x−23y+z3)3(3x – \frac{2}{3}y + \frac{z}{3})^3

(x)

4x2+9y2+36z2+12xz+36yz+24xy4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy

Using identity

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

We match:4x2=(2x)24x^2 = (2x)^29y2=(3y)29y^2 = (3y)^236z2=(6z)236z^2 = (6z)^2

So,a=2x,b=3y,c=6za = 2x,\quad b = 3y,\quad c = 6z

Check middle terms:2ab=2(2x)(3y)=12xy2ab = 2(2x)(3y) = 12xy2bc=2(3y)(6z)=36yz2bc = 2(3y)(6z) = 36yz2ca=2(2x)(6z)=24xz2ca = 2(2x)(6z) = 24xz

All terms match.

Hence,=(2x+3y+6z)2= (2x + 3y + 6z)^2

Ans: (2x+3y+6z)2(2x + 3y + 6z)^2

(xi)

27u3−1216−9u22+u427u^3 – \frac{1}{216} – \frac{9u^2}{2} + \frac{u}{4}

Using identity

(a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3

We match:27u3=(3u)327u^3 = (3u)^31216=(16)3\frac{1}{216} = \left(\frac{1}{6}\right)^3

So,a=3u,b=16a = 3u,\quad b = \frac{1}{6}

Now check middle terms:−3a2b=−3(3u)2(16)=−3⋅9u2⋅16=−27u26=−9u22-3a^2b = -3(3u)^2\left(\frac{1}{6}\right) = -3 \cdot 9u^2 \cdot \frac{1}{6} = -\frac{27u^2}{6} = -\frac{9u^2}{2}3ab2=3(3u)(16)2=3⋅3u⋅136=9u36=u43ab^2 = 3(3u)\left(\frac{1}{6}\right)^2 = 3 \cdot 3u \cdot \frac{1}{36} = \frac{9u}{36} = \frac{u}{4}

All terms match.

Hence,=(3u−16)3= (3u – \frac{1}{6})^3

Ans: (3u−16)3(3u – \frac{1}{6})^3

Q4. Simplify the following

(i)

4x2+4x+14x2−1\frac{4x^2 + 4x + 1}{4x^2 – 1}

Step 1: Factor the numerator

Using identity

a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^24x2+4x+1=(2x)2+2(2x)(1)+124x^2 + 4x + 1 = (2x)^2 + 2(2x)(1) + 1^2=(2x+1)2= (2x + 1)^2

Step 2: Factor the denominator

Using identity

a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b)

=(2x−1)(2x+1)= (2x – 1)(2x + 1)

Step 3: Simplify

(2x+1)2(2x−1)(2x+1)\frac{(2x + 1)^2}{(2x – 1)(2x + 1)}

Cancel common factor (2x+1)(2x+1)=2x+12x−1= \frac{2x + 1}{2x – 1}

Ans:2x+12x−1\frac{2x + 1}{2x – 1}

(ii)

9(3a3−24b3)9a2−36b2\frac{9(3a^3 – 24b^3)}{9a^2 – 36b^2}

Step 1: Factor the numerator

9(3a3−24b3)=9⋅3(a3−8b3)9(3a^3 – 24b^3) = 9 \cdot 3(a^3 – 8b^3)=27(a3−(2b)3)= 27(a^3 – (2b)^3)

Using identity

a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)=27(a−2b)(a2+2ab+4b2)= 27(a – 2b)(a^2 + 2ab + 4b^2)

Step 2: Factor the denominator

9a2−36b2=9(a2−4b2)9a^2 – 36b^2 = 9(a^2 – 4b^2)

Using identitya2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b)=9(a−2b)(a+2b)= 9(a – 2b)(a + 2b)

Step 3: Simplify

27(a−2b)(a2+2ab+4b2)9(a−2b)(a+2b)\frac{27(a – 2b)(a^2 + 2ab + 4b^2)}{9(a – 2b)(a + 2b)}

Cancel common factors:

  • a−2ba – 2b
  • 27/9=327/9 = 3

=3(a2+2ab+4b2)a+2b= \frac{3(a^2 + 2ab + 4b^2)}{a + 2b}

Ans:3(a2+2ab+4b2)a+2b\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}

(iii)

s3+125t3s2−2st−35t2\frac{s^3 + 125t^3}{s^2 – 2st – 35t^2}

Step 1: Factor the numerator

s3+125t3=s3+(5t)3s^3 + 125t^3 = s^3 + (5t)^3

Using identity

a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)=(s+5t)(s2−5st+25t2)= (s + 5t)(s^2 – 5st + 25t^2)

Step 2: Factor the denominator

s2−2st−35t2s^2 – 2st – 35t^2

Split middle term:=s2−7st+5st−35t2= s^2 – 7st + 5st – 35t^2=s(s−7t)+5t(s−7t)= s(s – 7t) + 5t(s – 7t)=(s+5t)(s−7t)= (s + 5t)(s – 7t)

Step 3: Simplify

(s+5t)(s2−5st+25t2)(s+5t)(s−7t)\frac{(s + 5t)(s^2 – 5st + 25t^2)}{(s + 5t)(s – 7t)}

Cancel common factor (s+5t)(s+5t)=s2−5st+25t2s−7t= \frac{s^2 – 5st + 25t^2}{s – 7t}

Ans:s2−5st+25t2s−7t\frac{s^2 – 5st + 25t^2}{s – 7t}

  1. Find possible expressions for the length and breadth of each of
    The following rectangles whose areas are given by the following
    expressions in square units.
    (i) 25a2 – 30ab + 9b2 (ii) 36s2 – 49t2

Solutions:

(i) 25a2−30ab+9b225a^2 – 30ab + 9b^2

Step 1: Identify identity

a2−2ab+b2=(a−b)2a^2-2ab+b^2=(a-b)^2

Step 2: Match terms

25a2=(5a)225a^2 = (5a)^29b2=(3b)29b^2 = (3b)^2

Check middle term:

Step 3: Factorise

25a2−30ab+9b2=(5a−3b)225a^2 – 30ab + 9b^2 = (5a – 3b)^2

Area = Length × Breadth=(5a−3b)(5a−3b)= (5a – 3b)(5a – 3b)

Ans:
Length = 5a−3b5a – 3b5a−3b
Breadth = 5a−3b5a – 3b5a−3b

(ii) 36s2−49t236s^2 – 49t^2

Step 1: Identify identity

a2−b2=(a−b)(a+b)a^2 – b^2 = (a-b)(a+b)

Step 2: Match terms

36s2=(6s)236s^2 = (6s)^249t2=(7t)249t^2 = (7t)^2

Step 3: Factorise

36s2−49t2=(6s−7t)(6s+7t)36s^2 – 49t^2 = (6s – 7t)(6s + 7t)

Area = Length × Breadth

Ans:
Length = 6s−7t6s – 7t
Breadth = 6s+7t6s + 7t

  1. Find possible expressions for the length, breadth, and height of
    each of the following cuboids whose volumes are given by the
    following expressions in cubic units.
    (i) 6a2 – 24b2 (ii) 3ps2 – 15ps + 12p

Solutions:

(i) 6a2−24b26a^2 – 24b^26a2−24b2

Step 1: Take the common factor

=6(a2−4b2)= 6(a^2 – 4b^2)

Step 2: Identify identity

a2−b2=(a−b)(a+b)a^2 – b^2 = (a-b)(a+b)

Step 3: Apply identity

a2−4b2=(a−2b)(a+2b)a^2 – 4b^2 = (a – 2b)(a + 2b)

Step 4: Combine

6a2−24b2=6(a−2b)(a+2b)6a^2 – 24b^2 = 6(a – 2b)(a + 2b)

Step 5: Express as a product of 3 factors

6=2×36 = 2 \times 3=2×3×(a−2b)(a+2b)= 2 \times 3 \times (a – 2b)(a + 2b)

Step 6: Interpret as dimensions

Volume = length × breadth × height

Ans:
Length = 2
Breadth = 3
Height = (a−2b)(a+2b)(a – 2b)(a + 2b)

(ii) 3ps2−15ps+12p3ps^2 – 15ps + 12p

Step 1: Take the common factor

=3p(s2−5s+4)= 3p(s^2 – 5s + 4)

Step 2: Factor the quadratic

Using identityx2−(a+b)x+ab=(x−a)(x−b)x^2 – (a+b)x + ab = (x-a)(x-b)

We need numbers whose:a+b=5,ab=4a+b = 5,\quad ab = 4

Numbers: 1 and 4

Step 3: Factorise

s2−5s+4=(s−1)(s−4)s^2 – 5s + 4 = (s – 1)(s – 4)

Step 4: Combine

3ps2−15ps+12p=3p(s−1)(s−4)3ps^2 – 15ps + 12p = 3p(s – 1)(s – 4)

Volume = product of 3 dimensions

Ans:
Length = 3p3p
Breadth = (s−1)(s – 1)
Height = (s−4)(s – 4)

  1. The village playground is shaped as a square of side 40 metres.
    A path of width s metres is created around the playground for
    people to walk. Find an expression for the area of the path in
    terms of s.

Solution:

Side of playground = 4040
Width of path = ss

Step 1: Outer side

=40+2s= 40 + 2s

Step 2: Use identity

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

Step 3: Area of path

=(40+2s)2−402= (40 + 2s)^2 – 40^2=(1600+160s+4s2)−1600= (1600 + 160s + 4s^2) – 1600=4s2+160s= 4s^2 + 160s

Ans:4s2+160s4s^2 + 160s

  1. If a number plus its reciprocal equals 10/3, find the number.

Solution:

Given,x+1x=103x + \frac{1}{x} = \frac{10}{3}

Step 1: Multiply by xx

x2+1=103xx^2 + 1 = \frac{10}{3}x

Bring to one side

x2−103x+1=0x^2 – \frac{10}{3}x + 1 = 0

Multiply by 3:3x2−10x+3=03x^2 – 10x + 3 = 0

Factorise

3x2−10x+3=3x2−9x−x+33x^2 – 10x + 3 = 3x^2 – 9x – x + 3=3x(x−3)−1(x−3)= 3x(x – 3) -1(x – 3)=(3x−1)(x−3)= (3x – 1)(x – 3)

Solve by equating to zero

3x−1=0⇒x=133x – 1 = 0 \Rightarrow x = \frac{1}{3}x−3=0⇒x=3x – 3 = 0 \Rightarrow x = 3

Final Answer

x=3or13x = 3 \quad \text{or} \quad \frac{1}{3}

  1. A rectangular pool has an area 2x2 + 7x + 3 square hastas. If its width
    is 2x + 1 hastas, find its length. Hasta was a unit used to measure
    length.

Solution :

Given,
Area =2x2+7x+3= 2x^2 + 7x + 3
Width =2x+1= 2x + 1

Use the formula

Area=Length×Width\text{Area} = \text{Length} \times \text{Width}Length=AreaWidth\text{Length} = \frac{\text{Area}}{\text{Width}}

Substitute

Length=2x2+7x+32x+1\text{Length} = \frac{2x^2 + 7x + 3}{2x + 1} Factor the numerator

2x2+7x+3=2x2+6x+x+32x^2 + 7x + 3 = 2x^2 + 6x + x + 3=2x(x+3)+1(x+3)= 2x(x + 3) + 1(x + 3)=(2x+1)(x+3)= (2x + 1)(x + 3)

Simplify

Length=(2x+1)(x+3)2x+1\text{Length} = \frac{(2x + 1)(x + 3)}{2x + 1}=x+3= x + 3

Final Answer

Length=x+3 hastas\text{Length} = x + 3 \text{ hastas}

  1. If both x – 2 and x – 1/2 are factors of px2 + 5x + r, show that p = r.

Given:

(x−2)(x – 2)(x−2) and (x−12)(x – \tfrac{1}{2})px2+5x+rpx^2 + 5x + r

Use Factor Theorem

If (x−a)(x – a)(x−a) is a factor, then f(a)=0f(a) = 0

Put x=2x = 2

p(2)2+5(2)+r=0p(2)^2 + 5(2) + r = 04p+10+r=0(1)4p + 10 + r = 0 \quad \text{(1)}

Put x=12x = \tfrac{1}{2}

p(12)2+5(12)+r=0p\left(\tfrac{1}{2}\right)^2 + 5\left(\tfrac{1}{2}\right) + r = 0p4+52+r=0(2)\frac{p}{4} + \frac{5}{2} + r = 0 \quad \text{(2)}

Multiply (2) by 4

p+10+4r=0(3)p + 10 + 4r = 0 \quad \text{(3)}

Solve (1) and (3)

From (1):4p+r=−104p + r = -10

From (3):p+4r=−10p + 4r = -10

Subtract equations

(4p+r)−(p+4r)=0(4p + r) – (p + 4r) = 03p−3r=03p – 3r = 0p=rp = r

Final Result

p=rp = r

  1. If a + b + c = 5 and ab + bc + ca = 10, then prove that
    a3 + b3 + c3 –3abc = – 25.

Solution:3a2+4ab+43b23a^2 + 4ab + \frac{4}{3}b^2

Now, using identitya2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2

We have,3a2+4ab+43b2=13(9a2+12ab+4b2)3a^2 + 4ab + \frac{4}{3}b^2 = \frac{1}{3}(9a^2 + 12ab + 4b^2)=13[(3a)2+2×3a×2b+(2b)2]= \frac{1}{3}\left[(3a)^2 + 2 \times 3a \times 2b + (2b)^2\right]

So,a=3a,b=2ba = 3a,\quad b = 2b

Hence,3a2+4ab+43b2=13(3a+2b)2 Ans3a^2 + 4ab + \frac{4}{3}b^2 = \frac{1}{3}(3a + 2b)^2 \ \text{Ans}


12. By factoring the expression, check that n3−n is always divisible by 6 for all natural numbers n. Give reasons.

Solution :

We have n3−n

Factor:n3−n=n(n2−1)=n(n−1)(n+1)

For any natural number n, n−1,n,n+1are three consecutive integers.

  • At least one of them is even → divisible by 2.
  • Exactly one of them is divisible by 3.

So the product (n−1)n(n+1) is divisible by 2×3=6

Thus, n3−n is always divisible by 6 for all natural numbers n.

13. Find the value of:

(i) x3+y3−12xy+64, when x+y=−4
(ii) x3−8y3−36xy−216, when x=2y+6

Solutions:

(i) x3+y3−12xy+64, when x+y=−4

Use the identity:x3+y3=(x+y)3−3xy(x+y)

Substitute x+y=−4:x3+y3=(−4)3−3xy(−4)=−64+12xy

Then:x3+y3−12xy+64=(−64+12xy)−12xy+64=−64+12xy−12xy+64=0

(ii)

x3−8y3−36xy−216,when x=2y+6x^3 – 8y^3 – 36xy – 216,\quad \text{when } x = 2y + 6

Rewrite:8y3=(2y)3,216=638y^3 = (2y)^3,\quad 216 = 6^3

So expression becomes:=x3−(2y)3−36xy−63= x^3 – (2y)^3 – 36xy – 6^3

Use identity:a3−b3=(a−b)(a2+ab+b2)a^3 – b^3 = (a-b)(a^2 + ab + b^2)

Let x=2y+6x = 2y + 6

Then,x−2y=6x – 2y = 6

So,x3−(2y)3=(x−2y)(x2+2xy+4y2)x^3 – (2y)^3 = (x-2y)(x^2 + 2xy + 4y^2)=6(x2+2xy+4y2)= 6(x^2 + 2xy + 4y^2)

Now expression becomes:=6(x2+2xy+4y2)−36xy−216= 6(x^2 + 2xy + 4y^2) – 36xy – 216

Expand:=6x2+12xy+24y2−36xy−216= 6x^2 + 12xy + 24y^2 – 36xy – 216=6x2−24xy+24y2−216= 6x^2 – 24xy + 24y^2 – 216

Factor:=6(x2−4xy+4y2−36)= 6(x^2 – 4xy + 4y^2 – 36)=6[(x−2y)2−36]= 6[(x-2y)^2 – 36]

Since x−2y=6x – 2y = 6:=6(62−36)= 6(6^2 – 36)=6(36−36)=0 Ans= 6(36 – 36) = 0 \ \text{Ans}

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