WBBSE Solutions For Class 9 Maths Chapter 8 Factorisation

Class IX Maths Solutions WBBSE Chapter 8 Factorisation: Formulae of Factorisation

1. a² – b² = (a + b) (a – b)
2. a³+b³ = (a + b) (a² – ab + b²)
3. a³-b³ (a – b) (a² + ab + b²)
4. a³ + b + c³-3abc = (a+b+c) (a²+ b²+c²-ab-bc-ca)

Read and Learn More WBBSE Solutions For Class 9 Maths

Ganit Prakash Class 9 Solutions Chapter 8 Factorisation Exercise 8.1

 

Let us factorise the following polynomials

Question 1. x³-3x+2

Solution: x³-3x+2

\(\begin{aligned}
& =x^3-1-3 x+3 \\
& =(x)^3-(1)^3-3(x-1) \\
& =(x-1)\left(x^2+x+1\right)-3(x-1) \\
& =(x-1)\left(x^2+x+1-3\right) \\
& =(x-1)\left(x^2+x-2\right) \\
& =(x-1)\left(x^2+2 x-x-2\right) \\
& =(x-1)\{x(x+2)-1(x+2)\} \\
& =(x-1)(x+2)(x-1) \\
& =(x-1)^2(x+2)
\end{aligned}\)

 

Question 2. x³ + 2x + 3

Solution: x³ + 2x + 3

\(\begin{aligned}
& =x^3+1+2 x+2 \\
& =(x)^3+(1)^3+2(x+1) \\
& =(x+1)\left(x^2-x+1\right)+2(x+1) \\
& =(x+1)\left(x^2-x+1+2\right) \\
& =(x+1)\left(x^2-x+3\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Question 3.a³-12a-16

Solution: a³-12a-16

\(\begin{aligned}
& =a^3+8-12 a-24 \\
& =(a)^3+(2)^3-12(a+2) \\
& =(a+2)\left(a^2-2 a+4\right)-12(a+2) \\
& =(a+2)\left(a^2-2 a+4-12\right) \\
& =(a+2)\left(a^2-2 a-8\right) \\
& =(a+2)\left(a^2-4 a+2 a-8\right) \\
& =(a+2)\{a(a-4)+2(a-4)\} \\
& =(a+2)(a+2)(a-4) \\
& =(a+2)^2(a-4)
\end{aligned}\)

 

Question 4. x³– 6x + 4

Solution: x³– 6x + 4

\(\begin{aligned}
& =x^3-8-6 x+12 \\
& =(x)^3-(2)^3-6(x-2) \\
& =(x-2)\left(x^2+2 x+4\right)-6(x-2) \\
& =(x-2)\left(x^2+2 x+4-6\right) \\
& =(x-2)\left(x^2+2 x-2\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Question 5. x³-19x-30

Solution: x³-19x-30

\(\begin{aligned}
& =x^3+8-19 x-38 \\
& =(x)^3+(2)^3-19(x+2) \\
& =(x+2)\left(x^2-2 x+4\right)-19(x+2) \\
& =(x+2)\left(x^2-2 x+4-19\right) \\
& =(x+2)\left(x^2-2 x-15\right) \\
& =(x+2)\left(x^2-5 x+3 x-15\right) \\
& =(x+2)\{x(x-5)+3(x-5)\} \\
& =(x+2)(x+3)(x-5)
\end{aligned}\)

 

Question 6. 4a³-9a²+ 3a +2

Solution: 4a³-9a²+ 3a +2
If a =1, 4a³-9a²+ 3a +2=0
∴ (a-1) is a factor of 4a³-9a²+ 3a +2

\(\begin{aligned}
& =4 a^3-9 a^2+3 a+2 \\
& =4 a^3-4 a^2-5 a^2+5 a-2 a+2 \\
& =4 a^2(a-1)-5 a(a-1)-2(a-1) \\
& =(a-1)\left(4 a^2-5 a-2\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Question 7. x³-9x²+23x-15

Solution: If x = 1, x³-9x²+23x-15=0
∴ (x-1) is a factor of  x³-9x²+23x-15

\(\begin{aligned}
& =x^3-x^2-8 x^2+8 x+15 x-15 \\
& =x^2(x-1)-8 x(x-1)+15(x-1) \\
& =(x-1)\left(x^2-8 x+15\right) \\
& =(x-1)\left(x^2-5 x-3 x+15\right) \\
& =(x-1)\{x(x-5)-3(x-5)\} \\
& =(x-1)(x-3)(x-5)
\end{aligned}\)

 

Question 8. 5a³ +11a²+ 4a-2

Solution: If a=-1,5a³ +11a²+ 4a-2=0
∴ (a + 1) is a factor of 5a³ +11a²+ 4a-2

\(\begin{aligned}
& =5 a^3+5 a^2+6 a^2+6 a-2 a-2 \\
& =5 a^2(a+1)+6 a(a+1)-2(a+1) \\
& =(a+1)\left(5 a^2+6 a-2\right)
\end{aligned}\)

Class 9 Math Solution WBBSE In English

Question 9. 2x³– x² + 9x + 5

Solution: If x = -1/2, 2x³– x² + 9x + 5=0
∴(2x + 1) is a factor of 2x³– x² + 9x + 5

\(\begin{aligned}
& =2 x^3+x^2-2 x^2-x+10 x+5 \\
& =x^2(2 x+1)-x(2 x+1)+5(2 x+1) \\
& =(2 x+1)\left(x^2-x+5\right)
\end{aligned}\)

 

Question 10. 2y³-5y²-19y+ 42

Solution: Putting, y = 2 then 2y³-5y²-19y+ 42= 0
∴ (y-2) is a factor of 2y³-5y²-19y+ 42

\(\begin{aligned}
& =2 y^3-4 y^2-y^2+2 y-21 y+42 \\
& =2 y^2(y-2)-y(y-2)-21(y-2) \\
& =(y-2)\left(2 y^2-y-21\right) \\
& =(y-2)\left(2 y^2-7 y+6 y-21\right) \\
& =(y-2)\{y(2 y-7)+3(2 y-7)\} \\
& =(y-2)(y+3)(2 y-7)
\end{aligned}\)

Class 9 Math Solution WBBSE In English Chapter 8 Factorisation Exercise 8.2

Question 1. \(\frac{x^4}{16}-\frac{y^4}{81}\)

Solution: \(\frac{x^4}{16}-\frac{y^4}{81}\)

\(\begin{aligned}
& =\left(\frac{x^2}{4}\right)^2-\left(\frac{y^2}{9}\right)^2 \\
& =\left(\frac{x^2}{4}+\frac{y^2}{9}\right)\left(\frac{x^2}{4}-\frac{y^2}{9}\right) \\
& =\left(\frac{x^2}{4}+\frac{y^2}{9}\right)\left\{\left(\frac{x}{2}\right)^2-\left(\frac{y}{3}\right)^2\right\} \\
& =\left(\frac{x^2}{4}+\frac{y^2}{9}\right)\left(\frac{x}{2}+\frac{y}{3}\right)\left(\frac{x}{2}-\frac{y}{3}\right)
\end{aligned}\)

 

Question 2. \(m^2+\frac{1}{m^2}+2-2 m-\frac{2}{m}\)

Solution: \(\begin{aligned}
& \left(m^2+\frac{1}{m^2}+2-2 m-\frac{2}{m}\right. \\
& (m)^2+\left(\frac{1}{m}\right)^2+2-2\left(m+\frac{1}{m}\right)
\end{aligned}\)

\(\begin{aligned}
& =\left(m+\frac{1}{m}\right)^2-2 m \cdot \frac{1}{m}+2-2\left(m+\frac{1}{m}\right) \\
& =\left(m+\frac{1}{m}\right)^2-2+2-2\left(m+\frac{1}{m}\right) \\
& =\left(m+\frac{1}{m}\right)^2-2\left(m+\frac{1}{m}\right) \\
& =\left(m+\frac{1}{m}\right)\left(m+\frac{1}{m}-2\right)
\end{aligned}\)

 

Question 3. \(9 p^2-24 p q+16 q^2+3 a p-4 a q\)

Solution: \(9 p^2-24 p q+16 q^2+3 a p-4 a q\)

\(\begin{aligned}
& =(3 p)^2-2.3 p .4 q+(4 q)^2+a(3 p-4 q) \\
& =(3 p-4 q)^2+a(3 p-4 q) \\
& =(3 p-4 q)(3 p-4 q+a)
\end{aligned}\)

Class 9 Maths WBBSE

Question 4. 4x⁴+81

Solution: 4x⁴+81

\(\begin{aligned}
& =\left(2 x^2\right)^2+(9)^2 \\
& =\left(2 x^2+9\right)^2-2.2 x^2 .9 \\
& =\left(2 x^2+9\right)^2-36 x^2 \\
& =\left(2 x^2+9\right)^2-(6 x)^2 \\
& =\left(2 x^2+9+6 x\right)\left(2 x^2+9-6 x\right) \\
& =\left(2 x^2+6 x+9\right)\left(2 x^2-6 x+9\right)
\end{aligned}\)

 

Question 5. x⁴ – 7x² + 1

Solution: x⁴ – 7x² + 1

\(\begin{aligned}
& =\left(x^2\right)^2+(1)^2-7 x^2 \\
& =\left(x^2+1\right)^2-2 \cdot x^2 \cdot 1-7 x^2 \\
& =\left(x^2+1\right)^2-9 x^2 \\
& =\left(x^2+1\right)^2-(3 x)^2 \\
& =\left(x^2+1+3 x\right)\left(x^2+1-3 x\right) \\
& =\left(x^2+3 x+1\right)\left(x^2-3 x+1\right)
\end{aligned}\)

 

Question 6. p⁴-11p²q²+q⁴

Solution: p⁴-11p²q²+q⁴

\(\begin{aligned}
& =\left(p^2\right)^2+\left(q^2\right)^2-11 p^2 q^2 \\
& =\left(p^2-q^2\right)^2+2 \cdot p^2 \cdot q^2-11 p^2 q^2 \\
& =\left(p^2-q^2\right)^2-9 p^2 q^2 \\
& =\left(p^2-q^2\right)^2-(3 p q)^2 \\
& =\left(p^2-q^2+3 p q\right)\left(p^2-q^2-3 p q\right) \\
& =\left(p^2+3 p q-q^2\right)\left(p^2-3 p q-q^2\right)
\end{aligned}\)

Class 9 Maths WBBSE

Question 7. a² + b² -c² -2ab

Solution: a² + b² -c² -2ab

\(\begin{aligned}
& =a^2-2 a b+b^2-c^2 \\
& =(a-b)^2-(c)^2 \\
& =(a-b+c)(a-b-c)
\end{aligned}\)

 

Question 8. 3a(3a+2c) – 4b(b + c)

Solution: 3a(3a+2c) – 4b(b + c) = 9a2+6ac4b2-4bc

\(\begin{aligned}
& =9 a^2+6 a c-4 b^2-4 b c \\
& =9 a^2-4 b^2+6 a c-4 b c \\
& =(3 a)^2-(2 b)^2+2 c(3 a-2 b) \\
& =(3 a+2 b)(3 a-2 b)+2 c(3 a-3 b) \\
& =(3 a-2 b)(3 a+2 b+2 c)
\end{aligned}\)

 

Question 9. a²-6ab + 12bc-4c²

Solution: a²-6ab + 12bc-4c²

\(\begin{aligned}
& =(a)^2-(2 c)^2-6 a b+12 b c \\
& =(a+2 c)(a-2 c)-6 b(a-2 c) \\
& =(a-2 c)(a+2 c-6 b) \\
& =(a-2 c)(a-6 b+2 c)
\end{aligned}\)

Class 9 Maths WBBSE

Question 10. 3a²+4ab+b²-2ac- c²

Solution: 3a²+4ab+b²-2ac- c²

\(\begin{aligned}
& =4 a^2+4 a b+b^2-a^2-2 a c-c^2 \\
& =(2 a)^2+2 \cdot 2 a \cdot b+(b)^2-\left(a^2+2 a c+c^2\right) \\
& =(2 a+b)^2-(a+c)^2 \\
& =(2 a+b+a+c)(2 a+b-a-c) \\
& =(3 a+b+c)(a+b-c)
\end{aligned}\)

Class 9 Maths WBBSE

Question 11. x² -y²-6ax + 2ay + 8a²

Solution: x² -y²-6ax + 2ay + 8a²

\(\begin{aligned}
& =x^2-6 a x+9 a^2-a^2+2 a y-y^2 \\
& =(x)^2-2 \cdot x \cdot 3 a+(3 a)^2-\left(a^2-2 a y+y^2\right) \\
& =(x-3 a)^2-(a-y)^2 \\
& =(x-3 a+a-y)(x-3 a-a+y) \\
& =(x-2 a-y)(x-4 a+y)
\end{aligned}\)

 

Question 12. a²-9b²+4c²-25d²-4ac+30bd

Solution: a²-9b²+4c²-25d²-4ac+30bd

\(\begin{aligned}
& =a^2-4 a c+4 c^2-9 b^2+30 b d-25 d^2 \\
& =(a)^2-2 \cdot a \cdot 2 c+(2 c)^2-\left(9 b^2-30 b d+25 d^2\right) \\
& =(a-2 c)^2-\left\{(3 b)^2-2 \cdot 3 b \cdot 5 d+(5 d)^2\right\} \\
& =(a-2 c)^2-(3 b-5 d)^2 \\
& =(a-2 c+3 b-5 d)(a-2 c-3 b+5 d) \\
& =(a+3 b-2 c-5 d)(a-3 b-2 c+5 d)
\end{aligned}\)

WBBSE Class 9 Maths Solutions

Question 13. 3a²b²-c²+2ab-2bc + 2ca

Solution: 3a²b²-c²+2ab-2bc + 2ca

\(\begin{aligned}
& =3 a^2-a b-a c+3 a b-b^2-b c+3 a c-b c-c^2 \\
& =a(3 a-b-c)+b(3 a-b-c)+c(3 a-b-c) \\
& =(3 a-b-c)(a+b+c)
\end{aligned}\)

 

Question 14. x² -2x-22499

Solution: x²-2x-22499

\(\begin{aligned}
& =x^2-(151-149) x-22499 \\
& =x^2-151 x+149 x-22499 \\
& =x(x-151)+149(x-151) \\
& =(x-151)(x+149)
\end{aligned}\)

WBBSE Class 9 Maths Solutions

Question 15. (x² – y²) (a² – b²) + 4abxy

Solution: (x² – y²) (a² – b²) + 4abxy

\(\begin{aligned}
& =a^2 x^2-b^2 x^2-a^2 y^2+b^2 y^2+4 a b x y \\
& =a^2 x^2+2 a b x y+b^2 y^2-b^2 x^2+2 a b x y-a^2 y^2 \\
& =(a x)^2+2 \cdot a x \cdot b y+(b y)^2-\left(b^2 x^2-2 a b x y+a^2 y^2\right) \\
& =(a x+b y)^2-(b x-a y)^2 \\
& =(a x+b y+b x-a y)(a x+b y-b x+a y)
\end{aligned}\)

WBBSE Class 9 Maths Solutions Chapter 8 Factorisation Exercise 8.3

Question 1. t⁹-512

Solution: t⁹-512

\(\begin{aligned}
& =\left(t^3\right)^3-(8)^3 \\
& =\left(t^3-8\right)\left\{\left(t^3\right)^2+t^3 .8+(8)^2\right\} \\
& =\left\{(t)^3-(2)^3\right\}\left(t^6+8 t^3+64\right) \\
& =(t-2)\left(t^2+2 t+4\right)\left(t^6+8 t^3+64\right)
\end{aligned}\)

Question 2. 729p⁶-q⁶

Solution: 729p⁶-q⁶

\(\begin{aligned}
& =\left(27 p^3\right)^2-\left(q^3\right)^2 \\
& =\left(27 p^3+q^3\right)\left(27 p^3-q^3\right) \\
& =\left\{(3 p)^3+(q)^3\right\}\left\{(3 p)^3-(q)^3\right\} \\
& =(3 p+q)\left\{(3 p)^2-3 p \cdot q+(q)^2\right\}(3 p-q)\left\{(3 p)^2+3 p . q+(q)^2\right\} \\
& =(3 p+q)\left(9 p^2-3 p q+q^2\right)(3 p-q)\left(9 p^2+3 p q+q^2\right) \\
& =(3 p+q)(3 p-q)\left(9 p^2-3 p q+q^2\right)\left(9 p^2+3 p q+q^2\right)
\end{aligned}\)

Question 3. 8(p-3)³+343

Solution: 8(p-3)³+343

\(\begin{aligned}
& =\left\{2(p-3\}^3+(7)^3\right. \\
& =(2 p-6)^3+(7)^3 \\
& =(2 p-6+7)\left\{(2 p-6)^2-(2 p-6) 7+(7)^2\right\} \\
& =(2 p+1)\left\{(2 p)^2-2.2 p \cdot 6+(6)^2-14 p+42+49\right\} \\
& =(2 p+1)\left(4 p^2-24 p+36-14 p+42+49\right) \\
& =(2 p+1)\left(4 p^2-38 p+127\right)
\end{aligned}\)

Question 4. \(\frac{1}{8 a^3}+\frac{8}{b^3}\)

Solution: \(\frac{1}{8 a^3}+\frac{8}{b^3}\)

\(\begin{aligned}
& =\left(\frac{1}{2 a}\right)^3+\left(\frac{2}{b}\right)^3 \\
& =\left(\frac{1}{2 a}+\frac{2}{b}\right)\left\{\left(\frac{1}{2 a}\right)^2-\frac{1}{2 a} \cdot \frac{2}{b}+\left(\frac{2}{b}\right)^2\right\} \\
& =\left(\frac{1}{2 a}+\frac{2}{b}\right)\left(\frac{1}{4 a^2}-\frac{1}{a b}+\frac{4}{b^2}\right)
\end{aligned}\)

Question 5.(2a³-b³)³-b⁹

Solution: (2a³-b³)³-b⁹

\(\begin{aligned}
& =\left(2 a^3-b^3\right)^3-\left(b^3\right)^3 \\
& =\left(2 a^3-b^3-b^3\right)\left\{\left(2 a^3-b^3\right)^2+\left(2 a^3-b^3\right) b^3+\left(b^3\right)^2\right\} \\
& =\left(2 a^3-2 b^3\right)\left\{\left(2 a^3\right)^2-2 \cdot 2 a^3 \cdot b^3+\left(b^3\right)^2+2 a^3 b^3-b^6+b^6\right\} \\
& =2\left(a^3-b^3\right)\left(4 a^6-4 a^3 b^3+b^6+2 a^3 b^3\right) \\
& =2(a-b)\left(a^2+a b+b^2\right)\left(4 a^6-2 a^3 b^3+b^6\right)
\end{aligned}\)

Question 6. AR³-Ar³+ AR²h – Ar²h

Solution: AR³-Ar³+ AR²h – Ar²h

\(\begin{aligned}
& =A\left(R^3-r^3\right)+A h\left(R^2-r^2\right) \\
& =A(R-r)\left(R^2+R r+r^2\right)+A h(R+r)(R-r) \\
& =A(R-r)\left\{R^2+R r+r^2+h(R+r)\right\} \\
& =A(R-r)\left(R^2+R r+r^2+h R+h r\right)
\end{aligned}\)

Question 7. \(a^3+3 a^2 b+3 a b^2+b^3-8\)

Solution:

\(\begin{aligned}
&a^3+3 a^2 b+3 a b^2+b^3-8 \\
& =(a+b)^3-(2)^3 \\
& =(a+b-2)\left\{(a+b)^2+(a+b) \cdot 2+(2)^2\right\} \\
& =(a+b-2)\left(a^2+2 a b+b^2+2 a+2 b+4\right)
\end{aligned}\)

Question 8. \(32 x^4-500 x\)

Solution:

\(\begin{aligned}
& 32 x^4-500 x \\
& =4 x\left(8 x^3-125\right) \\
& =4 x\left\{(2 x)^3-(5)^3\right\} \\
& =4 x(2 x-5)\left\{(2 x)^2+2 x .5+(5)^2\right\} \\
& =4 x(2 x-5)\left(4 x^2+10 x+25\right)
\end{aligned}\)

Class 9 Mathematics West Bengal Board

Question 9. \(8 a^3-b^3-4 a x+2 b x\)

Solution:

\(\begin{aligned}
&8 a^3-b^3-4 a x+2 b x \\
& =(2 a)^3-(b)^3-2 x(2 a-b) \\
& =(2 a-b)\left\{(2 a)^2+2 a \cdot b+(b)^2\right\}-2 x(2 a-b) \\
& =(2 a-b)\left(4 a^2+2 a b+b^2-2 x\right)
\end{aligned}\)

 

Question 10. \(x^3-6 x^2+12 x-35\)

Solution:

\(\begin{aligned}
&x^3-6 x^2+12 x-35 \\
& =(x)^3-3 \cdot x^2 \cdot 2+3 \cdot x \cdot(2)^2-(2)^3-27 \\
& =(x-2)^3-(3)^3 \\
& =(x-2-3)\left\{(x-2)^2+(x-2) 3+(3)^2\right\} \\
& =(x-5)\left(x^2-4 x+4+3 x-6+9\right) \\
& =(x-5)\left(x^2-x+7\right)
\end{aligned}\)

Class 9 Mathematics West Bengal Board Chapter 8 Factorisation Exercise 8.4

 

Question 1. \(8 x^3-y^3+1+6 x y\)

Solution:

\(\begin{aligned}
& =(2 x)^3+(-y)^3+(1)^3-3.2 x(-y) \cdot 1 \\
& =(2 x-y+1)\left\{(2 x)^2+(-y)^2+(1)^2-2 x(-y)-(-y) \cdot 1-1.2 x\right\} \\
& =(2 x-y+1)\left(4 x^2+y^2+1+2 x y+y-2 x\right) \\
& =(2 x-y+1)\left(4 x^2+y^2+1-2 x+y+2 x y\right)
\end{aligned}\)

 

Question 2. \(8 a^3-27 b^3-1-18 a b\)

Solution:

\(\begin{aligned}
& 8 a^3-27 b^5-1-18 a d \\
& =(2 a)^3+(-3 b)^3+(-1)^3-3 \cdot 2 a \cdot(-3 b)(-1) \\
& =(2 a-3 b-1)\left\{(2 a)^2+(-3 b)^2+(-1)^2-2 a(-3 b)-(-3 b)(-1)-(-1) \cdot 2 a\right\} \\
& =(2 a-3 b-1)\left(4 a^2+9 b^2+1+6 a b-3 b+2 a\right) \\
& =(2 a-3 b-1)\left(4 a^2+9 b^2+1+2 a-3 b+6 a b\right) \\
&
\end{aligned}\)

Class 9 Mathematics West Bengal Board

Question 3. \(1+8 x^3+18 x y-27 y^3\)

Solution:

\(\begin{aligned}
& =8 x^3-27 y^3+1+18 x y \\
& =(2 x)^3+(-3 y)^3+(1)^3-3 \cdot 2 x(-3 y) .1 \\
& =(2 x-3 y+1)\left\{(2 x)^2+(-3 y)^2+(1)^2-2 x \cdot(-3 y)-(-3 y) .1-1.2 x\right\} \\
& =(2 x-3 y+1)\left(4 x^2+9 y^2+1+6 x y+3 y-2 x\right) \\
& =(2 x-3 y+1)\left(4 x^2+9 y^2+1-2 x+3 y+6 x y\right)
\end{aligned}\)

 

Question 4. \(x^3+y^3-12 x y+64\)

Solution:

\(\begin{aligned}
& =x^3+y^3+64-12 x y \\
& =(x)^3+(y)^3+(4)^3-3 \cdot x \cdot y \cdot 4 \\
& =(x+y+4)\left\{(x)^2+(y)^2+(4)^2-x \cdot y-y \cdot 4-4 \cdot x\right\} \\
& =(x+y+4)\left(x^2+y^2+16-x y-4 y-4 x\right) \\
& =(x+y+4)\left(x^2+y^2+16-4 x-4 y-x y\right)
\end{aligned}\)

Class 9 Mathematics West Bengal Board

Question 5. \((3 a-2 b)^3+(2 b-5 c)^3+(5 c-3 a)^3\)

Solution:

 

 

Question 6. \((2 x-y)^3-(x+y)^3+(2 y-x)^3\)

Solution:

 

Let a = 2x-y, b= -(x + y), c = 2y –x

a + b + c = 2x – y – x – y + 2y – x =0

a³ + b³ = c³ = 3abc

= 3(2x –y) {-(x + y)} (2y –x)

= 3(2x –y) (x + y) (x–2y)

 

Class 9 Mathematics West Bengal Board

Question 7. \(a^6+32 a^3-64\)

Solution:

\(\begin{aligned}
& a^6+32 a^3-64 \\
& =\left(a^2\right)^3+(2 a)^3+(-4)^3-3 a^2 .2 a(-4) \\
& =\left(a^2+2 a-4\right)\left\{\left(a^2\right)^2+(2 a)^2+(-4)^2-a^2 .2 a-2 a(-4)-(-4) a^2\right\} \\
& =\left(a^2+2 a-4\right)\left(a^4+4 a^2+16-2 a^3+8 a+4 a^2\right) \\
& =\left(a^2+2 a-4\right)\left(a^4-2 a^3+8 a^2+8 a+16\right)
\end{aligned}\)

 

Question 8. \(a^6-18 a^3+125\)

Solution:

\(\begin{aligned}
&a^6-18 a^3+125 \\
& =\left(a^2\right)^3+(3 a)^3+(5)^3-3 \cdot a^2 \cdot 3 a \cdot 5 \\
& =\left(a^2+3 a+5\right)\left\{\left(a^2\right)^2+(3 a)^2+(5)^2-a^2 \cdot 3 a-3 a \cdot 5-5 \cdot a^2\right\} \\
& =\left(a^2+3 a+5\right)\left(a^4+9 a^2+25-3 a^3-15 a-5 a^2\right) \\
& =\left(a^2+3 a+5\right)\left(a^4-3 a^3+4 a^2-15 a+25\right) \\
&
\end{aligned}\)

 

Question 9. \(p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3\)

Solution:

\begin{aligned}
& p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3 \\
& =\{p(q-r)\}^3+\{q(r-p)\}^3+\{r(p-q)\}^3 \\
& \text { Let } a=p(q-r), b=q(r-p), c=r(p-q) \\
& ∴ a+b+c=p(q-r)+q(r-p)+r(p-q) \\
& =p q-p r+q r-p q+p r-q r \\
& =0 \\
& ∴a^3+b^3+c^3-3 a b c=0 \\
& ∴ a^3+\dot{b}^3+c^3=3 a b c \\
&∴ \{p(q-r)\}^3+\{q(r-p)\}^3+\{r(p-q)\}^3 \text {. } \\
& =3 p(q-r) \cdot q(r-p) r(p-q) \\
& =3 p q r(p-q)(q-r)(r-p) \\
&
\end{aligned}

Class 9 Mathematics West Bengal Board

Question 10. \(p^3+\frac{1}{p^3}+\frac{26}{27}\)

Solution:

\(\begin{aligned}
& p^3+\frac{1}{p^3}+\frac{26}{27} \\
& =p^3+\frac{1}{p^3}+\frac{27-1}{27} \\
& =p^3+\frac{1}{p^3}+\frac{27}{27}-\frac{1}{27} \\
& =p^3+\frac{1}{p^3}+1-\frac{1}{27}
\end{aligned}\)

Class 9 Maths WBBSE

\(\begin{aligned}
& =(p)^3+\left(\frac{1}{p}\right)^0+\left(-\frac{1}{3}\right)^0-3 \cdot p \cdot \frac{1}{p}\left(-\frac{1}{3}\right) \\
& =\left(p+\frac{1}{p}-\frac{1}{3}\right)\left\{(p)^2+\left(\frac{1}{p}\right)^2+\left(-\frac{1}{3}\right)^2-p \cdot \frac{1}{p}-\frac{1}{p} \cdot\left(-\frac{1}{3}\right)-\left(-\frac{1}{3}\right) \cdot p\right\}
\end{aligned}\)

 

\(\begin{aligned}
& =\left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^2+\frac{1}{p^2}+\frac{1}{9}-1+\frac{1}{3 p}+\frac{p}{3}\right) \\
& =\left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^2+\frac{1}{p^2}+\frac{p}{3}+\frac{1}{3 p}-\frac{8}{9}\right)
\end{aligned}\)

Class 9 Maths WBBSE Chapter 8 Factorisation Exercise 8.5

Question 1. \((a+b)^2-5 a-5 b+6\)

Solution:

\(
\begin{aligned}
& (a+b)^2-5 a-5 b+6 \\
& =(a+b)^2-5(a+b)+6
\end{aligned}
Let a+b=x\)

 

\(\begin{aligned}
& ∴ x^2-5 x+6 \\
& =x^2-(3+2) x+6 \\
& =x^2-3 x-2 x+6 \\
& =x(x-3)-2(x-3) \\
& =(x-3)(x-2)
\end{aligned}\)

pitting the value of x the given expression becomes (a+b-3)(a+b-2)

Question 2. (x+1)(x+2)(3x-1)(3x-4) +12

Solution: (x+1)(x+2)(3x-1)(3x-4) +12
=(x+1)(3x-1)(x+2)(3x-4)+12

\(\begin{aligned}
& =\left(3 x^2-x+3 x-1\right)\left(3 x^2-4 x+6 x-8\right)+12 \\
& =\left(3 x^2+2 x-1\right)\left(3 x^2+2 x-8\right)+12
\end{aligned}\)

 

\(\begin{aligned}
& \text { Let } 3 x^2+2 x=a \\
& \qquad \begin{array}{l}
therefore(a-1)(a-8)+12 \\
\quad=a^2-8 a-a+8+12 \\
\quad=a^2-9 a+20 \\
\quad=a^2-5 a-4 a+20 \\
\quad=a(a-5)-4(a-5) \\
\quad=(a-5)(a-4)
\end{array}
\end{aligned}\)

putting the value of a,

\(\begin{aligned}
& \left(3 x^2+2 x-5\right)\left(3 x^2+2 x-4\right) \\
& =\left(3 x^2+5 x-3 x-5\right)\left(3 x^2+2 x-4\right) \\
& =\{x(3 x+5)-1(3 x+5)\}\left(3 x^2+2 x-4\right) \\
& =(3 x+5)(x-1)\left(3 x^2+2 x-4\right)
\end{aligned}\)

Class 9 Maths WBBSE

Question 3. \(x\left(x^2-1\right)(x+2)-8\)

Solution:

\(\begin{aligned}
& x\left(x^2-1\right)(x+2)-8 \\
& =x(x+1)(x-1)(x+2)-8 \\
& =\left(x^2+x\right)\left(x^2+2 x-x-2\right)-8 \\
& =\left(x^2+x\right)\left(x^2+x-2\right)-8
\end{aligned}\)

 

Let,\(\begin{aligned}
& x^2+x=a \\
& therefore \mathrm{a}(\mathrm{a}-2)-8 \\
& =\mathrm{a}^2-2 \mathrm{a}-8 \\
& =a^2-(4-2) a-8 \\
& =a^2-4 a+2 a-8 \\
& =a(a-4)+2(a-4) \\
& =(a-4)(a+2) \\
&
\end{aligned}\)

 

Pitting the value of a, the given expression becomes \(\left(x^2+x-4\right)\left(x^2+x+2\right)\)

Question 4. \(7\left(a^2+b^2\right)^2-15\left(a^4-b^4\right)+8\left(a^2-b^2\right)^2\)

Solution:

 

\(\begin{aligned}
& =7 x^2-7 x y-8 x y+8 y^2 \\
& =7 x(x-y)-8 y(x-y) \\
& =(x-y)(7 x-8 y)
\end{aligned}\)

Putting the value of x and y, the given expression becomes

\(\begin{aligned}
& \left(a^2+b^2-a^2+b^2\right)\left(7 a^2+7 b^2-8 a^2+8 b^2\right) \\
& =2 b^2\left(15 b^2-a^2\right)
\end{aligned}\)

Class 9 Maths WBBSE

Question 5. \(\left(x^2-1\right)^2+8 x\left(x^2+1\right)+19 x^2\)

Solution:

\(\begin{aligned}
&\left(x^2-1\right)^2+8 x\left(x^2+1\right)+19 x^2 \\
& =\left(x^2+1\right)^2-4 \cdot x^2 \cdot 1+8 x\left(x^2+1\right)+19 x^2 \\
& =\left(x^2+1\right)^2+8 x\left(x^2+1\right)+15 x^2
\end{aligned}\)

 

Let,\(\begin{aligned}
& \text { Let } x^2+1=a \\
& \quad therefore a^2+8 a x+15 x^2 \\
& =a^2+(5+3) a x+15 x^2 \\
& =a^2+5 a x+3 a x+15 x^2 \\
& =a(a+5 x)+3 x(a+5 x) \\
& =(a+5 x)(a+3 x)
\end{aligned}\)

 

Putting the value of a the given expression becomes

\(\begin{aligned}
& \left(x^2+1+5 x\right)\left(x^2+1+3 x\right) \\
& =\left(x^2+5 x+1\right) \cdot\left(x^2+3 x+1\right)
\end{aligned}\)

 

Question 6. \((a-1) x^2-x-(a-2)\)

Solution:

\(\begin{aligned}
& (a-1) x^2-x-(a-2) \\
&=(a-1) x^2-\{(a-1)-(a-2)\} x-(a-2) \\
&=(a-1) x^2-(a-1) x+(a-2) x-(a-2) \\
&=(a-1) x(x-1)+(a-2)(x-1) \\
&=(x-1)\{(a-1) x+(a-2)\} \\
&=(x-1)(a x-x+a-2)
\end{aligned}\)

Class 9 Maths WBBSE

Question 7. \((a-1) x^2+a^2 x y+(a+1) y^2\)

Solution:

\(\begin{aligned}
&(a-1) x^2+a^2 x y+(a+1) y^2 \\
& =(a-1) x^2+\left\{\left(a^2-1\right)+1\right\} x y+(a+1) y^2 \\
& =(a-1) x^2+\left(a^2-1\right) x y+x y+(a+1) y^2 \\
& =(a-1) x^2+(a+1)(a-1) x y+x y+(a+1) y^2 \\
& =(a-1) x\{x+(a+1) y\}+y\{x+(a+1) y\} \\
& =\{x+(a+1) y\}\{(a-1) x+y\} \\
& =(x+a y+y)(a x-x+y)
\end{aligned}\)

 

Question 8. \(x^2-q x-p^2+5 p q-6 q^2\)

Solution:

\(\begin{aligned}
& x^2-q x-p^2+5 p q-6 q^2 \\
& =x^2-q x-\left(p^2-5 p q+6 q^2\right) \\
& =x^2-q x-\left(p^2-3 p q-2 p q+6 q^2\right) \\
& =x^2-q x-\{p(p-3 q)-2 q(p-3 q)\} \\
& =x^2-q x-(p-3 q)(p-2 q) \\
& =x^2-\{(p-2 q)-(p-3 q)\} x-(p-3 q)(p-2 q) \\
& =x^2-(p-2 q) x+(p-3 q) x-(p-3 q)(p-2 q) \\
& =x(x-p+2 q)+(p-3 q)(x-p+2 q) \\
& =(x-p+2 q)(x+p-3 q)
\end{aligned}\)

 

Question 9. \(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7\)

Solution:

\(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7
\begin{aligned}
& =2\left\{(a)^2+\left(\frac{1}{a}\right)^2\right\}-\left(a-\frac{1}{a}\right)-7 \\
& =2\left\{\left(a-\frac{1}{a}\right)^2+2 \cdot a \cdot \frac{1}{a}\right\}-\left(a-\frac{1}{a}\right)-7
\end{aligned}
\)

 

\(\begin{aligned}
& \text { Let } a-\frac{1}{a}=x \\
& \qquad \begin{aligned}
therefore & 2\left(x^2+2\right)-x-7 \\
= & 2 x^2+4-x-7 \\
= & 2 x^2-x-3 \\
= & 2 x^2-3 x+2 x-3 \\
= & x(2 x-3)+1(2 x-3) \\
= & (2 x-3)(x+1)
\end{aligned}
\end{aligned}\)

 

Putting the value of a, the given expression becomes

\(\begin{aligned}
& =\left\{2\left(a-\frac{1}{a}\right)-3\right\}\left(a-\frac{1}{a}+1\right) \\
& =\left(2 a-\frac{2}{a}-3\right)\left(a-\frac{1}{a}+1\right)
\end{aligned}\)

 

Question 10. \(\left(x^2-x\right) y^2+y-\left(x^2+x\right)\)

Solution:

\(\begin{aligned}
& \left(x^2-x\right) y^2+y-\left(x^2+x\right) \\
& =x(x-1) y^2+y-x(x+1) \\
& =x(x-1) y^2+\left\{x^2-\left(x^2-1\right)\right\} y-x(x+1) \\
& =x(x-1) y^2+x^2 y-\left(x^2-1\right) y-x(x+1) \\
& =x y\{(x-1) y+x\}-(x+1)\{(x-1) y+x\} \\
& =\{(x-1) y+x\}\{x y-(x+1)\} \\
& =(x y-y+x)(x y-x-1)
\end{aligned}\)

 

Question 11. Multiple Choice Questions

1)If a² – b²= 11 x 9; a and b are positive integers (a> b) then

(1) a = 11, b = 9
(2) a = 33, b = 3
(3) a = 10, b = 1
(4) a = 100, b = 1

Solution: a² – b²= 11 x 9
or, (a + b) (a – b) (10+ 1) (10-1)
∴ a = 10, b = 1

∴ (3) a = 10, b = 1

2) If \(\frac{a}{b}+\frac{b}{a}=1\) then the value of a³ + b³ is

(1)1
(2)a
(3)b
(4)0

Solution: \(\frac{a}{b}+\frac{b}{a}=1\)

or, \(\frac{a^2+b^2}{a b}=1\)

 

or,\(\begin{aligned}
& a^2+b^2=a b \\
& a^2-a b+b^2=0 \\
& therefore a^3+b^3 \\
& =(a+b)\left(a^2-a b+b^2\right) \\
& =(a+b) \times 0 \\
& =0
\end{aligned}\)

∴ (4) 0

3)The value of \(25^3-75^3+50^3+3 \times 25 \times 75 \times 50\) is

(1) 150
(2) 0
(3) 25
(4) 50

Solution: Let a = 25, b=-75 and c =
∴ a+b+c=25-75+ 50 = 0

\(\begin{aligned}
& therefore a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =0 \times\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =0 \\
& therefore 25^3-75^3+50^3+3 \times 25 \times 75 \times 50 \\
& =(25)^3+(-75)^3+(50)^3-3.25 .(-75) .50 \\
& =0
\end{aligned}\)

4) If a + b + c = 0, then the value of \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) is

(1) 0
(2) 1
(3) 1
(4) 3

Solution: a+b+c=0
∴ a³+b³+ c³-3abc = 0
or, a³+b³+ c³= 3abc

or, \(\frac{a^3+b^3+c^3}{a b c}=\frac{3 a b c}{a b c}\)

or, \(\frac{a^3}{a b c}+\frac{b^3}{a b c}+\frac{c^3}{a b c}=3\)

or, \(\frac{a^2}{b c}+\frac{b^2}{a c}+\frac{a^2}{a b}=3\)

∴ (4)3

5) If p²– px + 12 = (x-3) (x – a) is an identity, then the values of a and p are respectively
(1) a = 4, p = 7
(2) a = 7, p = 4
(3) a = 4, p = -7
(4) a = 4, p = 7

Soltion:  p²-px+12= (x-3) (x – a)
or,  p²-px+12=x²-ax-3x + 3a
or, p²-px+12= x²– (a + 3)x + 3a
∴ 3a= 12

or, a =12/3 =4
Again, a + 3 = p
or, 4+ 3 = p
or, p=7

(1) a = 4, p = 7

Question 12. Short answer type questions:

(1) Let us write the simplest value of

\(\frac{\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3}{(b-c)^3+(c-a)^3+(a-b)^3}\)

Solution:

 

 

(2) Let us write the relation between a, b, and c if a³ + b³ + c³-3abc = 0 and a+b+c ≠ 0

Solution:

 

 

(3 ) If a²-b² = 224 and a and b are negative integers (a<b), then let us write the values of a and b.

Solution :

 

 

(4) Let us write the value of \((x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)\) if 3x= a+b+c.

Solution: 3x=a+b+c
or, 3x-a-b-c=0
or, x-a+x-b+x-c=0
Let x-a = p, x-b=q and x-c=r
∴ p+q+r=0

\(\begin{aligned}
& therefore p^3+q^3+r^3-3 p q r=0 \\
& therefore(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)=0
\end{aligned}\)

 

(5) Let us write the values of a and p if 2x²+ px + 6 = (2x – a) (x-2) is an identity.

Solution:

\(2 x^2+p x+6=(2 x-a)(x-2)\)

 

\(\text { or, } 2 x^2+p x+6=2 x^2-4 x-a x+2 a\)

 

\(\text { or, } 2 x^2+p x+6=2 x^2-(4+a) x+2 a\)

 

\(\begin{aligned}
& therefore 2 a=6 \\
& \text { or, } a=\frac{6}{2}=3
\end{aligned}\)

 

Again,\(p=-(4+a)=-(4+3) or, p=-7
therefore a=3, p=-7
\)