Chapter – 8 : Factorisation | Chapter Solution Class 9

Let us work out – 8.1

1. Let us factorise the following polynomial: 

1. x^{3}-3 x+2\\

Solution:

Let us suppose

f(x)=x^{3}-3 x+2\\

We are observing that,

f(1) =1^{3}-3(1)+2 \\

=1-3+2 \\

=3-3=0 \\

\therefore From factor Theorem we get (x – 1) is a factor of f(x)

x^{3}-3 x+2 \\

= x^{3}-x^{2}+x^{2}-x-2 x+2 \\

= x^{2}(x-1)+x(x-1)-2(x-1) \\

= (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) \quad(\text { Ans })\\

2. x^{3}+2 x+3 \\

Solution:

Let us suppose, f(x) = x^{3}+2 x+3 \\

We are observing that,

f(-1) =(-1)^{3}+2(-1)+3 \\

=-1-2+3=0 \\

\therefore From Factor Theorem, we get (\mathrm{x}+1) is a factor of f(x)

\therefore x^{3}+2 x+3 \\

= x^{3}+x^{2}-x^{2}-x+3 x+3 \\

= x^{2}(x+1)-x(x+1)+3(x+1) \\

= (x+1)\left(x^{2}-x+3\right)(\text { Ans })\\

3. a^{3}-12 a-16 \\

Solution:

Let us suppose f(x) = a^{3}-12 a-16 \\

We are observing that

f(-2) =(-2)^{3}-12(-2)-16 \\

=-8+24-16 \\

=24-24=0\\

\therefore From Factor Theorem, we get (a + 2) is a factor of f(a)

a^{3}-12 a-16 =a^{3}+2 a^{2}-2 a^{2}-4 a-8 a-16 \\

=a^{2}(a+2)-2 a(a+2)-8(a+2) \\

=(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-4)(a+2) \quad(Ans)\\

4. x^{3}-6 x+4 \\

Solution:

Let us suppose f(x)=x^{3}-6 x+4 \\

We are observing that,

f(2) =(2)^{3}-6(2)+4 \\

=8-12+4 \\

=12-12=0 \\

\therefore From Factor Theorem we get (x – 2) is a factor of f(x)

x^{3}-6 x+4 \\

= x^{3}-2 x^{2}+2 x^{2}-4 x-2 x+4 \\

= x^{2}(x-2)+2 x(x-2)-2(x-2) \\

= (x-2)\left(x^{2}+2 x-2\right)(\text { Ans })\\

5. \mathrm{x}^{3}-19 \mathrm{x}-30 \\

Solution:

Let us suppose f(x)=x^{3}-19 x-30 \\

We are observing that,

f(-2) =(-2)^{3}-19(-2)-30 \\

=-8+38-30 \\

=38-38 \\

=0\\

\therefore From Factor Theorem, we get (x + 2) is a factor of f(x)

x^{3}-19 x-30 \\

= x^{3}+2 x^{2}-2 x-4 x-15 x-30 \\

= (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-5)(x+3)(\text { Ans }) \\

6. 4 {a}^{3}-9 {a}^{2}+{3 a}+2\\

Solution:

Let us suppose,

f(a)=4 a^{3}-9 a^{2}+3 a+2\\

We are observing that,

f(1) =4(1)^{3}-9(1)^{2}+3(1)+2 \\

=4-9+3+2 \\

=9-9\\

=0 \\

\therefore From the Factor Theorem, we get (a – 1) is a factor of f(a)

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)(\text { Ans })\\

7 . x^{3}-9 x^{2}+23 x-15\\

Solution:

Let us suppose

f(x)=x^{3}-9 x^{2}+23 x-15\\

We are observing that,

f(1) =(1)^{3}-9(1)+23(1)-15 \\

=1-9+23-15 \\

=24-24\\

=0 \\

\therefore From the Factor Theorem we get (x – 1) is a factor of

f(x) x^{3}-9 x^{2}+23 x-15 \\

= 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-5)(x-3)(\text { Ans })\\

8. 5 a^{3}+11 a^{2}+4 a-2 \\

Solution:

Let us suppose,

f(a)=5 a^{3}+11 a^{2}+4 a-2\\

We are observing that,

f(-1) =5(-1)^{3}+11(-1)^{2}+4(-1)-2 \\

=-5+11-4-2 \\

=11-11 \\

=0 \\

\therefore From the Factor Theorem, we get (a + 1) is a factor

f(a) 5 a^{3}+11 a^{2}+4 a-2 \\

= 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) \quad(\text { Ans })\\

9. 2 x^{3}-x^{2}+9 x+5\\

Solution:

Let us suppose,

f(x)=2 x^{3}-x^{2}+9 x+5\\

We are observing that,

f\left(-\frac{1}{2}\right) =2\left(-\frac{1}{2}\right)^{3}-\left(-\frac{1}{2}\right)^{2}+9\left(-\frac{1}{2}\right)+5 \\

=-2 \times \frac{1}{8}-\frac{1}{4}-9 \times \frac{1}{2}+5 \\

=-\frac{1}{4}-\frac{1}{4}-\frac{9}{2}+5 \\

=\frac{-1-1-18+20}{4}=\frac{20-20}{4}=0 \\

\therefore From the factor Theorem we get (2x + 1) is a factor of f(x)

2 x^{3}-x^{2}+9 x+5 \\

= 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) \quad(\text { Ans })\\

10. 2 y^{3}-5 y^{2}-19 y+42 \\

Solution:

Let us suppose,

f(y)=2 y^{3}-5 y^{2}-19 y+42\\

We are observing that,

\therefore f(2) =2(2)^{3}-5(2)^{2}-19(2)+42 \\

=2(8)-5(4)-19(2)+42 \\

=16-20-38+42 \\

=58-58=0 \\

\therefore From the Factor theorem we get (y – 2) is a factor f(y)

2 y^{3}-5 y^{2}-19 y+42 \\

= 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)(\text { Ans })\\

Let us work out – 8.2

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

Solution: 

\frac{x^{4}}{16}-\frac{y^{4}}{81} \\

=\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) (Ans)\\

2. m^{2}+\frac{1}{m^{2}}+2-2 m-\frac{2}{m}\\

Solution: 

m^{2}+\frac{1}{m^{2}}+2-2 m-\frac{2}{m} \\

= \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)(Ans)\\

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 \\

= (3 p)^{2}-2 \cdot 3 p \cdot 4 q+(4 q)^{2}+a(3 p-4 q) \\

= (3 p-4 q)^{2}+a \cdot(3 p-4 q) \\

= (3 p-4 q)(3 p-4 q+a)(A n s) \\

4. 4 x^{4}+81 4.4 x^{4}+81 \\

Solution: 

4 x^{4}+81 \\

= \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)(\text { Ans }) \\

5. x^{4}-7 x^{2}+1\\

Solution: 

x^{4}-7 x^{2}+1 \\

= x^{4}+1-7 x^{2} \\

= \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)(Ans)

6. p^{4}-11 p^{2} q^{2}+q^{4}\\

Solution:

p^{4}-11 p^{2} q^{2}+q^{4} \\

= p^{4}+q^{4}-11 p^{2} q^{2} \\

= \left(p^{2}\right)^{2}+\left(q^{2}\right)^{2}-11 p^{2} q^{2} \\

= \left(p^{2}-q^{2}\right)^{2}+2 p^{2} 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)(A n s) \\

7. {a}^{2}+{b}^{2}-{c}^{2}-2 {a b}\\

Solution:

a^{2}+b^{2}-c^{2}-2 a b \\

= a^{2}-2 a b+b^{2}-c^{2} \\

= (a-b)^{2}-(c)^{2} \\

= (a-b-c)(a-b+c)(A n s) \\

8. 3 {a}(3 {a}+2 {c})-{4 b} {( b + c )}\\

Solution:

3 a(3 a+2 c)-4 b(b+c) \\

= 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-2 b) \\

= (3 a-2 b)(3 a+2 b+2 c)(\text { Ans) }\\

9. a^{2}-6 a b+12 b c-4 c^{2} \\

Solution:

a^{2}-6 a b+12 b c-4 c^{2} \\

= a^{2}-4 c^{2}-6 a b+12 b c \\

= (a)^{2}-(2 c)^{2}-6 b(a-2 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) \quad(Ans) \\

10. 3 a^{2}+4 a b+b^{2}-2 a c-c^{2}\\

Solution:

3 a^{2}+4 a b+b^{2}-2 a c-c^{2} \\

= 4 a^{2}-a^{2}+4 a b+b^{2}-2 a c-c^{2} \\

= 4 a^{2}+4 a b+b^{2}-\left(a^{2}+2 a c+c^{2}\right) \\

= (2 a)^{2}+2 \cdot 2 a \cdot b+(b)^{2}-\left\{(a)^{2}+2 \cdot a \cdot c+(c)^{2}\right\} \\

= (2 a+b)^{2}-(a+c)^{2} \\

= \{(2 a+b)+(a+c)\}\{(2 a+b)-(a+c)\} \\

= (2 a+b+a+c)(2 a+b-a-c) \\

= (3 a+b+c)(a+b-c)(Ans)\\

11. x^{2}-y^{2}-6 a x+2 a y+8 a^{2} \\

Solution:

x^{2}-y^{2}-6 a x+2 a y+8 a^{2} \\

= x^{2}-y^{2}-6 a x+2 a y+9 a^{2}-a^{2} \\

= x^{2}-6 a x+9 a^{2}-y^{2}+2 a y-a^{2} \\

= x^{2}-6 a x+9 a^{2}-\left(y^{2}-2 \cdot y \cdot a+a^{2}\right) \\

= (x)^{2}-2 \cdot x a+(3 a)^{2}-\left\{(y)^{2}-2 y a+(a)^{2}\right\} \\

= (x-3 a)^{2}-(y-a)^{2} \\

= \{(x-3 a)+(y-a)\}\{(x-3 a)-(y-a)\} \\

= (x-3 a+y-a)(x-3 a-y+a) \\

= (x+y-4 a)(x-y-2 a)(\text { Ans) } \\

12. a^{2}-9 b^{2}+4 c^{2}-25 d^{2}-4 a c+30 b d\\

Solution: 

a^{2}-9 b^{2}+4 c^{2}-25 d^{2}-4 a c+20 b d \\

= a^{2}-4 a c+4 c^{2}-9 b^{2}+30 b d-25 d^{2} \\

= a^{2}-4 a c+4 c^{2}-\left(9 b^{2}-30 b d+25 d^{2}\right) \\

= (a)^{2}-4 a c+(2 c)^{2}-\left\{(3 b)^{2}-2.53 b 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-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) \quad \text { (Ans) }\\

13. 3a a^{2}-b^{2}-c^{2}+2 a b-2 b c+2 c a \\

Solution:

3 a^{2}-b^{2}-c^{2}+2 a b-2 b c+2 c a \\

= 4 a^{2}-a^{2}-b^{2}-c^{2}+2 a b-2 b c+2 c a \\

= 4 a^{2}-\left(a^{2}+b^{2}+c^{2}-2 a b+2 b c-2 c a\right) \\

= (2 a)^{2}-(a-b-c)^{2} \\

= \{2 a+(a-b-c)\}\{2 a-(a-b-c)\} \\

= (2 a+a-b-c)(2 a-a+b+c) \\

= (3 a-b-c)(a+b+c) \quad(\text { Ans }) \\

14. x^{2}-2 \mathrm{x}-22499\\

Solution: 

x^{2}-2 x-22499 \\

= x^{2}-(151-149) x-22499 \\

= x^{2}-151 x+149 x-22499 \\

= x(x-151)+149(x-151) \\

= (x-151)(x+149)(\text { Ans })

15. \left(x^{2}-y^{2}\right)\left(a^{2}-b^{2}\right)+4 a b x y \\

Solution: 

\left(x^{2}-y^{2}\right)\left(a^{2}-b^{2}\right)+4 a b x y \\

= x^{2} a^{2}-x^{2} b^{2}-a^{2} y^{2}+b^{2} y^{2}+2 a b x y+2 a b x y \\

= x^{2} a^{2}+2 a b x y+b^{2} y^{2}-x^{2} b^{2}+2 a b x y-a^{2} y^{2} \\

= x^{2} a^{2}+2 a b x y+b^{2} y^{2}-\left(x^{2} b^{2}-2 a b x y+a^{2} y^{2}\right) \\

= (x a)^{2}+2 \cdot x a b y+(b y)^{2}-\{(x b)^{2}-2 \cdot x b a y+a y) 2\} \\

= (x a+b y)^{2}-(x b-a y)^{2} \\

= \{(x a+b y)+(x b-a y)\}\{(x a+b y)-(x b-a y)\} \\

= (x a+b y+x b-a y)(x a+b y-x b+a y)(A n s)\\

Let us work out – 8.3

1. t^{9}-512 \\

Solution:

= \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}+t .2+(2)^{2}\right\}\left(t^{6}+8 t^{3}+64\right) \\

= \left.(t-2)\left(t^{2}+2 t+4\right)\left(t^{6}\right)+8 t^{3}+64\right)(\text { Ans }) \\

2. 729 p^{6}-q^{6} \\

Solution:

729 p^{6}-q^{6} \\

=\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)\{(3 p)^{2}-3 p \cdot q+(q)^{2}\}[(3 p-q)\{(3 p)^{2}+3 p \cdot q+(q)^{2}\}]\\

=(3 p+q)(9 p^{2}-3 p q+q^{2})(3 p-q)(9 p^{2}+3 p q+q^{2})(A n s)\\

3. 8(\mathbf{p}-3)^{3}+343\\

Solution: 

8(p-3)^{3}+343 \\

= \{2(p-3)\}^{3}+(7)^{3} \\

= (2 p-6)^{3}+(7)^{3} \\

= (2 p-6+7)\left\{(2 p-6)^{2}-(2 p-6) \cdot 7+(7)^{2}\right\} \\

= (2 p+1)\{(2 p)^{2}-2 \cdot 2 p \cdot 6+(6)^{2}-14 p+42+49\} \\

= (2 p+1)\left\{4 p^{2}-24 p+36-14 p+91\right\} \\

= (2 p+1)\left(4 p^{2}-38 p+127\right) \cdot(\text { Ans })\\

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

Solution :

\frac{1}{8 a^{3}}+\frac{8}{b^{3}} \\

= \left(\frac{1}{2 a}\right)^{3}+\left(\frac{2}{b}\right)^{3} \\

= \left(\frac{1}{2 a}+\frac{2}{a}\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^{2}}+\frac{4}{b^{2}}\right)(A n s)\\

5. \left(2 a^{3}-b^{3}\right)^{3}-b^{9} \\

Solution:

\left(2 a^{3}-\mathbf{b}^{3}\right)^{3}-\left(\mathbf{b}^{3}\right)^{3} \\

= = (2 a^{3}-b^{3}-b^{3})\{(2 a^{3}-b^{3})^{2}+(2 a^{3}-b^{3})(b^{3})(b^{3})^{2}\} \\

= (2 a^{3}-2 b^{3})\{(2 a^{3})^{2}-22 a^{3} \cdot b^{3}+(b^{3})^{2}+2 a^{3} \cdot b^{3}-b^{6}+b^{6}\} \\

= 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)(\text { Ans) }\\

6. A R^{3}-A r^{3}+A R^{2} h-A r^{2} h \\

Solution:

\mathbf{A R}^{3}-\mathbf{A} \mathbf{r}^{3}+\mathbf{A} \mathbf{R}^{2} \mathbf{h}-\mathbf{A} \mathbf{r}^{2} \mathbf{h} \\

= A\left[R^{3}-r^{3}+R^{2} h-r^{2} h\right] \\

= A\left[(R-r)\left(R^{2}+R r+r^{2}\right)+h\left(R^{2}-r^{2}\right)\right] \\

= A\left[(R-r)\left(R^{2}+R r+r^{2}\right)+h(R-r)(R+r)\right] \\

= A \cdot(R-r)\left[R^{2}+R r+r^{2}+h(R+r)\right] \\

= A(R-r)\left(R^{2}+R r+r^{2}+R h+r h\right)(A n s) \\

7. a^{3}+3 a^{2} b+3 a b^{2}+b^{3}-8\\

Solution:

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)(A n s)\\

8. 32 \mathrm{x}^{4}-500 \mathrm{x} \\

Solution: 

32 {x}^{4}-500 \mathbf{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)(A n s)\\

9. 8 a^{3}-b^{3}-4 a x+2 b x \\

Solution: 

8 a^{3}-b^{3}-4 \mathbf{a x}+2 \mathbf{b x} \\

= (2 a)^{3}-(b)^{3}-2 x(2 a-b) \\

= (2 a-b)\left\{(2 a)^{2}+2 a b+(b)^{2}\right\}-2 x(2 a-b) \\

= (2 a-b)\left(4 a^{2}+2 a b+b^{2}\right)-2 x(2 a-b) \\

= (2 a-b)\left(4 a^{2}+2 a b+b^{2}-2 x\right)(A n s) \\

10. x^{3}-6 \mathbf{x}^{2}+12 \mathbf{x}-35\\

Solution:

x^{3}-6 x^{2}+12 x-35 \\

= x^{3}-6 x^{2}+12 x-8-27 \\

= (x)^{3}-3 \cdot x^{2} \cdot 2+3 \cdot x(2)^{2}-(2)^{3}-(3)^{3} \\

= (x-2)^{3}-(3)^{3} \\

= (x-2-3)\left\{(x-2)^{2}+(x-2) \cdot 3+(3)^{2}\right\} \\

= (x-5)\left\{(x)^{2}-2 \cdot x \cdot 2+(2)^{2}+3 x-6+9\right\} \\

= (x-5)\left(x^{2}-4 x+4+3 x+3\right) \\

= (x-5)\left(x^{2}-x+7\right)(\text { Ans })\\

Let us work out – 8.4

1. 8 x^{3}-y^{3}+1+6 x y \\

Solution: 

8 x^{3}-y^{3}+1+6 x y \\

= (2 x)^{3}-(y)^{3}+(1)^{3}+3 \cdot 2 x \cdot y \cdot 1 \\

= (2 x-y+1)\{(2 x)^{2}+(-y)^{2}+(1)^{2}-2 x(-y)-(-y) \cdot 1-1 \cdot 2 x\}\\

= (2 x-y+1)\left(4 x^{2}+y^{2}+1+2 x y+y-2 x\right)(\text { Ans })

2. 8 a^{3}-27 b^{3}-1-18 a b \\

Solution: 

8^{3}-27 \mathbf{b}^{3}-1-18 \mathbf{a b} \\

= (2 a)^{3}-(3 b)^{3}-(1)^{3}-3 \cdot 2 a \cdot 3 b \cdot 1 \\

= (2 a-3 b-1)\{(2 a)^{2}+(-3 b)^{2}+(-1)^{2}-(2 a) \cdot(-3 b)-(-3 b) \cdot(-1)-(-1)(2a)\}\\ \\

= (2 a-3 b-1)(4 a^{2}+9 b^{2}+1+6 a b-3 b+2 a)(\text { Ans })

3. 1+8 x^{3}+18 x y-27 y^{3} \\

Solution: 

1+8 x^{3}+18 x y-27 y^{3} \\

= 1+8 x^{3}-27 y^{3}+18 x y \\

= (1)^{3}+(2 x)^{3}-(3 y)^{3}-3.1 .2 x(-3 y) \\

= (1+2 x-3 y)\{(1)^{2}+(2 x)^{2}+(-3 y)^{2}-1.2 x-2 x(-3 y)-1 .(-3 y)\} \\

= (1+2 x-3 y)(1+4 x^{2}+9 y^{2}-2 x+6 x y+6 y)(\text { Ans })\\

4. x^{3}+y^{3}-12 x y+64 \\

Solution: 

\mathbf{x}^{3}+y^{3}-12 \mathbf{x y}+64 \\

= (x)^{3}+(y)^{3}+(4)^{3}-3 x \cdot y \cdot 4 \\

= (x+y+4)\{(x)^{2}+(y)^{2}+(4)^{2}-x \cdot y-y \cdot 4-4 \cdot x\} \\

= (x+y+4)(x^{2}+y^{2}+16-x y-4 y-4 x)(A n s)\\

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

Solution:

(3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3}

Let us suppose,

3 a-2 b=x, 2 b-5 c=y, 5 c-3 a=z \\

\therefore x+y+z=3 a-2 b+2 b-5 c+5 c-3 a \\

= 0

\therefore (3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3} \\

= x^{3}+y^{3}+z^{3} \\

= 3 x y z \cdot[\because x+y+z=0, \text { So } x^{3}+y^{3}+z^{3}=3 x y z] \\

= 3(3 a-2 b)(2 b-5 c)(5 c-3 a)(\text { Ans })\\

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

Solution: 

(2 x-y)^{3}-(x+y)^{3}+(2 y-x)^{3} \\

= (2 x-y)^{3}+(-x-y)^{3}+(2 y-x)^{3}\\

Let us suppose,

2 x-y=a,-x-y=b, 2 y-x=c \\

\therefore a+b+c=2 x-y-x-y+2 y-x \\

=0 \\

\therefore(2 x-y)^{3}+(-x-y)^{3}+(2 y-x)^{3} \\

=a^{3}+b^{3}+c^{3} \\

=3 a b c\left[\therefore a+b+c =0 \text { So, } a^{3}+b^{3}+c^{3}=3 a b c\right] \\

=3 \cdot(2 x-y)(-x-y)(2 y-x) \\

=-3(2 x-y)(x+y)(2 y-x)(\text { Ans })\\

7. a^{6}+32 a^{3}-64 \\

Solution: 

a^{6}+32 a^{3}-64 \\

= a^{6}+8 a^{3}-64+24 a^{3} \\

= \left(a^{2}\right)^{3}+(2 a)^{3}+(-4)^{3}-3 \cdot a^{2} \cdot 2 a(-4) \\

= (a^{2}+2 a-4)\{(a^{2})^{2}+(2 a)^{2}+(-4)^{2}-a^{2} \cdot 2 a \cdot-2 a(-4)-(-4) \cdot a^{2}\} \\

= \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)(\text { Ans })\\

8. a^{6}-18 a^{3}+125 \\

Solution: 

{a}^{6}-18 a^{3}+125 \\

= a^{6}+27 a^{3}+125-45 a^{3} \\

= \left(a^{2}\right)^{3}+(3 a)^{3}+(5)^{3}-3 \cdot a^{2} \cdot 3 a \cdot 5 \\

= (a^{2}+3 a+5)\{(a^{2})^{2}+(3 a)^{2}+(5)^{2}-a^{2} \cdot 3 a-3 a \cdot 5-5 a^{2}\}\\

= \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) \\

9. \mathbf{p}^{3}(\mathbf{q}-\mathbf{r})^{3}+\mathbf{q}^{3}(\mathbf{r}-\mathbf{p})^{3}+\mathbf{r}^{3}(\mathbf{p}-\mathbf{q})^{3}\\

Solution: 

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} \\

= (p q-p r)^{3}+(q r-p q)^{3}+(p r-q r)^{3}

Let us suppose,

p q-p r=a, q r-p q=b, p r-q r=c \\

\therefore a+b+c=p q-p r+q r-p q+p r-q r \\

= 0 \\

\therefore (p q-p r)^{3}+(q r-p q)^{3}+(p r-q r)^{3} \\

= a^{3}+b^{3}+c^{3} \\

= 3 a b c\left[\therefore a+b+c=0, \quad \text { So } a^{3}+b^{3}+c^{3}=3 a b c\right] \\

= 3 \cdot(p q-p r)(q r-p q)(p r-q r) \\

= 3 p q r(q-r)(r-p)(p-q)(A n s)\\

10. \mathbf{p}^{3}+\frac{1}{\mathbf{p}^{3}}+\frac{\mathbf{2 6}}{\mathbf{2 7}}\\

Solution:

\mathbf{p}^{3}+\frac{1}{\mathbf{p}^{3}}+\frac{\mathbf{2 6}}{\mathbf{2 7}} \\

= (p)^{3}+\left(\frac{1}{p}\right)^{3}+1-\frac{1}{27} \\

= (p)^{3}+\left(\frac{1}{p}\right)^{3}+\left(-\frac{1}{3}\right)^{3}-3 \cdot p \cdot \frac{1}{p} \cdot\left(-\frac{1}{3}\right) \\

= (p+\frac{1}{p}-\frac{1}{3})\{(p)^{2}+(\frac{1}{p})^{2}+(-\frac{1}{3})^{2}-p \cdot \frac{1}{p}-\frac{1}{p}(-\frac{1}{3})-p(-\frac{1}{3})\} \\

= (p+\frac{1}{p}-\frac{1}{3})(p^{2}+\frac{1}{p^{2}}+\frac{1}{9}-1+\frac{1}{3 p}+\frac{p}{3}) \\

= \left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^{2}+\frac{1}{p^{2}}-\frac{8}{9}+\frac{1}{3 p}+\frac{p}{3}\right)(A n s) .\\

Let us work out – 8.5

1. Let us factorise following polynomials:

(i) (a+b)^{2}-5 a-5 b+6 \\

Solution:

(a+b)^{2}-5 a-5 b+6 \\

=(a+b)^{2}-5(a+b)+6 \\

\text { Let } x=a+b \text {, } \\

x^{2}-5 x+6 \\

=x^{2}-3 x-2 x+6 \\

=x(x-3)-2(x-3) \\

=(x-3)(x-2) \\

putting the value of x (a+b-3)(a+b-2) \text { (Ans) }\\

(ii) (x+1)(x+2)(3 x-1)(3 x-4)+12 \\

Solution:

(\mathbf{x}+1)(\mathbf{x}+2)(3 \mathbf{x}-1)(3 \mathbf{x}-4)+12 \\

=(x+1)(3 x-1)(x+2)(3 x-4)+12 \\

=(3 x^{2}+3 x-x-1)(3 x^{2}+6 x-4 x-8)+12 \\

=(3 x^{2}+2 x-1)(3 x^{2}+2 x-8)+12 \\

\text { Let } a=3 x^{2}+2 x \\

(a-1)(a-8)+12 \\

= a^{2}-8 a-a+8+12 \\

= a^{2}-9 a+20\\

=a^{2}-5 a-4 a+20 \\

=a(a-5)-4(a-5) \\

=(a-5)(a-4)\\

Putting the value of a (3 x^{2}+2 x-5)(3 x^{2}+2 x-4) \\

= (3 x^{2}+5 x-3 x-5)(3 x^{2}+2 x-4) \\

= \{x(3 x+5)-1(3 x+5)\}(3 x^{2}+2 x-4) \\

= (3 x+5)(x-1)(3 x^{2}+2 x - 4 ) Ans\\

(iii) x(x^{2}-1)(x+2)-8 \\

Solution: 

\mathbf{x}(x^{2}-1)(x+2)-8 \\

= x(x+1)(x-1)(x+2)-8 \\

= (x^{2}+x)(x^{2}+2 x-x-2)-8 \\

= (x^{2}+x)(x^{2}+x-2)-8

Let a=x^{2}+x

=a(a-2)-8 \\

=a^{2}-2 a-8 \\

=a^{2}-4 a+2 a-8 \\

=a(a-4)+2(a-4) \\

=(a-4)(a+2)\\

Putting the value of a,

(x^{2}+x-4)(x^{2}+x+2)(\text { Ans })\\

(iv) 7(a^{2}+b^{2})^{2}-15(a^{4}-b^{4})+8(a^{2}-b^{2})^{2} \\

Solution:

7(a^{2}+b^{2})^{2}-15(a^{4}-b^{4})+8(a^{2}-b^{2})^{2} \\

=7(a^{2}+b^{2})^{2}-15(a^{2}+b^{2})(a^{2}-b^{2})+8(a^{2}-b^{2})^{2}\\

Let x=a^{2}+b^{2}, y=a^{2}-b^{2} \\

7 x^{2}-15 x y+8 y^{2} \\

= 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)\\

putting the value of \mathrm{x} \ and \ \mathrm{y} \\

\{(a^{2}+b^{2})-(a^{2}-b^{2})\}\{7(a^{2}+b^{2})-8(a^{2}-b^{2})\} \\

=(\cancel{a}^{2}+b^{2}-\cancel{a}^{2}+b^{2})(7 a^{2}+7 b^{2}-8 a^{2}+8 b^{2})\\

=2 b^{2}(15 b^{2}-a^{2})(\text { Ans) }\\

(v) (x^{2}-1)^{2}+8 x(x^{2}+1)+19 x^{2} \\

Solution:

(x^{2}-1)^{2}+8 x(x^{2}+1)+19 x^{2} \\

=(x^{2}+1)^{2}-4 x^{2}+8 x(x^{2}+1)+19 x^{2} \\

=(x^{2}+1)^{2}+8 x(x^{2}+1)+15 x^{2}\\

Let a=x^{2}+1 \\

a^{2}+8 x a+15 x^{2} \\

= a^{2}+5 x a+3 x a+15 x^{2} \\

= a(a+5 x)+3 x(a+5 x)\\

=(a+5 x)(a+3 x)\\

putting the value of a,

=(x^{2}+1+5 x)(x^{2}+1+3 x) \\

=(x^{2}+5 x+1)(x^{2}+3 x+1) \text { (Ans) }\\

(vi) (a-1) x^{2}-x-(a-2) \\

Solution: 

(a-1) \mathbf{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 \cdot(x-1)+(a-2)(x-1) \\

= (x-1)\{(a-1) x+(a-2)\} \\

= (x-1)(a x-x+a-2)(\text { Ans })\\

(vii) (a-1) x^{2}+a^{2} x y+(a+1) y^{2} \\

Solution:

(\mathbf{a - 1}) \mathbf{x}^{2}+\mathbf{a}^{2} \mathbf{x y}+(\mathbf{a}+1) \mathbf{y}^{2} \\

= (a-1) x^{2}+(a^{2}-1) x y+x y+(a+1) y^{2} \\

= (a-1) x^{2}+(a+1)(a-1) x y+x y \quad(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)(\text { Ans) }\\

(viii) x^{2}-q x-p^{2}+5 p q-6 q^{2} \\

Solution: 

\mathbf{x}^{2}-\mathbf{q} \mathbf{x}-\mathbf{p}^{2}+5 \mathbf{p q}-6 q^{2} \\

= x^{2}-q x-(p^{2}-5 p q+6 q^{2}) \\

= x^{2}-q x-(p^{2}-3 p q-2 p q+6 q^{2}) \\

= x^{2}-q x-\{p(p-3 q)-2 q(p-3 q)\} \\

= x^{2}-q x-(p-3 y)(p-2 q) \\

= x^{2} -(1 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)\} \\

= (x-p+2 q)(x+p-3 q) \quad(Ans)\\

(ix) 2(a^{2}+\frac{1}{a^{2}})-(a-\frac{1}{a})-7 \\

Solution: 

2(\mathbf{a}^{2}+\frac{1}{\mathbf{a}^{2}})-(a-\frac{1}{a})-7 \\

= 2\{(a-\frac{1}{a})^{2}+2 \cdot d \cdot \frac{1}{d}\}-(a-\frac{1}{a})-7 \\

= 2\{(a-\frac{1}{a})^{2}+2\}-(a-\frac{1}{a})-7 \\

= (a-\frac{1}{a})^{2}+4-(a-\frac{1}{a})-7 \\

= 2(a-\frac{1}{a})^{2}-(a-\frac{1}{a})-3\\

Let x=a-\frac{1}{a} \\

Then,

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)\\

putting the value of x,

=\{2(a-\frac{1}{a})-3\}\{a-\frac{1}{a}+1\} \\

=(2 a-\frac{2}{a}-3)(a-\frac{1}{a}+1)(\text { Ans })\\

(x) (x^{2}-x) y^{2}+y-(x^{2}+x) \\

Solution: 

(\mathbf{x}^{2}-\mathbf{x}) y^{2}+y-(\mathbf{x}^{2}+\mathbf{x}) \\

= x(x-1) y^{2}+y-x(x+1) \\

= x(x-1) y^{2}+\{x^{2}-(x^{2}-1)\} y-x(x+1) \\

= x(x-1) y^{2}+x^{2} y^{2}-(x^{2}-1) y-x(x+1) \\

= x(x-1) y^{2}+x^{2} y-(x+1)(x-1) 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)(\text { Ans })\\

2. M.C.Q

(i) If a^{2}-b^{2}=11 \times 9 \ and \ a & b are positive integers (a  > b) then

(a) a=11, b=9 \\

(b) a=33, b=3 \\

(c) a=10, b=1 \\

(d) a=100, b=1 \\

Solution:

a^{2}-b^{2}=11 \times 9\\

or, (a+b)(a-b)=(10+1)(10-1) \\

By compairing,

a=10, b=1\\

(c) is correct option

(ii) If \frac{a}{b}+\frac{b}{a}=1 then the value of a^{3}+b^{3} is

(a) 1

(b) a

(c) b

(d) 0

Solution:

Given,

\frac{a}{b}+\frac{b}{a}=1 \\

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

\text { or, } a^{2}+b^{2}=a b \\

\text { or, } a^{2}-a b+b^{2}=0\\

Now,

a^{3}+b^{3} \\

= (a+b)(a^{2}-a b+b^{2}) \\

= (a+b) \times 0 \\

= 0 <p dir="auto">\therefore (d)  is correct option

(iii) The value 25^{3}-75^{3}+50^{3}+3 \times 25 \times 75 \times 50 is

(a) 150

(b) 0

(c) 25

(d) 50

Solution:

We have,

=25^{3}-75^{3}+50^{3}+3 \times 25 \times 75 \times 50 \\

=(25-75+50)\{(25)^{2}+(-75)^{2}+(50)^{2}-25(-75)-(-75) 50-(25)(50)\} \\

= 0

\therefore (b) is correct option

(iv) 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

(a) 0

(b) 1

(c) -1

(d) 3

Solution:

if a+b+c=0 \\

Then a^{3}+b^{3}+c^{3}=3 a b c \\

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

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}}{c a}+\frac{c^{2}}{a b}=3 \\

\therefore(d) is correct option.

(v) If x^{2}-p x+12=(x-3)(x-a) is an identity, then the values of a and p are respectively

(a) a=4, p=7 \\

(b) a=7, p=4\\

(c) a=4, p=7 \\

(d) a=-4, p=7 \\

Solution:

x^{2}-p x+12=(x-3)(x-a) \\

\text { or, } x^{2}-p x+12=x^{2}-x a-3 x+3 a\\

or, x^{2}-p x+12=x^{2}-(a+3) x+3 a \\

By compairing,

or,3 a=12 ,

or p=(a+3)\\

\text { or, } a=\frac{12}{3} =4+3 \\

\therefore a=4, \quad =7\\

(a) is correct option

Short answer type question:

(i) Let us write the simplest value of \frac{(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}+(a^{2}-b^{2})^{3}}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}} \\

Solution:

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

=\frac{3(b^{2}-c^{2})(c^{2}-a^{2})(a^{2}-b^{2})}{3(b-c)(c-a)(a-b)} \\

=\frac{(b+c)(b-c)(c-a)(c+a)(a-b)(a+b)}{(b-c)(c-a)(a-b)} \\

=(b+c)(c+a)(a+b)(\text { Ans })\\

(ii) Let us write the relation of a, b and c if a^{3}+b^{3}+c^{3}-3 a b c=0 and a+b+c \neq 0 \\

Solution:

a^{3}+b^{3}+c^{3}-3 a b c=0\\

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

It holds onlya=b=c \\

(iii) a^{2}-b^{2}=224 and a and b are negative integers (a < b), then let us write the value of a and b 

Solution:

a^{2}-b^{2}=224\\

or, a^{2}-b^{2}=255-1 \\

or, a^{2}-b^{2}=(15)^{2}-(1)^{2} \\

or, a^{2}-b^{2}=(-15)^{2}-(-1)^{2} \\

By compairing,

a=-15, b=1 \text { (Ans) }\\

(iv)Let us write the value of (x-a)^{3}+(x-b)^{3}+(x-c)^{3}-3(x-a)(x-b)(x-c) if 3 x=a+b+c \\

Solution: 

3 x=a+b+c \\

(x-a)+(x-b)+(x-c)=0 \\

\therefore(x-a)^{3}+(x-b)^{3}+(x-c)^{3}=3(x-a)(x-b)(x-c)\\

or, (x-a)^{3}+(x-b)^{3}+(x-c)^{3}-3(x-a)(x-b)(x-c)=0 (Ans)\\

(v) Let us write the value of a and p if 2 x^{2}+p x+6=(2 x-a)(x-2) is an identity.

Solution:

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

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}-(a+4) x+2 a\\

By compairing,

2 a=6 \quad p=-(a+4) \\

\text { or, } a=\frac{\cancel{6}}{\cancel{3}}=3 \quad=-(3+4) \\

\therefore a=3 \quad=-7 \\

\therefore a=3, p=7 \text { (Ans) }\\