Question 13 Marks
Factorize: $6ab - b^2 + 12ac - 2bc$
Answer$6ab - b^2 + 12ac - 2bc$
Taking $b$ common in $(6ab - b^2)$ and $2c$ in $(12ac - 2bc) $
$= b(6a - b) + 2c (6a - b)$
Taking $(6a - b)$ common in the terms$ = (6a - b)(b + 2c)$
$\therefore 6ab - b^2 + 12ac - 2bc = (6a - b)(b + 2c)$
View full question & answer→Question 23 Marks
Factorize: $x^2 + y - xy - x$
Answer$x^2 + y - xy - x$ On rearranging $x^2 - xy - x + y$
Taking $x$ common in the $(x^2 - xy)$ and $-1$ in $(-x + y) = x(x - y) - 1(x - y)$
Taking $(x - y)$ common in the terms $= (x - y)(x - 1)$
$\therefore$ $x^2 + y - xy - x = (x - y)(x - 1)$
View full question & answer→Question 33 Marks
Find the value of $x^3+y^3-12 x y+64$, when $x+y=-4$
Answer$\because x+y=-4$
$\therefore x+y+4=0$$...(1)$
Now, $x^3+y^3-12 x y+64$
$=x^3+y^3+64-12 x y$
$=(x)^3+y^3+4^3-3 \times x \times y \times 4$
$=(x+y+4)\left(x^2+y^2+16-x y-4 y-4 x\right)$
$=0\left(x^2+y^2+16-x y-4 y-4 x\right)[\text { from }(1)]$
$=0$
$\therefore x^3+y^3-12 x y+64=0 \text { when } x+y=-4$
View full question & answer→Question 43 Marks
Factorize: $a^2 + 4b^2 - 4ab - 4c^2$
AnswerThe given expression to be factorized is: $a^2+4 b^2-4 a b-4 c^2$
This can be arrange in the form $a^2+4 b^2-4 a b-4 c^2$
$= (a^2 - 4ab + 4b^2) - 4c^2$
$= {(a)^2 - 2.a.2b + (2b)^2} - 4c^2$
$= (a - 2b)^2 - 4c^2$
Substitute $x = (a - 2b) a^2 + 4b^2 - 4ab - 4c^2$
$= x^2 - 4c^2 = x^2 - (2c)^2$
$= (x + 2c)(x - 2c)$
$Put x = (a - 2b) a^2 + 4b^2 - 4ab - 4c^2$
$= {(a - 2b) + 2c}{(a - 2b) - 2c}$
$= (a - 2b + 2c)(a - 2b - 2c)$
we cannot further factorize the expresion.
So, the required factorization of $a^2 + 4b^2 - 4ab - 4c^2 is (a - 2b + 2c)(a - 2b - 2c)$
View full question & answer→Question 53 Marks
Factorize: $x^3 + x - 3x^2 - 3$
Answer$x^3+x-3 x^2-3$
Taking $x$ common in $x^3+x=x\left(x^2+1\right)-3 x^2-3$
Taking $-3$ common in $-3 x^2-3=x\left(x^2+1\right)-3\left(x^2+1\right)$
Now, we take $\left(x^2+1\right)$
common $= (x^2 + 1)(x - 3)$
$\therefore x^3 + x - 3y^2 - 3 = (x^2 + 1)(x - 3)$
View full question & answer→Question 63 Marks
Factorize:
$x\left(x^3-y^3\right)+3 x y(x-y)$
Answer$x\left(x^3-y^3\right)+3 x y(x-y)$
Elaborating $x^3-y^3$ using the identity
$x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
$=x(x-y)\left(x^2+x y+y^2\right)+3 x y(x-y)$
Taking common $\mathrm{x}(\mathrm{x}-\mathrm{y})$ in both the terms
$=x(x-y)\left(x^2+x y+y^2+3 y\right)$
$\therefore x\left(x^3-y^3\right)+3 x y(x-y)$
$=x(x-y)\left(x^2+x y+y^2+3 y\right)$
View full question & answer→Question 73 Marks
Factorize the following expressions:
$1029-3 x^3$
Answer$1029-3 x^3$
$=3\left(343-x^3\right)$
$=3\left((7)^3-x^3\right)$
$=3(7-x)\left(72+7 x+x^2\right)$
${\left[\therefore a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]}$
$=3(7-x)\left(49+7 x+x^2\right)$
$\therefore 1029-3 x^3=3(7-x)\left(49+7 x+x^2\right)$
View full question & answer→Question 83 Marks
Factorize: $x^3 - 2x^2y + 3xy^2 - 6y^3$
Answer$x^3-2 x^2 y+3 x y^2-6 y^3$
Taking $x^2$ common in $\left(x^3-2 x^2 y\right)$ and $+3 y^2$
common in $\left(3 x y^2-6 y^3\right)=x^2(x-2 y)+3 y^2(x-2 y)$
Taking $(x-2 y)$ common in the terms $=(x-2 y)\left(x^2+3 y^2\right)$
$\therefore x^3-2 x^2 y+3 x y^2-6 y^3=(x-2 y)\left(x^2+3 y^2\right)$
View full question & answer→Question 93 Marks
Factorize: $7(x - 2y)^2 - 25(x - 2y) + 12$
AnswerLet $x - 2y = P = 7P^2 - 25P + 12$
Splitting the middle term, $= 7P^2 - 21P - 4P + 12$
$= 7P(P - 3) - 4(P - 3) = (P - 3)(7P - 4)$
Substituting $P = x - 2y = (x - 2y - 3)(7(x - 2y) - 4) $
$= (x - 2y - 3)(7x - 14y - 4)$
$\therefore$$ 7(x - 2y)^2 - 25(x - 2y) + 12$
$ = (x - 2y - 3)(7x - 14y - 4)$
View full question & answer→Question 103 Marks
Factorize the following expressions: $a^3 + 3a^2b + 3ab^2 + b^3 - 8$
Answer$=(a+b)^3-8\left[\therefore a^3+3 a^2 b+3 a b^2+b^3=(a+b) 3\right]$
$=(a+b)^3-23=(a+b-2)\left((a+b)^2+(a+b) \times 2+2^2\right)$
$=(a+b-2)\left(a^2+2 a b+b^2+2 a+2 b+4\right)$
$\therefore a^3+3 a^2 b+3 a b^2+b^3-8$
$=(a+b-2)\left(a^2+2 a b+b^2+2 a+2 b+4\right)$
View full question & answer→Question 113 Marks
Factorize the following expressions: $32a^3 + 108b^3$
Answer$32a^3 + 108b^3 = 4(8a^3 + 27b^3) = 4((2a)^3 + (3b)^3) = 4[(2a + 3b)((2a)^2 - 2a \times 3b + (3b)^2$
$\therefore [a^3 + b^3 = (a + b)(a^2 - ab + b^2)] = 4(2a + 3b)(4a^2 - 6ab + 9b^2)$
$\therefore 32a^3 + 108b^3 = 4(2a + 3b)(4a^2 - 6ab + 9b^2)$
View full question & answer→Question 123 Marks
Factorize: $x^4 + x^2 + 25.$
AnswerThe given expression to be factorized is $x^4+x^2+25$
This can be written in the form $x^4+x^2+25=\left(x^2\right)^2+2 \cdot x^2 .5+(5)^2-9 x^2$
$=\left\{\left(x^2\right)^2+2 \cdot x^2 \cdot 5+(5)^2\right\}-(3 x)^2=\left(x^2+5\right)^2-(3 x)^2=\left(x^2+5+3 x\right)\left(x^2+5-3 x\right)$
We cannot further factorize the expression.
So, the required factorization is $x^4+x^2+25=\left(x^2+5+3 x\right)\left(x^2+5-3 x\right)$
View full question & answer→Question 133 Marks
Factorize: $xy^9 - yx^9$
AnswerThe given expression to be factorized is $xy^9 - yx^9$
This can be wriiten in the form $xy^9 - yx^9 = x.y.y^8 - y.x.x^8$
Take common xy from the two terms of the above expression $xy^9 - yx^9 = xy(y^8 - x^8) = xy(y^8 - x^8)$
$= {xy(y^4)^2 - (x^4)^2)} = xy(y^4 + x^4)(y^4 - x^4) xy^9 - yx^9$
$= xy(y^4 + x^4){(y^2)^2 - (x^2)^2} = xy(y^4 + x^4)(y^2 + x^2)(y^2 - x^2)$
$= xy(y^4 + x^4)(y^2 + x^2){(y)^2 - (x)^2}$
$= xy(y^4 + x^4)(y^2 + x^2)(y + x)(y - x)$
We cannot further factorize the expression.
So, the required factorization of $xy^9 - yx^9 is xy(y^4 + x^4)(y^2 + x^2)(y + x)(y - x)$
View full question & answer→Question 143 Marks
Factorize: $2\text{x}^2+3\sqrt{5}\text{x}+5$
Answer$2\text{x}^2+3\sqrt{5}\text{x}+5$
Splitting the middle term, $=2\text{x}^2+2\sqrt{5}\text{x}+\sqrt{5}\text{x}+5$
$=2\text{x}\big(\text{x}+\sqrt{5}\big)+\sqrt{5}\big(\text{x}+\sqrt{5}\big)$
$=\big(\text{x}+\sqrt{5}\big)\big(2\text{x}+\sqrt{5}\big)$
$\therefore2\text{x}^2+3\sqrt{5}\text{x}+5$
$=\big(\text{x}+\sqrt{5}\big)\big(2\text{x}+\sqrt{5}\big)$
View full question & answer→Question 153 Marks
Factorize the following expressions: $a^3 + b^3 + a + b$
Answer$a^3+b^3+a+b=\left(a^3+b^3\right)+1(a+b)$
$=(a+b)\left(a^2-a b+b^2\right)+1(a+b)$
${\left[\therefore a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]}$
$=(a+b)\left(a^2-a b+b^2+1\right)$
$\therefore a^3+b^3+a+b$
$=(a+b)\left(a^2-a b+b^2+1\right)$
View full question & answer→Question 163 Marks
Factorize: $x^2 - y^2 - 4xz + 4z^2$
Answer$x^2-y^2-4 x z+4 z^2$
On rearranging the terms $=x^2-4 x z+4 z^2-y^2$
$=(x)^2-2 \times x \times 2 z+(2 z)^2-y^2$
Using the identity $x^2-2 x y+y^2$
$=(x-y)^2=(x-2 z)^2-y^2$
Using the identity $p^2-q^2=(p+q)(p-q)$
$=(x-2 z+y)(x-2 z-y)$
$\therefore x^2-y^2-4 x z+4 z^2$
$=(x-2 z+y)(x-2 z-y)$
View full question & answer→Question 173 Marks
Factorize: $(a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a)$
Answer$(a-b+c)^2+(b-c+a)^2+2(a-b+c)(b-c+a)$
Let $(a-b+c)=x$ and $(b-c+a)=y=x^2+y^2+2 x y$
Using the identity $(a+b)^2=a^2+b^2+2 a b=(x+y)^2$
Now, substituting $x$ and $y(a-b+c+b-c+a)^2$
Cancelling $- b ,+ b \&+ c ,- c =(2 a )^2=4 a ^2$
$\therefore(a-b+c)^2+(b-c+a)^2+2(a-b+c)(b-c+a)=4 a^2$
View full question & answer→Question 183 Marks
Factorize: $5\sqrt{5}\text{x}^2+20\text{x}+3\sqrt{5}$
Answer$5\sqrt{5}\text{x}^2+20\text{x}+3\sqrt{5}$
Splitting the middle term, $=5\sqrt{5}\text{x}^2+15\text{x}+5\text{x}+3\sqrt{5}$
$\big[\therefore20=15+5 \ \text{and} \ 15\times5=5\sqrt{5}\times3\sqrt{5}\big]$
$=5\text{x}\big(\sqrt{5}\text{x}+3\big)+\sqrt{5}\big(\sqrt{5}\text{x}+3\big)$
$=\big(\sqrt{5}\text{x}+3\big)\big(5\text{x}+\sqrt{5}\big)$
$\therefore5\sqrt{5}\text{x}^2+20\text{x}+3\sqrt{5}$
$=\big(\sqrt{5}\text{x}+3\big)\big(5\text{x}+\sqrt{5}\big)$
View full question & answer→Question 193 Marks
Factorize the following expressions:$ x^4y^4 - xy$
Answer$x^4 y^4-x y=x y\left(x^3 y^3-1\right)=x y\left((x y)^3-1^3\right)$
$=x y(x y-1)\left((x y)^2+x y \times 1+12\right)$
${\left[\therefore x^3-y^3=(x-y)\left(x^2+x y+y^2\right)\right]}$
$=x y(x y-1)\left(x^2 y^2+x y+1\right)$
$\therefore x^4 y^4-x y=x y(x y-1)\left(x^2 y^2+x y+1\right)$
View full question & answer→Question 203 Marks
Factorize: $x^2 - 1 - 2a - a^2$
AnswerThe given expression to be factorized is $x^2 - 1 - 2a - a^2$
Take common $-1$ from the last three terms and then we have $x^2 - 1 - 2a - a^2$
$= x^2 - (1 + 2a + a^2)$
$= x^2 - {(1)^2 + 2.1.a + (a)^2}$
$= x^2 - (1 + a)^2$
$= (x)^2 - (1 + a)^2$
$= {x + (1 + a)}{x - (1 + a)}$
$= (x + 1 + a)(x - 1 - a)$
We cannot further factorize the expression.
So, the required factorization is $x^2 - 1 - 2a - a^2$
$= (x + 1 + a)(x - 1 - a).$
View full question & answer→Question 213 Marks
Simplify: $\frac{155\times155\times155-55\times55\times55}{155\times155+155\times55+55\times55}$
Answer$\frac{155\times155\times155-55\times55\times55}{155\times155+155\times55+55\times55}$
$=\frac{155^3-55^3}{155^2+155\times55+55^2}$
$=\frac{(155-55)(155^2+155\times55+55^2)}{155^2+155\times55+55^2}$
$\big[\therefore a^3 - b^3 = (a - b)(a^2 + ab + b^2)$\big]$ = (155 - 55) = 100$
View full question & answer→Question 223 Marks
Factorize:
$a^2 x^2+\left(a x^2+1\right) x+a$
Answer$a^2 x^2+\left(a x^2+1\right) x+a$
We multiply $\mathrm{x}\left(\mathrm{ax} \mathrm{x}^2+1\right)=\mathrm{ax}{ }^3+\mathrm{x}$
$=a^2 x^2+a x^3+x+a$
Taking common $a x^2$ in $\left(a^2 x^2+a x^3\right)$ and $1$ in $(x+a)$
$=a x^2(a+x)+1(x+a)$
$=a x^2(a+x)+1(a+x)$
Taking $(\mathrm{a}+\mathrm{x})$ common in both the terms
$=(a+x)\left(a x^2+1\right)$
$\therefore a^2 x^2+\left(a x^2+1\right) x+a$
$=(a+x)\left(a x^2+1\right)$
View full question & answer→Question 233 Marks
Factorize:
$(x+2)\left(x^2+25\right)-10 x^2-20 x$
Answer$(x+2)\left(x^2+25\right)-10 x^2-20 x$
$(x+2)\left(x^2+25\right)-10 x(x+2)$
Taking $(x+2)$ common in both the terms
$=(x+2)\left(x^2+25-10 x\right)$
$=(x+2)\left(x^2-10 x+25\right)$
Splitting the middle term of $\left(x^2-10 x+25\right)$
$=(x+2)\left(x^2-5 x-5 x+25\right)$
$=(x+2)\{x(x-5)-5(x-5)\}$
$=(x+2)(x-5)(x-5)$
$\therefore(x+2)\left(x^2+25\right)-10 x^2-20 x=(x+2)(x-5)(x-5)$
View full question & answer→Question 243 Marks
Factorize:
$a^3-3 a^2 b+3 a b^2-b^3+8$
Answer$a^3-3 a^2 b+3 a b^2-b^3+8$
$\left.=(a-b)^3+8\right)\left[\because a^3-b^3+3 a^2 b+3 a b^2=(a-b)^3\right]$
$=(a-b)^3+2^3$
$=(a-b+2)\left((a-b)^2-(a-b)^2+2^2\right)\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=(a-b+2)\left(a^2+b^2-2 a b-2 a+2 b+4\right)$
$\therefore a^3-3 a^2 b+3 a b^2-b^3+8$
$=(a-b+2)\left(a^2+b^2-2 a b-2 a+2 b+4\right)$
View full question & answer→Question 253 Marks
Factorize:
$9(2 a-b)^2-4(2 a-b)-13$
Answer$\text { Let } 2 \mathrm{a}-\mathrm{b}=\mathrm{x}$
$=9 \mathrm{x}^2-4 \mathrm{x}-13$
Splitting the middle term,
$=9 x^2-13 x+9 x-13$
$=x(9 x-13)+1(9 x-13)$
$=(9 x-13)(x+1)$
Substituting $\mathrm{x}=2 \mathrm{a}-\mathrm{b}$
$=[9(2 a-b)-13](2 a-b+1)$
$=(18 a-9 b-13)(2 a-b+1)$
$\therefore 9(2 a-b)^2-4(2 a-b)-13$
$=(18 a-9 b-13)(2 a-b+1)$
View full question & answer→Question 263 Marks
Factorize the following expressions:
$8 x^3-125 y^3+180 x y+216$
Answer$8 x^3-125 y^3+180 x y+216$
$\text { or, } 8 x^3-125 y^3+216+180 x y$
$=(2 x)^3+(-5 y)^3+6^3-3 x(2 x)(-5 y)(6)$
$=(2 x+(-5 y)+6)\left((2 x)^2+(-5 y)^2+6^2-2 x(-5 y)-(-5 y) 6-6(2 x)\right)$
$=(2 x-5 y+6)\left(4 x^2+25 y^2+36+10 x y+30 y-12 x\right)$
$\therefore 8 x^3-125 y^3+180 x y+216$
$=(2 x-5 y+6)\left(4 x^2+25 y^2+36+10 x y+30 y-12 x\right)$
View full question & answer→Question 273 Marks
Factorize: $2\text{x}^2-\frac{5}{6}\text{x}+\frac{1}{12}$
Answer$2\text{x}^2-\frac{5}{6}\text{x}+\frac{1}{12}$
Splitting the middle term, $=2\text{x}^2-\text{x}^2-\text{x}^3+\frac{1}{12}$
$\Big[\therefore-\frac{5}{6}=-\frac{1}{2}-\frac{1}{3} \ \text{also} \ -\frac{1}{2}\times-\frac{1}{3}=2\times\frac{1}{12}\Big]$
$=\text{x}\Big(2\text{x}-\frac{1}{2}\Big)-\frac{1}{6}\Big(2\text{x}-\frac{1}{2}\Big)$
$=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
$\therefore2\text{x}^2-56\text{x}+\frac{1}{12}=\Big(2\text{x}-\frac{1}{2}\Big)\Big(\text{x}-\frac{1}{6}\Big)$
View full question & answer→Question 283 Marks
Factorize the following expressions:
$125+8 x^3-27 y^3+90 x y$
Answer$125+8 x^3-27 y^3+90 x y$
$=5^3+(2 x)^3+(-3 y)^3-3 \times 5 \times 2 x \times(-3 y)$
$=(5+2 x+(-3 y))\left(5^2+(2 x)^2+(-3 y)^2-5(2 x)-2 x(-3 y)-(-3 y) 5\right)$
$=(5+2 x+-3 y)\left(25+4 x^2+9 y^2-10 x+6 x y+15 y\right)$
$\therefore 125+8 x^3-27 y^3+90 x y$
$=(5+2 x+-3 y)\left(25+4 x^2+9 y^2-10 x+6 x y+15 y\right)$
View full question & answer→Question 293 Marks
Factorize:
$2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
Answer$2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
$=\big(\sqrt{2}\text{a}\big)^2+2\times\sqrt{2}\text{a}\times\sqrt{3}\text{b}+\big(\sqrt{3}\text{b}\big)^2$
Using the identity $(p+q)^2=p^2+q^2+2 p q$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)^2$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)$
$\therefore2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)$
View full question & answer→Question 303 Marks
Factorize the following expressions:
$\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\text{a}+\frac{2}{\text{a}}$
Answer$=\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}^3-\Big(\frac{1}{\text{a}^3}\Big)\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\Big(\frac{1}{\text{a}}\Big)^2\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$\left[\therefore \mathrm{a}^3-\mathrm{b}^3=(\mathrm{a}-\mathrm{b})\left(\mathrm{a}^2+\mathrm{ab}+\mathrm{b}^2\right)\right]$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
$\therefore\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+2\text{a}$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
View full question & answer→Question 313 Marks
Factorize: $4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2$
Answer$4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^{2$}$
$Let(x - y) = x,(x + y) = y = 4x^2 - 12xy + 9y^2$
Splitting the middle term $-12 = -6 - 6$
also $4 \times 9 = -6 \times -6 = 4x^2 - 6xy - 6xy + 9y^2$
$= 2x(2x - 3y) - 3y(2x - 3y)$
$= (2x - 3y)(2x - 3y) = (2x - 3y)^2$
Substituting $x = x - y y = x + y$
$= [2(x - y) - 3(x + y)]^2$
$= [2x - 2y - 3x - 3y]^2$
$= (2x - 3x - 2y - 3y)²$
$= [-x - 5y]^2 = [(-1)(x + 5y)]^2$
$= (x + 5y)^2 [(-1)^2 = 1]$
$\therefore$ $4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2 = (x + 5y)^2$
View full question & answer→Question 323 Marks
Simplify: $\frac{173\times173\times173+127\times127\times127}{173\times173-173\times127+127\times127}$
Answer$\frac{173\times173\times173+127\times127\times127}{173\times173-173\times127+127\times127}$
$=\frac{173^3+127^3}{173^2-173\times127+127^2}$
$=\frac{(173+127)(173^2-173\times127+127^2)}{173^2-173\times127+127^2}$
$\left[\therefore a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]=(173+127)$
View full question & answer→Question 333 Marks
Simplify:
$\frac{1.2\times1.2\times1.2-0.2\times0.2\times0.2}{1.2\times1.2+1.2\times0.2+0.2\times0.2}$
Answer$\frac{1.2\times1.2\times1.2-0.2\times0.2\times0.2}{1.2\times1.2+1.2\times0.2+0.2\times0.2}$
$=\frac{1.2^3-0.23}{1.2^2+1.2\times0.2+0.2^2}$
$=\frac{(1.2-0.2)((1.2)^2+1.2\times0.2+(0.2)^2)}{1.2^2+1.2\times0.2+0.2^2}$
${\left[\therefore a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]}$
$=(1.2-0.2)$
$=1.0$
View full question & answer→Question 343 Marks
Factorize:
$21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer$21 x^2-2 x+\frac{1}{21}$
$=(\sqrt{21 x})^2-2 \sqrt{21} x \times \frac{1}{\sqrt{21}}+\left(\frac{1}{\sqrt{21}}\right)^2$
Using the identity $(x-y)^2=x^2+y^2-2 x y$
$\left(\sqrt{21} x-\frac{1}{\sqrt{21}}\right)^2$
$\therefore 21 x^2-2 x+\frac{1}{21}=\left(\sqrt{21} x-\frac{1}{\sqrt{21}}\right)^2$
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Factorize the following expressions:
$8 a^3-b^3-4 a x+2 b x$
Answer$=(2 a)^3-b^3-2 x(2 a-b)$
$=(2 a-b)\left((2 a)^2+2 a \times b+b^2\right)-2 x(2 a-b)$
${\left[\therefore a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]}$
$=(2 a-b)\left(4 a^2+2 a b+b^2-2 x\right)$
$\therefore 8 a^3-b^3-4 a x+2 b x=(2 a-b)\left(4 a^2+2 a b+b^2-2 x\right)$
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Multiply: $(9x^2 + 25y^2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)$
Answer$=(3 x-5 y+4)\left(9 x^2+25 y^2+15 x y+20 y-12 x+16\right)$
$=(3 x+(5 y)+4)\left\{(3 x)^2+(-5 y)^2+4^2-3 x(-5 y)-(-5 y) 4-4(3 x)\right\}$
${\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]}$
$\text { Here, } a=3 x, b=-5 y, c=4$
$=(3 x)^3+(-5 y)^3+4^3-3(3 x)(-5 y)(4)$
$=27 x^3-125 y^3+64+180 x y$
$\therefore(3 x-5 y+4)\left(9 x^2+25 y^2+15 x y+20 y-12 x+16\right)$
$=27 x^3-125 y^3+64+180 x y$
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Factorize the following expressions:
$54 x^6 y+2 x^3 y^4$
Answer$54 x^6 y+2 x^3 y^4$
$=2 x^3 y\left(27 x^3+y^3\right)$
$=2 x^3 y\left((3 x)^3+y^3\right)$
$=2 x^3 y(3 x+y)\left((3 x)^2-3 x \times y+y^2\right)$
$\therefore\left[a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=2 x^3 y(3 x+y)\left(9 x^2-3 x y+y^2\right)$
$\therefore 54 x^6 y+2 x^3 y^4=2 x^3 y(3 x+y)\left(9 x^2-3 x y+y^2\right)$
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Factorize the following expressions: $x^3y^3 + 1$
Answer$x^3y^3 + 1 = (xy)^3 + 1^3 = (xy + 1)((xy)^2 + xy + 1^2)$
$\big[\therefore$ $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
$\big]$ $= (xy + 1)(x^2y^2 - xy + 1)$
$\therefore$ $x^3y^3 + 1 = (xy + 1)(x^2y^2 - xy + 1)$
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Factorize the following expressions: $(x + 2)^3 + (x - 2)^3$
Answer$=(x+2+x-2)((x+2)^2-(x+2)(x-2)+(x-2)^2)[\therefore[a^3+b^3=(a+b)(a^2-a b+b^2)]$
$=2 x(x^2+4 x+4-(x+2)(x-2)+x^2-4 x+..4)$
$=2 x(2 x^2+8-(x^2-2^2))[\therefore(a+b)(a-b)=a^2-b^2]$
$=2 x(2 x^2+8-x^2+4)=2 x(x^2+12)$
$\therefore(x+2)^3+(x-2)^3=2 x(x^2+12)$
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Factorize the following expressions:
$8 x^2 y^3-x^5$
Answer$8 x^2 y^3-x^5$
$=x^2\left((2 y)^3-x^3\right)$
$=x^2(2 y-x)\left((2 y)^2+2 y \times x+x^2\right)$
${\left[\therefore x^3-y^3=(x-y)\left(x^2+x y+y^2\right)\right]}$
$=x^2(2 y-x)\left(4 y^2+2 x y+x^2\right)$
$\therefore 8 x^2 y^3-x^5=x^2(2 y-x)\left(4 y^2+2 x y+x^2\right)$
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Factorize the following expressions: $\frac{1}{27}\text{x}^3-\text{y}^3+125\text{z}^3+5\text{xyz}$
Answer$\frac{1}{27}\text{x}^3-\text{y}^3+125\text{z}^3+5\text{xyz}$
$=\Big(\frac{\text{x}}{3}\Big)^3+(-\text{y})^3+(5\text{z})^3-3\times\frac{\text{x}}{3}(-\text{y})(5\text{z})$
$=\Big(\frac{\text{x}}{3}+(-\text{y})+5\text{z}\Big)\Big(\frac{\text{x}}{3}\Big)^2+(-\text{y})^2+(5\text{z})^2-\frac{\text{x}}{3}(-\text{y})-(-\text{y})5\text{z}-5\text{z}\Big(\frac{\text{x}}{3}\Big)\Big)$
$=\Big(\frac{\text{x}}{3}-\text{y}+5\text{z}\Big)\Big(\frac{\text{x}^2}{9}+\text{y}^2+25\text{z}^2+\frac{\text{xy}}{3}+5\text{yz}-\frac{5}{3}\text{zx}\Big)$
$\therefore\frac{1}{27}\text{x}^3-\text{y}^3+125\text{z}^3+5\text{xyz}$
$=\Big(\frac{\text{x}}{3}-\text{y}+5\text{z}\Big)\Big(\frac{\text{x}^2}{9}+\text{y}^2+25\text{z}^2+\frac{\text{xy}}{3}+5\text{yz}-\frac{5}{3}\text{zx}\Big)$
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