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 $x^3 + y^3 - 12xy + 64,$ when $ x + y = -4$
Answer$\because$ $x + y = -4$$\therefore$ $x + y + 4 = 0 ...(1)$
Now,$ x^3 + y^3 - 12xy + 64$
$= x^3 + y^3 + 64 - 12xy$
$= (x)^3 + y^3 + 4^3 - 3 \times x \times y \times 4$
$= (x + y + 4)(x^2 + y^2 + 16 - xy - 4y - 4x)$
$= 0(x^2 + y^2 + 16 - xy - 4y - 4x) [from(1)]$
$= 0$
$\therefore$ $x^3 + y^3 - 12xy + 64 = 0$ when $x + y = -4$
View full question & answer→Question 43 Marks
Factorize:
$a^2 + 4b^2 - 4ab - 4c^2$
AnswerThe given expression to be factorized is:
$a^2 + 4b^2 - 4ab - 4c^2$
This can be arrange in the form
$a^2 + 4b^2 - 4ab - 4c^2$
$= (a^2 - 4ab + 4b^2) - 4c^2$
$= {(a)^2 - 2.a.2b + (2b)^2} - 4c^2$
$= (a - 2b)^2 - 4c^2$
Substitute $x = (a - 2b)$
$a^2 + 4b^2 - 4ab - 4c^2 = x^2 - 4c^2$
$= x^2 - (2c)^2$
$= (x + 2c)(x - 2c)$
$Put x = (a - 2b)$
$a^2 + 4b^2 - 4ab - 4c^2 = {(a - 2b) + 2c}{(a - 2b) - 2c}$
$= (a - 2b + 2c)(a - 2b - 2c)$
we cannot further factorize the expresion.
So, the required factorization of $a^2 + 4b^2 - 4ab - 4c^2 is (a - 2b + 2c)(a - 2b - 2c)$
View full question & answer→Question 53 Marks
Factorize:
$x^3 + x - 3x^2 - 3$
Answer$x^3 + x - 3x^2 - 3$
Taking x common in $x^3 + x$
$= x(x^2 + 1) - 3x^2 - 3$
Taking -3 common in $-3x^2 - 3$
$= x(x^2 + 1) - 3(x^2 + 1)$
Now,
we take $(x^2+ 1)$ common
$= (x^2 + 1)(x - 3)$
$\therefore$ $x^3 + x - 3y^2 - 3$
$= (x^2 + 1)(x - 3)$
View full question & answer→Question 63 Marks
Factorize:
$x(x^3 - y^3) + 3xy(x - y)$
Answer$x\left(x^3-y^3\right)+3 x y(x-y)$
Elaborating $x ^3- y ^3$
using the identity $x^3-y^3$
$=(x-y)\left(x^2+x y+y^2\right)$
$=x(x-y)\left(x^2+x y+y^2\right)+3 x y(x-y)$
Taking common $x(x-y)$ in both the terms $=x(x-y)\left(x^2+x y+y^2+3 y\right)$
$\therefore x\left(x^3-y^3\right)+3 x y(x-y)$
$=x(x-y)\left(x^2+x y+y^2+3 y\right)$
View full question & answer→Question 73 Marks
Factorize the following expressions:
$1029 - 3x^3$
Answer$1029 - 3x^3$
$= 3(343 - x^3)$
$= 3((7)^3 - x^3)$
$= 3(7 - x)(72 + 7x + x^2)$
$\big[\therefore$ $a^3- b^3= (a - b)(a^2 + ab + b^2)$$\big]$
$= 3(7 - x)(49 + 7x + x^2)$
$\therefore$ $1029 - 3x^3 = 3(7 - x)(49 + 7x + x^2)$
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$= (a + b)^3 - 8$
$\big[\therefore$ $a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)$3$\big]$
$= (a + b)^3 - 23$
$= (a + b - 2)((a + b)^2 + (a + b) \times 2 + 2^2)$
$= (a + b - 2)(a² + 2ab + b² + 2a + 2b + 4)$
$\therefore$ $a^3 + 3a^2b + 3ab^2 + b^3 - 8$
$= (a + b - 2)(a² + 2ab + b² + 2a + 2b + 4)$
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
$x^4 + x^2 + 25 = (x^2)^2 + 2.x^2.5 + (5)^2 - 9x^2$
$= {(x^2)^2 + 2.x^2.5 + (5)^2} - (3x)^2$
$= (x^2 + 5)^2 - (3x)^2$
$= (x^2 + 5 + 3x)(x^2 + 5 - 3x)$
We cannot further factorize the expression.
So, the required factorization is $x^4 + x^2 + 25 = (x^2 + 5 + 3x)(x^2 + 5 - 3x).$
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$a^3 + b^3 + a + b$
$= (a^3 + b^3) + 1(a + b)$
$= (a + b)(a^2 - ab + b^2) + 1(a + b)$
$\big[\therefore$ $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$\big]$
$= (a + b)(a^2 - ab + b^2 + 1)$
$\therefore$ $a^3 + b^3 + a + b = (a + b)(a^2 - ab + b^2 + 1)$
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 + 2xy$
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab = (x + y)^{2}$
Now, substituting $x$ and $y$ $(a - b + c + b - c + a)^2$
Cancelling $-b, +b + c, -c = (2a)^2 = 4a^2$
$\therefore$ $(a - b + c)^2 + (b - c + a)^2 + 2(a - b + c)(b - c + a) = 4a^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)$$\big]$
$= (155 - 55)$
$= 100$
View full question & answer→Question 223 Marks
Factorize:
$a^2x^2 + (ax^2 + 1)x + a$
Answer$a^2x^2 + (ax^2 + 1)x + a$
We multiply $x(ax^2 + 1) = ax^3 + x = a^2x^2 + ax^3 + x + a$
Taking common $ax^2$ in $(a^2x^2 + ax^3) and 1 in (x + a)$
$= ax^2(a + x) + 1(x + a)$
$= ax^2(a + x) + 1(a + x)$
Taking $(a + x)$ common in both the terms
$= (a + x)(ax^2 + 1)$
$\therefore$ $a^2x^2 + (ax^2+ 1)x + a$
$= (a + x)(ax^2 + 1)$
View full question & answer→Question 233 Marks
Factorize:
$(x + 2)(x^2 + 25) - 10x^2 - 20x$
Answer$(x + 2)(x^2 + 25) - 10x^2 - 20x (x + 2)(x^2 + 25) - 10x (x + 2)$
Taking (x + 2) common in both the terms =$ (x + 2)(x^2 + 25 - 10x)$
$= (x + 2)(x^2 - 10x + 25)$
Splitting the middle term of
$(x^2 - 10x + 25)$
$= (x + 2)(x^2 - 5x - 5x + 25)$
$= (x + 2){x(x - 5)-5 (x - 5)}$
$= (x + 2)(x - 5)(x - 5)$
$\therefore$ $(x + 2)(x^2 + 25) - 10x^2 - 20x = (x + 2)(x - 5)(x - 5)$
View full question & answer→Question 243 Marks
Factorize:
$a^3 - 3a^2b + 3ab^2 - b^3 + 8$
Answer$a^3 - 3a^2b + 3ab^2 - b^3 + 8 = (a - b)^3 + 8)$
$\big[\because$ $a^3 - b^3 + 3a^2b + 3ab^2 = (a - b)^3$$\big]$
$= (a - b)^3 + 2^3 = (a - b + 2)((a - b)^2 - (a - b)^2 + 2^2)$
$\big[\because$ $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$\big]$
$= (a - b + 2)(a^2 + b^2 - 2ab - 2a + 2b + 4)$
$\therefore$ $a^3 - 3a^2b + 3ab^2 - b^3 + 8$
$= (a - b + 2)(a^2 + b^2 - 2ab - 2a + 2b + 4)$
View full question & answer→Question 253 Marks
Factorize:
$9(2a - b)^2 - 4(2a - b) - 13$
AnswerLet $2a - b = x = 9x^2 - 4x - 13$
Splitting the middle term,
$= 9x^2 - 13x + 9x - 13$
$= x(9x - 13) + 1(9x - 13)$
$= (9x - 13)(x + 1)$
Substituting$ x = 2a - b = [9(2a - b) - 13](2a - b + 1) $
$= (18a - 9b - 13)(2a - b + 1)$
$\therefore$ $9(2a - b)^2 - 4(2a - b) - 13$
$= (18a - 9b - 13)(2a - b + 1)$
View full question & answer→Question 263 Marks
Factorize the following expressions:$8x^3 - 125y^3 + 180xy + 216$
Answer$8x^3 - 125y^3 + 180xy + 216 or, 8x^3 - 125y^3+ 216 + 180xy$
$= (2x)^3 + (-5y)^3 + 6^3 - 3 \times (2x)(-5y)(6)$
$= (2x + (-5y) + 6)((2x)^2 + (-5y)^2 +6^2 - 2x(-5y) - (-5y)6 - 6(2x))$
$= (2x - 5y + 6)(4x^2 + 25y^2 + 36 + 10xy + 30y -12x)$
$\therefore$$ 8x^3 - 125y^3 + 180xy + 216$
$= (2x - 5y + 6)(4x^2 + 25y^2 + 36 + 10xy + 30y -12x)$
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:
$125 + 8x^3 - 27y^3 + 90xy$
Answer$125 + 8x^3 - 27y^3 + 90xy$
$= 5^3 + (2x)^3 + (-3y)^3 - 3 \times 5 \times 2x \times (-3y)$
$= (5 + 2x + (-3y))(5^2 + (2x)^2 + (-3y)^2 - 5(2x) - 2x(-3y) - (-3y)5)$
$= (5 + 2x + -3y)(25 + 4x^2 + 9y^2 - 10x + 6xy + 15y)$
$\therefore$ $125 + 8x^3 - 27y^3 + 90xy$
$= (5 + 2x + -3y)(25 + 4x^2 + 9y^2 - 10x + 6xy + 15y)$
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$$=\big(\sqrt{2}\text{a}\big)^2+2\times\sqrt{2}\text{a}\times\sqrt{3}\text{b}+\big(\sqrt{3}\text{b}\big)^2$
Using the identity $(p + q)^2 = p^2 + q^2 + 2pq$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)^2$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)$
$\therefore2\text{a}^2+2\sqrt{6}\text{ab}+3\text{b}^2$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}\big)$
View full question & answer→Question 303 Marks
Factorize the following expressions:$\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\text{a}+\frac{2}{\text{a}}$
Answer$=\Big(\text{a}^3-\frac{1}{\text{a}^3}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$$=\Big(\text{a}^3-\Big(\frac{1}{\text{a}^3}\Big)\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\text{a}\times\frac{1}{\text{a}}+\Big(\frac{1}{\text{a}}\Big)^2\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$\big[\therefore$ $a^3 - b^3 = (a - b)(a^2+ ab + b^2)$$\big]$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}\Big)-2\Big(\text{a}-\frac{1}{\text{a}}\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+1+\frac{1}{\text{a}^2}-2\Big)$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
$\therefore\text{a}^3-\frac{1}{\text{a}^3}-2\text{a}+2\text{a}$
$=\Big(\text{a}-\frac{1}{\text{a}}\Big)\Big(\text{a}^2+\frac{1}{\text{a}^2}-1\Big)$
View full question & answer→Question 313 Marks
Factorize:
$4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2$
Answer$4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^{2}$
Let $(x - y) = x,(x + y) = y = 4x^2 - 12xy + 9y^2$
Splitting the middle term $-12 = -6 - 6$
also$ 4 \times 9 = -6 \times -6 = 4x^2 - 6xy - 6xy + 9y^2$
$= 2x(2x - 3y) - 3y(2x - 3y)$
$= (2x - 3y)(2x - 3y) = (2x - 3y)^2$
Substituting $x = x - y$ & $y$
$= x + y = [2(x - y) - 3(x + y)]^2 = [2x - 2y - 3x - 3y]^2$
$= (2x - 3x - 2y - 3y)² = [-x - 5y]^2$
$= [(-1)(x + 5y)]^2 = (x + 5y)^2 [(-1)^2 = 1]$
$\therefore$ $4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2 = (x + 5y)^2$
View full question & answer→Question 323 Marks
Simplify:$\frac{173\times173\times173+127\times127\times127}{173\times173-173\times127+127\times127}$
Answer$\frac{173\times173\times173+127\times127\times127}{173\times173-173\times127+127\times127}$$=\frac{173^3+127^3}{173^2-173\times127+127^2}$
$=\frac{(173+127)(173^2-173\times127+127^2)}{173^2-173\times127+127^2}$
$\big[\therefore$ $a^3 + b^3 = (a + b)(a^2 − ab + b^2)$$\big]$
$= (173 + 127)$
$= 300$
View full question & answer→Question 333 Marks
Simplify:$\frac{1.2\times1.2\times1.2-0.2\times0.2\times0.2}{1.2\times1.2+1.2\times0.2+0.2\times0.2}$
Answer$\frac{1.2\times1.2\times1.2-0.2\times0.2\times0.2}{1.2\times1.2+1.2\times0.2+0.2\times0.2}$$=\frac{1.2^3-0.23}{1.2^2+1.2\times0.2+0.2^2}$
$=\frac{(1.2-0.2)((1.2)^2+1.2\times0.2+(0.2)^2)}{1.2^2+1.2\times0.2+0.2^2}$
$\big[\therefore$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2$)$\big]$
$= (1.2 - 0.2)$
$= 1.0$
View full question & answer→Question 343 Marks
Factorize:$21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer$21\text{x}^2-2\text{x}+\frac{1}{21}$
$=\big(\sqrt{21\text{x}}\big)^2-2\sqrt{21}\text{x}\times\frac{1}{\sqrt{21}}+\Big(\frac{1}{\sqrt{21}}\Big)^2$
Using the identity $(x - y)^2 = x^2 + y^2 - 2xy$
$\Big(\sqrt{21}\text{x}-\frac{1}{\sqrt{21}}\Big)^2$
$\therefore21\text{x}^2-2\text{x}+\frac{1}{21}=\Big(\sqrt{21}\text{x}-\frac{1}{\sqrt{21}}\Big)^2$
View full question & answer→Question 353 Marks
Factorize the following expressions:
$8a^3 - b^3 - 4ax + 2bx$
Answer$= (2a)^3 - b^3 - 2x(2a - b)$
$= (2a - b)((2a)^2 + 2a \times b + b^2) - 2x(2a - b)$
$\big[\therefore$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$\big]$
$= (2a - b)(4a^2 + 2ab + b^2- 2x)$
$\therefore$ $8a^3 - b^3 - 4ax + 2bx$
$ = (2a - b)(4a^2 + 2ab + b^2 - 2x)$
View full question & answer→Question 363 Marks
Multiply:
$(9x^2 + 25y^2 + 15xy + 12x - 20y + 16) by (3x - 5y + 4)$
Answer$= (3x - 5y + 4)(9x^2 + 25y^2 + 15xy + 20y - 12x + 16)$
$= (3x + (5y) + 4){(3x)^2 + (-5y)^2+ 4^2 - 3x(-5y) - (-5y)4 - 4(3x)}$
$\big[\because$ $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)= a^3 + b^3 + c^3 - 3abc$$\big]$
Here, $a = 3x, b = -5y, c = 4$
$= (3x)^3 + (-5y)^3 + 4^3 - 3(3x)(-5y)(4)$
$= 27x^3 - 125y^3 + 64 + 180xy$
$\therefore$$ (3x - 5y + 4)(9x^2 + 25y^2 + 15xy + 20y - 12x + 16)$
$= 27x^3 - 125y^3 + 64 + 180xy$
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Factorize the following expressions:
$54x^6y + 2x^3y^4$
Answer$54x^6y + 2x^3y^4 = 2x^3y(27x^3 + y^3)$
$= 2x^3y((3x)^3+ y^3) = 2x^3y(3x + y)((3x)^2 - 3x \times y + y^2)$
$\therefore$ $[a^3+ b^3 = (a + b)(a^2 - ab + b^2)]$
$= 2x^3y(3x + y)(9x^2 - 3xy + y^2)$
$\therefore$ $54x^6y + 2x^3y^4= 2x^3y(3x + y)(9x^2 - 3xy + y^2)$
View full question & answer→Question 383 Marks
Factorize the following expressions:
$x^3 y^3+1$
Answer$x^3 y^3+1=(x y)^3+1^3$
$=(x y+1)\left((x y)^2+x y+1^2\right)\left[\therefore x^3+y^3=(x+y)\left(x^2-x y+y^2\right)\right]$
$=(x y+1)\left(x^2 y^2-x y+1\right) \therefore x^3 y^3+1=(x y+1)\left(x^2 y^2-x y+1\right)$
View full question & answer→Question 393 Marks
Factorize the following expressions:
$(x + 2)^3 + (x - 2)^3$
Answer$= (x + 2 + x - 2)((x + 2)^2- (x + 2)(x - 2) + (x - 2)^2)$
$\big[\therefore$ [$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$\big]$
$= 2x(x^2 + 4x + 4 - (x + 2)(x - 2) + x^2 - 4x + 4)$
$= 2x(2x^2 + 8 - (x^2 - 2^2))$
$\big[\therefore$ $(a + b)(a - b) = a^2 - b^2$$\big]$
$= 2x(2x^2 + 8 - x^2 + 4)$
$= 2x(x^2 + 12)$
$\therefore$ $(x + 2)^3 + (x - 2)^3 = 2x(x^2 + 12)$
View full question & answer→Question 403 Marks
Factorize the following expressions:
$8x^2y^3- x^5$
Answer$8x^2y^3- x^5$
$= x^2((2y)^3 - x^3)$
$= x^2(2y - x)((2y)^2 + 2y \times x + x^2)$
$\big[\therefore$ $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$\big]$
$= x^2(2y - x)(4y^2 + 2xy + x^2)$
$\therefore$ $8x^2y^3 - x^5 = x^2(2y - x)(4y^2 + 2xy + x^2)$
View full question & answer→Question 413 Marks
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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