Question 12 Marks
Factorize:$\text{x}^2-\sqrt{3}\text{x}-6$
Answer$\text{x}^2-\sqrt{3}\text{x}-6$Splitting the middle term,
$=\text{x}^2-2\sqrt{3}\text{x}+\sqrt{3}\text{x}-6$
$\big[\therefore-\sqrt{3}=-2\sqrt{3}+\sqrt{3} \ \text{also} \ -2\sqrt{3}\times\sqrt{3}=-6\big]$
$=\text{x}\big(\text{x}-2\sqrt{3}\big)+\sqrt{3}\big(\text{x}-2\sqrt{3}\big)$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
$\therefore\text{x}^2-\sqrt{3}\text{x}-6$
$=\big(\text{x}-2\sqrt{3}\big)\big(\text{x}+\sqrt{3}\big)$
View full question & answer→Question 22 Marks
Factorize:
$2(x+y)^2-9(x+y)-5$
AnswerLet $x+y=z$
$=2 z^2-9 z-5$
Splitting the middle term,
$=2 z^2-10 z+z-5$
$=2 z(z-5)+1(z-5)$
$=(z-5)(2 z+1)$
Substituting $z=x+y$
$=(x+y-5)(2(x+y)+1)$
$=(x+y-5)(2 x+2 y+1)$
$\therefore 2(x+y)^2-9(x+y)-5$
$=(x+y-5)(2 x+2 y+1)$
View full question & answer→Question 32 Marks
Factorize:$\text{x}^2+2\sqrt{3}\text{x}-24$
Answer$\text{x}^2+2\sqrt{3}\text{x}-24$Splitting the middle term,
$=\text{x}^2+4\sqrt{3}\text{x}-2\sqrt{3}\text{x}-24$
$\big[\therefore2\sqrt{3}=4\sqrt{3}-2\sqrt{3} \ \text{also} \ 4\sqrt{3}\big(-2\sqrt{3}\big)=-24\big]$
$=\text{x}\big(\text{x}+4\sqrt{3}\big)-2\sqrt{3}\big(\text{x}+4\sqrt{3}\big)$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
$\therefore\text{x}^2+2\sqrt{3}\text{x}-24$
$=\big(\text{x}+4\sqrt{3}\big)\big(\text{x}-2\sqrt{3}\big)$
View full question & answer→Question 42 Marks
If $a + b + c = 0$, then write the value of $a^3 + b^3 + c^3.$
AnswerRecall the formula
$a^3 + b^3 + c^3 - 3abc$ =$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
When, $(a + b + c) = 0,$ we have
$a^3 + b^3 + c^3 - 3abc = 0$.$(a^2 + b^2 + c^2 - ab - bc - ca)$
$= 0$
$a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$
View full question & answer→Question 52 Marks
Factorize:
$a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3$
Answer$a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3$
$= (ax)^3 - 3(ax)^2 \times b + 3(ax)b^2 - b^3$
$= (ax - b)^3$ $\big[\because$ $a^3 - 3a^2b + 3ab^2 - b^3= (a - b)^3$$\big]$
$= (ax - b)(ax - b)(ax - b)$$\therefore$ $a^3x^3 - 3a^2bx^2 + 3ab^2x - b^3$
$= (ax - b)(ax - b)(ax - b)$
View full question & answer→Question 62 Marks
Factorize the following expressions:
$(2x - 3y)^3 + (4z - 2x)^3+ (3y - 4z)^3$
Answer$(2x - 3y)^3 + (4z - 2x)^3+ (3y - 4z)^3$
Let $2x - 3y = a, 4z - 2x = b, 3y - 4z = c$
$\therefore$ $a + b + c = 2x - 3y + 4z - 2x + 3y - 4z = 0$
$\because$ $a + b + c = 0$
$\therefore$ $a^3 + b^3 + c^3 = 3abc$
$\therefore$ $(2x - 3y)^3 + (4z - 2x)^3+ (3y - 4z)^3$
$= 3(2x - 3y)(4z - 2x)(3y - 4z)$
View full question & answer→Question 72 Marks
Factorize the following expressions:$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$
Answer$2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$$=\big(\sqrt{2}\text{a}\big)^3+\big(\sqrt{3}\text{b}\big)^3+\text{c}^3-3\times\sqrt{2}\text{a}\times\sqrt{3}\text{b}\times\text{c}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\Big(\big(\sqrt{2}\text{a}\big)^2+\big(\sqrt{3}\text{b}\big)^2+\\\text{c}^2-(\sqrt{2}\text{a}\big)(\sqrt{3}\text{b}\big)-(\sqrt{3}\text{b}\big)\text{c}-(\sqrt{2}\text{a}\big)\text{c}\Big)$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6}\text{ab}-\sqrt{3}\text{bc}-\sqrt{2}\text{ac}\big)$
$\therefore2\sqrt{2}\text{a}^3+3\sqrt{3}\text{b}^3+\text{c}^3-3\sqrt{6}\text{abc}$
$=\big(\sqrt{2}\text{a}+\sqrt{3}\text{b}+\text{c}\big)\big(2\text{a}^2+3\text{b}^2+\text{c}^2-\sqrt{6}\text{ab}-\sqrt{3}\text{bc}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 82 Marks
Give the possible expression for the length & breadth of the rectangle having $35y^2 - 13y - 12$ as its area.
AnswerArea is given as $35y^2 - 13y - 12$ Splitting the middle term,
Area = $35y^2+ 218y - 15y - 12$
$= 7y(5y + 4) - 3(5y + 4)$
$= (5y + 4)(7y - 3)$
We also know that area of rectangle = length \times breadth
$\therefore$ Possible length =$(5y + 4)$ and breadth = $(7y - 3)$ Or
possible length = $(7y - 3)$ and breadth = $(5y + 4)$
View full question & answer→Question 92 Marks
Factorize:
$64a^3 + 125b^3 + 240a^2b + 300ab^2$
Answer$64 a^3+125 b^3+240 a^2 b+300 a b^2$
$=(4 a)^3+(5 b)^3+3 \times(4 a)^2 \times 5 b+3(4 a)(5 b)^2$
$=(4 a+5 b)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=(4 a+5 b)(4 a+5 b)(4 a+5 b)$
$\therefore 64 a^3+125 b^3+240 a^2 b+300 a b^2$
$=(4 a+5 b)(4 a+5 b)(4 a+5 b)$
View full question & answer→Question 102 Marks
Multiply:
$\left(x^2+4 y^2+z^2+2 x y+x z-2 y z\right)$ by $(x-2 y-z)$
Answer$=(x-2 y-z)\left(x^2+4 y^2+z^2+2 x y+x z-2 y z\right)$
$=(x+(-2 y)+(-z))\left(x^2+(-2 y)^2+(-z)^2-x(-2 y)-(-2 y)(-z)-(-z) x\right)$
$=x^3+(-2 y)^3+(-z)^3-3 \times x(-2 y)(-z)\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]$
$=x^3-8 y^3-z^3+3 \times x \times 2 y z$
$=x^3-8 y^3-z^3-6 x y z$
View full question & answer→Question 112 Marks
Factorize:
$8x^3 + 27y^3 + 36x^2y + 54xy^2$
Answer$8 x^3+27 y^3+36 x^2 y+54 x y^2$
$=(2 x)^3+(3 y)^3+3 \times(2 x)^2 \times 3 y+3 \times(2 x)(3 y)^2$
$=(2 x+3 y)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=(2 x+3 y)(2 x+3 y)(2 x+3 y)$
$\therefore 8 x^3+27 y^3+36 x^2 y+54 x y^2$
$=(2 x+3 y)(2 x+3 y)(2 x+3 y)$
View full question & answer→Question 122 Marks
Factorize the following expressions:
$1 - 27a^3$
Answer$1 - 27a^3$
$= (1)^3 - (3a)^{3}$
$= (1 - 3a)(1^2 + 1 \times 3a + (3a)^2)$
$\therefore$ $[a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= (1 - 3a)(1^2 + 3a + 9a^2)$
$\therefore$ $1 - 27a^3 = (1 - 3a)(1^2 + 3a + 9a^2)$
View full question & answer→Question 132 Marks
Factorize the following expressions:
$p^3 + 27$
Answer$p^3+27=p^3+3^3$
$\therefore\left[a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=(p+3)\left(p^2-3 p-9\right)$
$\therefore p^3+27=(p+3)\left(p^2-3 p-9\right)$
View full question & answer→Question 142 Marks
Factorize:$\text{x}^2+5\sqrt{5}\text{x}+30$
Answer$\text{x}^2+5\sqrt{5}\text{x}+30$Splitting the middle term,
$=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$
$\big[\therefore5\sqrt{5}=2\sqrt{5}+3\sqrt{5} \ \text{also} \ 2\sqrt{5}\times3\sqrt{5}=30\big]$
$=\text{x}\big(\text{x}+2\sqrt{5}\big)+3\sqrt{5}\big(\text{x}+2\sqrt{5}\big)$
$=\big(\text{x}+2\sqrt{5}\big)\big(\text{x}+3\sqrt{5}\big)$
$\therefore\text{x}^2+5\sqrt{5}\text{x}+30$
$=\big(\text{x}+2\sqrt{5}\big)\big(\text{x}+3\sqrt{5}\big)$
View full question & answer→Question 152 Marks
Factorize the following expressions:$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer$2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$$=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+(\text{c})^3-3\times\sqrt{2}\text{a}\times2\sqrt{2}\text{b}\times\text{c}$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b})+\text{c}\big)\Big(\big(\sqrt{2}\text{a}\big)^2+\big(2\sqrt{2}\text{b})^2+\text{c}^2\\\big(\sqrt{2}\text{a}\big)\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\text{c}-\big(\sqrt{2}\text{a}\big)\text{c}\Big)$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2}\text{bc}-\sqrt{2}\text{ac}\big)$
$\therefore2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
$=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2}\text{bc}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 162 Marks
Factorize:
$8a^3 + 27b^3 + 36a^2b + 54ab^2$
Answer$8 a^3+27 b^3+36 a^2 b+54 a b^2$
$=(2 a)^3+(3 b)^3+3 \times(2 a)^2 \times 3 b+3 \times(2 a)(3 b)^2$
$=(2 a+3 b)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=(2 a+3 b)(2 a+3 b)(2 a+3 b)$
$\therefore 8 a^3+27 b^3+36 a^2 b+54 a b^2$
$=(2 a+3 b)(2 a+3 b)(2 a+3 b)$
View full question & answer→Question 172 Marks
Factorize:
$8a^3 - 27b^3 - 36a^2b + 54ab^2$
Answer$8 a^3-27 b^3-36 a^2 b+54 a b^2$
$=(2 a)^3-(3 b)^3-3 \times(2 a)^2 \times 3 b+3 \times(2 a)(3 b)^2$
$=(2 a-3 b)^3\left[\because a^3-b^3-3 a^2 b+3 a b^2=(a-b)^3\right]$
$=(2 a-3 b)(2 a-3 b)(2 a-3 b)$
$\therefore 8 a^3-27 b^3-36 a^2 b+54 a b^2$
$=(2 a-3 b)(2 a-3 b)(2 a-3 b)$
View full question & answer→Question 182 Marks
Factorize the following expressions:
$x^3 - 8y^3 + 27z^3 + 18xyz$
Answer$x^3-8 y^3+27 z^3+18 x y z$
$=x^3+(-2 y)^3+(3 z) 3-3 \times x \times(-2 y)(3 z)$
$=(x+(-2 y)+3 z)\left(x^2+(-2 y)^2+(3 z)^2-x(-2 y)-(-2 y)(3 z)-3 z(x)\right)\left[\because a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-\right.\right.$
$b c-c a)]$
$=(x-2 y+3 z)\left(x^2+4 y^2+9 z^2+2 x y+6 y z-3 z x\right)$
$\therefore x^3-8 y^3+27 z^3+18 x y z=(x-2 y+3 z)\left(x^2+4 y^2+9 z^2+2 x y+6 y z-3 z x\right)$
View full question & answer→Question 192 Marks
Factorize:
$8x^3 + y^3 + 12x^2y + 6xy^2$
Answer$8 x^3+y^3+12 x^2 y+6 x y^2$
$=(2 x)^3+y^3+3 \times(2 x)^2 \times y+3(2 x) \times y^2$
$=(2 x+y)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=(2 x+y)(2 x+y)(2 x+y)$
$\therefore 8 x^3+y^3+12 x^2 y+6 x y^2$
$=(2 x+y)(2 x+y)(2 x+y)$
View full question & answer→Question 202 Marks
Factorize the following expressions:
$27x^3 - y^3 - z^3 - 9xyz$
AnswerWe know that $x^3 + y^3 + z^3 - 3xyz$
$= ( x + y + z)(x^2 + y^2 + z^2 - xy - yz -zx)$
$\therefore$ $27x^3 - y^3 - z^3 - 9xyz$
$= (3x)^3 + (-y)^3 + (-z)^3 - 3(3x)(-y)(-z)$
$= [3x +(-y) + (-z)][(3x)^2 + (-y)^2 + (-z)^2 - (3x)(-y)(-z)-(-z)(3x)]$
$= (3x - y - z)(9x^2 + y^2 + z^2 + 3xy - yz + 3zx)$
View full question & answer→Question 212 Marks
Factorize the following expressions:
$10x^4y - 10xy^4$
Answer$10x^4y - 10xy^4$
$= 10xy(x^3 - y^3)$
$= 10xy(x - y)(x^2 + xy + y^2)$
$\therefore$ $[x^3 - y^3= (x - y)(x^2 + xy + y^2)]$
$\therefore$ $10x^4y - 10xy^4 = 10xy(x - y)(x^2 + xy + y^2)$
View full question & answer→Question 222 Marks
Factorize the following expressions:
$y^3 + 125$
Answer$y^3 + 125 = y^3 + 5^3$
$\therefore$ $[a^3 + b^3 = (a + b)(a^2 - ab + b^2)]$
$= (y + 5)(y^2- 5y + 5^2)$
$= (y + 5)(y^2 - 5y + 25)$
$\therefore$ $y^3 + 125= (y + 5)(y^2 - 5y + 25)$
View full question & answer→Question 232 Marks
Factorize the following expressions:
$64a^3- b^3$
Answer$64a^3- b^3$
$= (4a)^3 - b^3$
$= (4a - b)((4a)^2 + 4a \times b + b^2)$
$\therefore$ $[a^3- b^3 = (a - b)(a^2 + ab + b^2)]$
$= (4a − b)(16a^2 + 4ab + b^2)$
$\therefore$ $64a^3 - b^3 = (4a - b)(16a^2+ 4ab + b^2)$
View full question & answer→Question 242 Marks
Factorize:$\text{x}^2+6\sqrt{2}\text{x}+10$
Answer$\text{x}^2+6\sqrt{2}\text{x}+10$Splitting the middle term,
$=\text{x}^2+5\sqrt{2}\text{x}+\sqrt{2}\text{x}+10$
$\big[\therefore6\sqrt{2}=5\sqrt{2}+\sqrt{2} \ \text{and} \ 5\sqrt{2}\times\sqrt{2}=10\big]$
$=\text{x}\big(\text{x}+5\sqrt{2}\big)+\sqrt{2}\big(\text{x}+5\sqrt{2}\big)$
$=\big(\text{x}+5\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)$
$\therefore\text{x}^2+6\sqrt{2}\text{x}+10$
$=\big(\text{x}+5\sqrt{2}\big)\big(\text{x}+\sqrt{2}\big)$
View full question & answer→Question 252 Marks
Factorize the following expressions:
$8x^3 + 27y^3 - 216z^3 + 108xyz$
Answer$8 x^3+27 y^3-216 z^3+108 x y z$
$=(2 x)^3+(3 y)^3+(-6 z)^3-3(2 x)(3 y)(-6 z)$
$=(2 x+3 y-6 z)\left((2 x)^2+(3 y)^2+(-6 z)^2-2 x \times 3 y-3 y(-6 z)-(-6 z) 2 x\right)$
$ =(2 x+3 y-6 z)\left(4 x^2+9 y^2+36 z^2-6 x y+18 y z+12 z x\right)$
$\therefore 8x^3 + 27y^3 - 216z^3 + 108xyz$
$=(2 x+3 y-6 z)\left(4 x^2+9 y^2+36 z^2-6 x y+18 y z+12 z x\right)$
View full question & answer→Question 262 Marks
Factorize:
$125x^3 - 27y^3 - 225x^2y + 135xy^2$
Answer$125 x^3-27 y^3-225 x^2 y+135 x y^2$
$=(5 x)^3-(3 y)^3-3 \times(5 x)^2 \times 3 y+3 \times(5 x)(3 y)^2$
$=(5 x-3 y)^3\left[\because a^3-b^3-3 a^2 b+3 a b^2=(a-b)^3\right]$
$=(5 x-3 y)(5 x-3 y)(5 x-3 y)$
$\therefore 125 x^3-27 y^3-225 a^2 y+135 x y^2$
$=(5 x-3 y)(5 x-3 y)(5 x-3 y)$
View full question & answer→Question 272 Marks
Multiply:
$\left(x^2+4 y^2+2 x y-3 x+6 y+9\right) \text { by }(x-2 y+3)$
Answer$=(x-2 y+3)\left(x^2+4 y^2+9+2 x y+6 y-3 x\right)$
$=(x+(-2 y)+3)\left(x^2+(-2 y)^2+3^2-x(-2 y)-(-2 y) 3-3 x\right)$
$=x^3+(-2 y)^3+3^3-3 x x(-2 y)^3\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3-8 y^3+27+18 x y$
View full question & answer→Question 282 Marks
Factorize the following expressions:
$(3x - 2y)^3 + (2y - 4z)^3+ (4z - 3x)^3$
Answer$(3 x-2 y)^3+(2 y-4 z)^3+(4 z-3 x)^3$
Let $(3 x-2 y)=a,(2 y-4 z)=b,(4 z-3 x)=c$
$\therefore a+b+c=3 x-2 y+2 y-4 z+4 z-3 x=0$
$\because a+b+c=0$
$\therefore a^3+b^3+c^3+3 a b c$
$\therefore(3 x-2 y)^3+(2 y-4 z)^3+(4 z-3 x)^3$
$=3(3 x-2 y)(2 y-4 z)(4 z-3 x)$
View full question & answer→Question 292 Marks
Factorize the following expressions:
$(a - 2b)^3 - 512b^3$
Answer$(a - 2b)^3 - 512b^{3}$
$= (a - 2b)^3 - (8b)^3$
$= (a - 2b - 8b)((a - 2b)^2 + (a - 2b)8b + (8b)^2)$\therefore$ [a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= (a - 10b)(a^2 + 4b^2 - 4ab + 8ab - 16b^2 + 64b^2)$
$= (a - 10b)(a^2 + 52b^2 + 4ab)$
$\therefore$ $(a - 2b)^3 − 512b^3$
$= (a - 10b)(a^2 + 52b^2 + 4ab)$
View full question & answer→Question 302 Marks
Factorize the following expressions:$\frac{\text{x}^3}{216}=8\text{y}^3$
Answer$\frac{\text{x}^3}{216}-8\text{y}^3$$=\frac{\text{x}^3}{6}-(2\text{y})^3$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\Big(\frac{\text{x}}{6}\Big)^2+\frac{\text{x}}{6}\times2\text{y}+(2\text{y})^2\Big)$
$\therefore\big[\text{x}^3-\text{y}^3=(\text{x}-\text{y})(\text{x}^2+\text{xy}+\text{y}^2)\big]$
$=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}^2}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
$\therefore\frac{\text{x} ^3}{216}-8\text{y} ^3=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
View full question & answer→Question 312 Marks
Factorize the following expressions:
$(a-3 b)^3+(3 b-c)^3+(c-a)^3$
Answer$(a-3 b)^3+(3 b-c)^3+(c-a)^3 \text { Let }(a-3 b)=x,(3 b-c)=y,(c-a)=z$
$x+y+z=a-3 b+3 b-c+c-a=0$
$\because x+y+z=0$
$\therefore x^3+y^3+z^3=3 x y z$
$\therefore(a-3 b)^3+(3 b-c)^3+(c-a)^3$
$=3(a-3 b)(3 b-c)(c-a)$
View full question & answer→Question 322 Marks
If $a + b + c = 9$ and $ab + bc + ca = 40$, find $a^2 + b^2 +c^2.$
AnswerRecall the formula $(a + b + c)^2$
$= a^2 + b^2 + c^2 + 2(ab + bc + ca)$
.Given that $(a + b + c) = 9$, $ab + bc + ca = 40,$
Then we have $(a + b + c)^2$
$= a^2 + b^2 + c^2 + 2(ab + bc + ca) (9)^2$
$= a^2 + b^2 + c^2 + 2.(40) a^2 + b^2 + c^2 + 80$
$= 81 a^2 + b^2 + c^2$
$= 81 - 80 a^2 + b^2 + c^2$
$= 1$
View full question & answer→Question 332 Marks
Factorize:$\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
Answer$\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$$=\Big(\frac{2}{3}\text{x}\Big)^3+(1)^3+3\times\Big(\frac{2}{3}\text{x}\Big)^2\times1+3(1)^2\times\Big(\frac{2}{3}\text{x}\Big)$
$=\Big(\frac{2}{3}\text{x}+1\Big)^3$$\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
$\therefore\frac{8}{27}\text{x}^3+1+\frac{4}{3}\text{x}^2+2\text{x}$
$=\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)\Big(\frac{2}{3}\text{x}+1\Big)$
View full question & answer→Question 342 Marks
Factorize the following expressions:$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
Answer$3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$$=\big(\sqrt{3}\text{a}\big)^3+(-\text{b})^3+\big(-\sqrt{5}\text{c}\big)^3-3\times\big(\sqrt{3}\text{a}\big)(-\text{b})\big(-\sqrt{5}\text{c}\big)$
$=\big(\sqrt{3}\text{a}+(-\text{b})+\big(-\sqrt{5}\text{c}\big)\big)\Big(\big(\sqrt{3}\text{a}\big)^2+(-\text{b})^2+\big(-\sqrt{5}\text{c}\big)^2\\-\sqrt{3}\text{a}(-\text{b})-(-\text{b})\big(-\sqrt{5}\text{c}\big)-\big(-\sqrt{5}\text{c}\big)\sqrt{3}\text{a}\Big)$
$=\big(\sqrt{3}\text{a}+(-\text{b})-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{ac})$
$\therefore3\sqrt{3}\text{a}^3-\text{b}^3-5\sqrt{5}\text{c}^3-3\sqrt{15}\text{abc}$
$=\big(\sqrt{3}\text{a}+(-\text{b})-\sqrt{5}\text{c}\big)\big(3\text{a}^2+\text{b}^2+5\text{c}^2+\sqrt{3}\text{ab}-\sqrt{5}\text{bc}+\sqrt{15}\text{ac})$
View full question & answer→Question 352 Marks
Factorize:
$x^4 + x^2y^2 + y^4$
Answer$x^4+x^2 y^2+y^4$
Adding $x^2 y^2$ and subtracting $x^2 y^2$ to the given equation
$=x^4+x^2 y^2+y^4+x^2 y^2-x^2 y^2$
$=x^4+2 x^2 y^2+y^4-x^2 y^2$
$=\left(x^2\right)^2+2 \times x^2 \times y^2+\left(y^2\right)^2-(x y)^2$
Using the identity $(p+q)^2=p^2+q^2+2 p q$
$=\left(x^2+y^2\right)^2-(x y)^2$
Using the identity $p^2-q^2=(p+q)(p-q)$
$=\left(x^2+y^2+x y\right)\left(x^2+y^2-x y\right)$
$\therefore x^4+x^2 y^2+y^4=\left(x^2+y^2+x y\right)\left(x^2+y^2-x y\right)$
View full question & answer→Question 362 Marks
Factorize:
$a^2 + b^2 + 2(ab + bc + ca)$
Answer$= a^2 + b^2 + 2ab + 2bc + 2ca$
Using the identity $(p + q)^2 = p^2 + q^2 + 2pq$
We get, $= (a + b)^2 + 2bc + 2ca$
$= (a + b)^2 + 2c(b + a) Or (a + b)^2 + 2c(a + b)$
Taking $(a + b)$ common = $(a + b)(a + b + 2c)$
$\therefore$ $a^2+ b^2 + 2(ab + bc + ca) = (a + b)(a + b + 2c)$
View full question & answer→Question 372 Marks
Factorize:
$x^3-12 x(x-4)-64$
Answer$x^3-12 x(x-4)-64$
$=x^3-12 x^2+48 x-64$
$=(x)^3-3 \times x^2 \times 4+3 \times 4^2 \times x-4^3$
${\left[\because a^3-3 a^2 b+3 a b^2-b^3=(a-b)^3\right]}$
$=(x-4)(x-4)(x-4)$
$\therefore x^3-12 x(x-4)-64$
$=(x-4)(x-4)(x-4)$
View full question & answer→Question 382 Marks
Factorize:
$a^2-b^2+2 b c-c^2$
Answer$a^2-b^2+2 b c-c^2 a^2-\left(b^2-2 b c+c^2\right)$
Using the identity $(a-b)^2$
$=a^2+b^2-2 a b$
$=a^2-(b-c)^2$
Using the identity $a^2-b^2$
$=(a+b)(a-b)$
$=(a+b-c)(a-(b-c))$
$=(a+b-c)(a-b+c)$
$\therefore a^2-b^2+2 b c-c^2$
$=(a+b-c)(a-b+c)$
View full question & answer→Question 392 Marks
What are the possible expression for the cuboid having volume $3 x^2-12 x$.
AnswerVolume $=3 x^2-12 x$
$=3 x(x-4)$
$=3 \times x(x-4)$
Also volume $=$ Length $\times$ Breadth $\times$ Height
$\therefore$ Possible expression for dimensions of cuboid are $=3, x,(x-4)$
View full question & answer→Question 402 Marks
If $a^2+b^2+c^2=20$ and $a+b+c=0$, find $a b+b c+c a$.
AnswerRecall the formula
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$a^2+b^2+c^2=20$
$(a+b+c)=0$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(0)^2=20+2(a b+b c+c a)$
$20+2(a b+b c+c a)=0$
$2(a b+b c+c a)=-20$
$(a b+b c+c a)=-10$
View full question & answer→Question 412 Marks
Factorize the following expressions:
$x^6+y^6$
Answer$=\left(x^2\right)^3+\left(y^2\right)^3$
$\left.=\left(x^2+y^2\right)\left(x^2\right)^2-x^2 y^2+\left(y^2\right)^2\right)$
$=\left(x^2+y^2\right)\left(x^4-x^2 y^2+y^4\right)\left[\therefore a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$\therefore x^6+y^6=\left(x^2+y^2\right)\left(x^4-x^2 y^2+y^4\right)$
View full question & answer→Question 422 Marks
Factorize the following expressions:
$a^3+8 b^3+64 c^3-24 a b c$
Answer$a^3 + 8b^3 + 64c^3 - 24abc$
$= (a)^3 + (2b)^3 + (4c)^3 - 3 \times 2b \times 4c$
$= (a + 2b + 4c)(a^2 + (2b)^2 + (4c)^2 - a \times 2b - 2b \times 4c - 4c \times a)$
$\big[\because$ $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca$)$\big]$
$= (a + 2b + 4c)(a^2 + 4b^2+ 16c^2 - 2ab - 8bc - 4ac)$
$\therefore$ $a^3 + 8b^3 + 64c^3 - 24abc$
$= (a + 2b + 4c)(a^2 + 4b^2 + 16c^2 - 2ab - 8bc - 4ac)$
View full question & answer→Question 432 Marks
Factorize the following expressions:
$8 x^3 y^3+27 a^3$
Answer$8 x^3 y^3+27 a^3$
$=(2 x y)^3+(3 a)^3$
$=(2 x y+3 a)\left((2 x y)^2-2 x y \times 3 a+(3 a)^2\right)$
$\therefore\left[a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]=(2 x y+3 a)\left(4 x^2 y^2-6 x y a+9 a^2\right)$
$\therefore 8 x^3 y^3+27 a^3=(2 x y+3 a)\left(4 x^2 y^2-6 x y a+9 a^2\right)$
View full question & answer→Question 442 Marks
Factorize:$\text{x}^2-2\sqrt{2}\text{x}-30$
Answer$\text{x}^2-2\sqrt{2}\text{x}-30$Splitting the middle term,
$=\text{x}^2=5\sqrt{2}\text{x}+3\sqrt{2}\text{x}-30$
$\big[\therefore-2\sqrt{2}=-5\sqrt{2}+3\sqrt{2} \ \text{also} \ -5\sqrt{2}\times3\sqrt{2}=-30\big]$
$=\text{x}\big(\text{x}-5\sqrt{2}\big)+3\sqrt{2}\big(\text{x}-5\sqrt{2}\big)$
$=\big(\text{x}-5\sqrt{2}\big)\big(\text{x}+3\sqrt{2}\big)$
$\therefore\text{x}^2-2\sqrt{2}\text{x}-30$
$=\big(\text{x}-5\sqrt{2}\big)\big(\text{x}+3\sqrt{2}\big)$
View full question & answer→Question 452 Marks
Multiply:
$\left(x^2+y^2+z^2-x y+x z+y z\right) \text { by }(x+y-z)$
Answer$\left(x^2+y^2+z^2-x y+x z+y z\right) \text { by }(x+y-z)$
$=(x+y-z)\left(x^2+y^2+z^2-x y+x z+y z\right)$
$=(x+y+(-z))\left(x^2+y^2+(-z)^2-x y-y(-z)-(-z) x\right)$
$=x^3+y^3+(-z)^3-3 x y z(-z)\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3+y^3-z^3+3 x y z$
View full question & answer→Question 462 Marks
Factorize the following expressions:
$a^{12}+b^{12}$
Answer$=\left(a^4\right)^3+\left(b^4\right)^3$
$=\left(a^4+b^4\right)\left(\left(a^4\right)^2-a^4 \times b^4+\left(b^4\right)^2\right)\left[\therefore a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=\left(a^4+b^4\right)\left(a^8-a^4 b^4+b^8\right)$
$\therefore a^{12}+b^{12}=\left(a^4+b^4\right)\left(a^8-a^4 b^4+b^8\right)$
View full question & answer→Question 472 Marks
Factorize:
$x^3 + 8y^3 + 6x^2y + 12xy^2$
Answer$x^3+8 y^3+6 x^2 y+12 x y^2$
$=(x)^3+(2 y)^3+3 \times x^2 \times 2 y+3 \times x \times(2 y)^2$
$=(x+2 y)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=(x+2 y)(x+2 y)(x+2 y)$
$\therefore x^3+8 y^3+6 x^2 y+12 x y^2$
$=(x+2 y)(x+2 y)(x+2 y)$
View full question & answer→Question 482 Marks
If $a^2+b^2+c^2=250$ and $a b+b c+c a=3$, find $a+b+c$.
AnswerRecall the formula
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$a^2+b^2+c^2=250, a b+b c+c a=3$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(a+b+c)^2=250+2 \cdot(3)$
$(a+b+c)^2=256$
$(a+b+c)= \pm 16$
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