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
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
When, $(a+b+c)=0$, we have
$a^3+b^3+c^3-3 a b c=0 \cdot\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=0$
$a^3+b^3+c^3-3 a b c=0$
$\Rightarrow a^3+b^3+c^3=3 a b c$
View full question & answer→Question 52 Marks
Factorize:
$a^3 x^3-3 a^2 b x^2+3 a b^2 x-b^3$
Answer$a^3 x^3-3 a^2 b x^2+3 a b^2 x-b^3$
$=(a x)^3-3(a x)^2 x b+3(a x) b^2-b^3$
$=(a x-b)^3\left[\because a^3-3 a^2 b+3 a b^2-b^3=(a-b)^3\right]$
$=(a x-b)(a x-b)(a x-b)$
$\therefore a^3 x^3-3 a^2 b x^2+3 a b^2 x-b^3$
$=(a x-b)(a x-b)(a x-b)$
View full question & answer→Question 62 Marks
Factorize the following expressions:
$(2 x-3 y)^3+(4 z-2 x)^3+(3 y-4 z)^3$
Answer$(2 x-3 y)^3+(4 z-2 x)^3+(3 y-4 z)^3$
$\text { Let } 2 x-3 y=a, 4 z-2 x=b, 3 y-4 z=c$
$\therefore a+b+c=2 x-3 y+4 z-2 x+3 y-4 z=0$
$\because a+b+c=0$
$\therefore a^3+b^3+c^3=3 a b c$
$\therefore(2 x-3 y)^3+(4 z-2 x)^3+(3 y-4 z)^3$
$=3(2 x-3 y)(4 z-2 x)(3 y-4 z)$
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 $35 y^2-13 y-12$ as its area.
AnswerArea is given as $35 y^2-13 y-12$
Splitting the middle term,
$\text { Area }=35 y^2+218 y-15 y-12$
$=7 y(5 y+4)-3(5 y+4)$
$=(5 y+4)(7 y-3)$
We also know that area of rectangle $=$ length $\times$ breadth
$\therefore$ Possible length $=(5 y+4)$ and breadth $=(7 y-3)$
Or possible length $=(7 y-3)$ and breadth $=(5 y+4)$
View full question & answer→Question 92 Marks
Factorize:
$64 a^3+125 b^3+240 a^2 b+300 a b^2$
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: $(x^2+4 y^2+z^2+2 x y+x z-2 y z)$ 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:
$8 x^3+27 y^3+36 x^2 y+54 x y^2$
Answer$8 x^3+27 y^3+36 x^2 y+54 x y^2$
$=(2 x)^3+(3 y)^3+3 x(2 x)^2 \times 3 y+3 \times(2 x)(3 y)^2$
$=(2 x+3 y)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=(2 x+3 y)(2 x+3 y)(2 x+3 y)$
$\therefore 8 x^3+27 y^3+36 x^2 y+54 x y^2$
$=(2 x+3 y)(2 x+3 y)(2 x+3 y)$
View full question & answer→Question 122 Marks
Factorize the following expressions:
$1-27 a^3$
Answer$1-27 a^3$
$=(1)^3-(3 a)^3$
$=(1-3 a)\left(1^2+1 \times 3 a+(3 a)^2\right)$
$\therefore\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=(1-3 a)\left(1^2+3 a+9 a^2\right)$
$\therefore 1-27 a^3=(1-3 a)\left(1^2+3 a+9 a^2\right)$
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:
$8 a^3+27 b^3+36 a^2 b+54 a b^2$
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:
$8 a^3-27 b^3-36 a^2 b+54 a b^2$
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:
$x^3-8 y^3+27 z^3+18 x y z$
Answer$x^3-8 y^3+27 z^3+18 x y z$
$=x^3+(-2 y)^3+(3 z) 3-3 x x \times(-2 y)(3 z)$
$=(x+(-2 y)+3 z)\left(x^2+(-2 y)^2+(3 z)^2-x(-2 y)-(-2 y)(3 z)-3 z(x)\right)$
${\left[\because a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]}$
$=(x-2 y+3 z)\left(x^2+4 y^2+9 z^2+2 x y+6 y z-3 z x\right)$
$\therefore x^3-8 y^3+27 z^3+18 x y z=(x-2 y+3 z)\left(x^2+4 y^2+9 z^2+2 x y+6 y z-3 z x\right)$
View full question & answer→Question 192 Marks
Factorize:
$8 x^3+y^3+12 x^2 y+6 x y^2$
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: $27 x^3-y^3-z^3-9 x y z$
AnswerWe know that
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$\therefore 27 x^3-y^3-z^3-9 x y z$
$=(3 x)^3+(-y)^3+(-z)^3-3(3 x)(-y)(-z)$
$=[3 x+(-y)+(-z)]\left[(3 x)^2+(-y)^2+(-z)^2-(3 x)(-y)(-z)-(-z)(3 x)\right]$
$=(3 x-y-z)\left(9 x^2+y^2+z^2+3 x y-y z+3 z x\right)$
View full question & answer→Question 212 Marks
Factorize the following expressions:
$10 x^4 y-10 x y^4$
Answer$10 x^4 y-10 x y^4$
$=10 x y\left(x^3-y^3\right)$
$=10 x y(x-y)\left(x^2+x y+y^2\right)$
$\therefore\left[x^3-y^3=(x-y)\left(x^2+x y+y^2\right)\right]$
$\therefore 10 x^4 y-10 x y^4=10 x y(x-y)\left(x^2+x y+y^2\right)$
View full question & answer→Question 222 Marks
Factorize the following expressions:
$y^3+125$
Answer$y^3+125$
$=y^3+5^3$
$\therefore\left[a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=(y+5)\left(y^2-5 y+5^2\right)$
$=(y+5)\left(y^2-5 y+25\right)$
$\therefore y^3+125$
$=(y+5)\left(y^2-5 y+25\right)$
View full question & answer→Question 232 Marks
Factorize the following expressions:
$64 a^3-b^3$
Answer$64 a^3-b^3$
$=(4 a)^3-b^3$
$=(4 a-b)\left((4 a)^2+4 a \times b+b^2\right)$
$\therefore\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=(4 a-b)\left(16 a^2+4 a b+b^2\right)$
$\therefore 64 a^3-b^3=(4 a-b)\left(16 a^2+4 a b+b^2\right)$
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:
$8 x^3+27 y^3-216 z^3+108 x y z$
Answer$8 x^3+27 y^3-216 z^3+108 x y z$
$=(2 x)^3+(3 y)^3+(-6 z)^3-3(2 x)(3 y)(-6 z)$
$=(2 x+3 y-6 z)\left((2 x)^2+(3 y)^2+(-6 z)^2-2 x \times 3 y-3 y(-6 z)-(-6 z) 2 x\right)$
$=(2 x+3 y-6 z)\left(4 x^2+9 y^2+36 z^2-6 x y+18 y z+12 z x\right)$
$\therefore 8 x^3+27 y^3-216 z^3+108 x y z$
$=(2 x+3 y-6 z)\left(4 x^2+9 y^2+36 z^2-6 x y+18 y z+12 z x\right)$
View full question & answer→Question 262 Marks
Factorize:
$125 x^3-27 y^3-225 x^2 y+135 x y^2$
Answer$125 x^3-27 y^3-225 x^2 y+135 x y^2$
$=(5 x)^3-(3 y)^3-3 x(5 x)^2 \times 3 y+3 \times(5 x)(3 y)^2$
$=(5 x-3 y)^3\left[\because a^3-b^3-3 a^2 b+3 a b^2=(a-b)^3\right]$
$=(5 x-3 y)(5 x-3 y)(5 x-3 y)$
$\therefore 125 x^3-27 y^3-225 a^2 y+135 x y^2$
$=(5 x-3 y)(5 x-3 y)(5 x-3 y)$
View full question & answer→Question 272 Marks
Multiply: $\left(x^2+4 y^2+2 x y-3 x+6 y+9\right)$ by $(x-2 y+3)$
Answer$=(x-2 y+3)\left(x^2+4 y^2+9+2 x y+6 y-3 x\right)$
$=(x+(-2 y)+3)\left(x^2+(-2 y)^2+3^2-x(-2 y)-(-2 y) 3-3 x\right)$
$=x^3+(-2 y)^3+3^3-3 \times x(-2 y)^3\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3-8 y^3+27+18 x y$
View full question & answer→Question 282 Marks
Factorize the following expressions:
$(3 x-2 y)^3+(2 y-4 z)^3+(4 z-3 x)^3$
Answer$(3 x-2 y)^3+(2 y-4 z)^3+(4 z-3 x)^3$
$\text { Let }(3 x-2 y)=a,(2 y-4 z)=b,(4 z-3 x)=c$
$\therefore a+b+c=3 x-2 y+2 y-4 z+4 z-3 x=0$
$\because a+b+c=0$
$\therefore a^3+b^3+c^3+3 a b c$
$\therefore(3 x-2 y)^3+(2 y-4 z)^3+(4 z-3 x)^3$
$=3(3 x-2 y)(2 y-4 z)(4 z-3 x)$
View full question & answer→Question 292 Marks
Factorize the following expressions:
$(a-2 b)^3-512 b^3$
Answer$(a-2 b)^3-512 b^3$
$=(a-2 b)^3-(8 b)^3$
$=(a-2 b-8 b)\left((a-2 b)^2+(a-2 b) 8 b+(8 b)^2\right)$
$\therefore\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=(a-10 b)\left(a^2+4 b^2-4 a b+8 a b-16 b^2+64 b^2\right)$
$=(a-10 b)\left(a^2+52 b^2+4 a b\right)$
$\therefore(a-2 b)^3-512 b^3=(a-10 b)\left(a^2+52 b^2+4 a b\right)$
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 $a b+b c+c a=40$, find $a^2+b^2+c^2$.
AnswerRecall the formula
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$(a+b+c)=9, a b+b c+c a=40$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(9)^2=a^2+b^2+c^2+2 \cdot(40)$
$a^2+b^2+c^2+80=81$
$a^2+b^2+c^2=81-80$
$a^2+b^2+c^2=1$
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$\frac{8}{27} x^3+1+\frac{4}{3} x^2+2 x$
$=\left(\frac{2}{3} x\right)^3+(1)^3+3 \times\left(\frac{2}{3} x\right)^2 \times 1+3(1)^2 \times\left(\frac{2}{3} x\right)$
$=\left(\frac{2}{3} x+1\right)^3\left[\because a^3+b^3+3 a^2 b+3 a b^2=(a+b)^3\right]$
$=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$
$\therefore \frac{8}{27} x^3+1+\frac{4}{3} x^2+2 x$
$=\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)\left(\frac{2}{3} x+1\right)$
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:
$x^4+x^2 y^2+y^4$
Answer$x^4+x^2 y^2+y^4$
Adding $x^2 y^2$ and subtracting $x^2 y^2$ to the given equation
$=x^4+x^2 y^2+y^4+x^2 y^2-x^2 y^2$
$=x^4+2 x^2 y^2+y^4-x^2 y^2$
$=\left(x^2\right)^2+2 x x^2 \times y^2+\left(y^2\right)^2-(x y)^2$
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: $a^2+b^2+2(a b+b c+c a)$
Answer$=a^2+b^2+2 a b+2 b c+2 c a$
Using the identity $(p+q)^2=p^2+q^2+2 p q$
We get,
$=(a+b)^2+2 b c+2 c a$
$=(a+b)^2+2 c(b+a)$
$\text { Or }(a+b)^2+2 c(a+b)$
Taking $(a+b)$ common
$=(a+b)(a+b+2 c)$
$\therefore a^2+b^2+2(a b+b c+c a)=(a+b)(a+b+2 c)$
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 $\mathrm{a}^2-\mathrm{b}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$
$=(a+b-c)(a-(b-c))$
$=(a+b-c)(a-b+c)$
$\therefore a^2-b^2+2 b c-c^2$
$=(a+b-c)(a-b+c)$
View full question & answer→Question 392 Marks
What are the possible expression for the cuboid having volume $3 x^2-12 x$.
Answer$\text { Volume }=3 x^2-12 x$
$=3 x(x-4)$
$=3 \times x(x-4)$
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$=\left(x^2\right)^3+\left(y^2\right)^3$
$=\left(x^2+y^2\right)\left(\left(x^2\right)^2-x^2 y^2+\left(y^2\right)^2\right)$
$=\left(x^2+y^2\right)\left(x^4-x^2 y^2+y^4\right)$
${\left[\therefore a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]}$
$\therefore x^6+y^6=\left(x^2+y^2\right)\left(x^4-x^2 y^2+y^4\right)$
View full question & answer→Question 422 Marks
Factorize the following expressions:
$a^3+8 b^3+64 c^3-24 a b c$
Answer$a^3+8 b^3+64 c^3-24 a b c$
$=(a)^3+(2 b)^3+(4 c)^3-3 \times 2 b \times 4 c$
$=(a+2 b+4 c)\left(a^2+(2 b)^2+(4 c)^2-a \times 2 b-2 b \times 4 c-4 c \times a\right)$
${\left[\because a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]}$
$=(a+2 b+4 c)\left(a^2+4 b^2+16 c^2-2 a b-8 b c-4 a c\right)$
$\therefore a^3+8 b^3+64 c^3-24 a b c=(a+2 b+4 c)\left(a^2+4 b^2+16 c^2-2 a b-8 b c-4 a c\right)$
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:
$\left(x^2+y^2+z^2-x y+x z+y z\right) b y(x+y-z)$
Answer$\left(x^2+y^2+z^2-x y+x z+y z\right) b y(x+y-z)$
$=(x+y-z)\left(x^2+y^2+z^2-x y+x z+y z\right)$
$=(x+y+(-z))\left(x^2+y^2+(-z)^2-x y-y(-z)-(-z) x\right)$
$=x^3+y^3+(-z)^3-3 x y z(-z)\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3+y^3-z^3+3 x y z$
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:
$x^3+8 y^3+6 x^2 y+12 x y^2$
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
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(a+b+c)^2=250+2 \cdot(3)$
$(a+b+c)^2=256$
$(a+b+c)= \pm 16$
View full question & answer→