Question 12 Marks
Factorise:
$8 x^2 y^3-x^5$
Answer$8 x^2 y^3-x^5$
$=x^2\left(8 y^3-x^3\right)$
$=x^2\left[(2 y)^3-x^3\right]$
$=x^2\left[(2 y-x)\left[(2 y)^2+(2 y)(x)+x^2\right]\right.$
$=x^2(2 y-x)\left(4 y^2+2 x y+x^2\right)$
View full question & answer→Question 22 Marks
Factorise:
$6 x^2+17 x+12$
Answer$6 x^2+17 x+12$
$=6 x^2+9 x+8 x+12$
$=3 x(2 x+3)+4(2 x+3)$
$=(2 x+3)(3 x+4)$
View full question & answer→Question 32 Marks
Factorise: $2\sqrt{3}\text{x}^2+\text{x}-5\sqrt{3}$
Answer$2\sqrt{3}\text{x}^2+\text{x}-5\sqrt{3}$ $=2\sqrt{3}\text{x}^2+6\text{x}-5\text{x}-5\sqrt{3}$ $=2\sqrt{3}\text{x}\big(\text{x}+\sqrt{3}\big)-5\big(\text{x}+\sqrt{3}\Big)$ $=\big(\text{x}+\sqrt{3}\big)\big(2\sqrt{3}\text{x}-5\big)$
View full question & answer→Question 42 Marks
Factorise:
$2 x^2+3 x-90$
Answer$2 x^2+3 x-90$
$=2 x^2-12 x+15 x-90$
$=2 x(x-6)+15(x-6)$
$=(x-6)(2 x+15)$
View full question & answer→Question 52 Marks
Factorise: $\text{x}^2-2\sqrt{3}\text{x}-24$
Answer$\text{x}^2-2\sqrt{3}\text{x}-24$ $=\text{x}^2-4\sqrt{3}\text{x}+2\sqrt{3}\text{x}-24$ $=\text{x}(\text{x}-4\sqrt{3})+2\sqrt{3}(\text{x}-4\sqrt{3})$ $=\text{x}(\text{x}-4\sqrt{3})(\text{x}+2\sqrt{3})$
View full question & answer→Question 62 Marks
Factorise:
$x^2-x-156$
Answer$x^2-x-156$
$=x^2-13 x+12 x-156$
$=x(x-13)+12(x-13)$
$=(x-13)(x+12)$
View full question & answer→Question 72 Marks
Factorise: $a^2-b^2-4 a c+4 c^2$
Answer$a^2-4 a c+4 c^2-b^2$
$=a^2-4 a c+4 c^2-b^2$
$=a^2-2 \times a \times 2 c+(2 c)^2-b^2$
$=(a-2 c)^2-b^2\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(a-2 c+b)(a-2 c-b)$
View full question & answer→Question 82 Marks
Factorise:
$x^2+7 x-98$
Answer$x^2+7 x-98$
$=x^2+14 x-7 x-98$
$=x(x+14)-7(x+14)$
$=(x+14)(x-7)$
View full question & answer→Question 92 Marks
Factorise:
$2 x^2+11 x-21$
Answer$2 x^2+11 x-21$
$=2 x^2+14 x-3 x-21$
$=2 x(x+7)-3(x+7)$
$=(x+7)(2 x-3)$
View full question & answer→Question 102 Marks
Factorise: $(3 a-2 b)^3+(2 b-5 c)^3+(5 c-3 a)^3$
AnswerPut $(3 a-2 b)=x,(2 b-5 c)=y$ and $(5 c-3 a)=z$.
We have: $x+y+z=3 a-2 b+2 b-5 c+5 c-3 a=0$
Now, $(3 a-2 b)^3$ $+(2 b-5 c)^3+(5 c-3 a)^3=x^3+y^3+z^3=3 x y z\left[\right.$
Here, $x+y+z=0$.
So, $\left.x^3+y^3+z^3\right]=(3 a-2 b)(2 b-5 c)(5 c-3 a)$
View full question & answer→Question 112 Marks
Factorise:
$2 x^4-32$
Answer$2 x^4-32$
$=2\left(x^4-16\right)$
$=2\left[\left(x^2\right)^2-(4)^2\right]$
$=2\left[\left(x^2-4\right)\left(x^2+4\right)\right]$
$=2\left[\left(x^2-2^2\right)\left(x^2+4\right)\right]$
$=2\left[(x-2)(x+2)\left(x^2+4\right)\right]$
$=2(x-2)(x+2)\left(x^2+4\right)$
View full question & answer→Question 122 Marks
Factorise:
$25 x^2+4 y^2+9 z^2-20 x y-12 y z+30 x z$.
AnswerWe have:
$25 x^2+4 y^2+9 z^2-20 x y-12 y z+30 x z$
$=(5 x)^2+(-2 y)^2+(3 z)^2+2(5 x)(-2 y)+2(-2 y)(3 z)+2(3 z)(5 x)$
$=[(5 x)+(-2 y)+(3 z)]^2$
$=(5 x-2 y+3 z)^2$
View full question & answer→Question 132 Marks
Expand: $\Big(3\text{a}+\frac{1}{4\text{b}}\Big)^3$
Answer$\Big(3\text{a}+\frac{1}{4\text{b}}\Big)^3$ $=(3\text{a})^3+\Big(\frac{1}{4\text{b}}\Big)^3+3(3\text{a})^2\Big(\frac{1}{4\text{b}}\Big)+3(3\text{a})\Big(\frac{1}{4\text{b}}\Big)^2$ $=27\text{a}^3+\frac{1}{64\text{b}^3}+\frac{27\text{a}^2}{4\text{b}}+\frac{9\text{a}}{16\text{b}^2}$
View full question & answer→Question 142 Marks
Factorise:
$64 a^3-343$
Answer$64 a^3-343$
$=(4 a)^3-(7)^3$
$=(4 a-7)\left[(4 a)^2+(4 a)(7)+(7)^2\right]$
$=(4 a-7)\left(16 a^2+28 a+49\right)$
$=64+112+196 a-112-196 a-343$
$=64 a^3-343$
View full question & answer→Question 152 Marks
Factorise:
$18 x^2+3 x-10$
Answer$18 x^2+3 x-10$
$=18 x^2-12 x+15 x-10$
$=6 x(3 x-2)+5(3 x-2)$
$=(6 x+5)(3 x-2)$
View full question & answer→Question 162 Marks
Factorise:
$a^2-b^2+2 b c-c^2$
Answer$a^2-b^2+2 b c-c^2$
$=a^2-\left(b^2-2 b c+c^2\right)$
$=a^2-(b-c)^2$
$=[a-(b-c)][a+(b-c)]$
$=(a-b+c)(a+b-c)$
View full question & answer→Question 172 Marks
Factorise:
$9 x^2+16 y^2+4 z^2-24 x y+16 y z-12 x z$
AnswerWe have:
$9 x^2+16 y^2+4 z^2-24 x y+16 y z-12 x z$
$=(2 x)^2+(3 y)^2+(-4 z)^2+2(2 x)(3 y)+2(3 y)(-4 z)+2(-4 z)(2 x)$
$=[(2 x)+(3 y)+(-4 z)]^2$
$=(2 x+3 y-4 z)^2$
View full question & answer→Question 182 Marks
Factorise: $x^2+18 x+32$
Answer$x^2+18 x+32$
$=x^2+16 x+2 x+32$
$=x(x+16)+2(x+16)$
$=(x+16)(x+2)$
View full question & answer→Question 192 Marks
Factorise:
$\text{x}^2+7\sqrt{6}+60$
Answer $\text{x}^2+7\sqrt{6}+60$
$=\text{x}^2+2\sqrt{6}\text{x}+5\sqrt{6}\text{x}+60$
$=\text{x}(\text{x}+2\sqrt{6})+5\sqrt{6}(\text{x}+2\sqrt{6})$
$=(\text{x}+2\sqrt{6})(\text{x}+5\sqrt{6})$
View full question & answer→Question 202 Marks
Factorise: $\frac{3}{5}\text{x}^2-\frac{19}{5}\text{x}+4$
Answer$\frac{3}{5}\text{x}^2-\frac{19}{5}\text{x}+4$ $=\frac{3}{5}\text{x}^2+\frac{4}{5}\text{x}-3\text{x}+4$ $=\frac{\text{x}}{5}(3\text{x}-4)-1(3\text{x}-4)$ $=\Big(\frac{\text{x}}{5}-1\Big)(3\text{x}-4)$
View full question & answer→Question 212 Marks
Factorise: $\text{x}^2-2\sqrt{2}\text{x}-30$
Answer$\text{x}^2-2\sqrt{2}\text{x}-30$ $=\text{x}^2-5\sqrt{2}\text{x}+3\sqrt{2}\text{x}-30$ $=\text{x}(\text{x}+5\sqrt{2})-3\sqrt{2}(\text{x}+5\sqrt{2})$ $=(\text{x}-5\sqrt{2})(\text{x}+3\sqrt{2})$
View full question & answer→Question 222 Marks
Factorise:
$2 a^3+16 b^3-5 a-10 b$
Answer$2 a^3+16 b^3-5 a-10 b$
$=2\left(a^3+8 b^3\right)-5(a+2 b)$
$=2\left[(a)^3+(2 b)^3\right]-5(a+2 b) \text { Since } a^3+b^3=\left(a+b\left(a^2-a \times b+b^2\right)\right.$
$=2(a+2 b)\left[(a)^2-a \times 2 b+(2 b)^2\right]-5(a+2 b)$
$=(a+2 b)\left[2\left(a^2-2 a b+4 b^2\right)-5\right]$
View full question & answer→Question 232 Marks
Evaluate: $(995)^2$
Answer$(995)^2=(1000-5)^2$
$=[(1000)+(-5)]^2$
$=(1000)^2+2 \times(1000) \times(-5)+(-5)^2$
$=1000000-10000+25$
$=990025$
View full question & answer→Question 242 Marks
Factorise:
$40+3 x-x^2$
Answer$40+3 x-x^2$
$=40+8 x-5 x-x^2$
$=8(5+x)-x(5+x)$
$=(5+x)(8-x)$
View full question & answer→Question 252 Marks
Factorise:
$x^6-7 x^3-8$
AnswerGiven equation is $x^6-7 x^3-8$.
Putting $x^3=y$, we get
$y^2-7 y-8$
$=y^2-8 y+y-8$
$=y(y-8)+1(y-8)$
$=(y-8)(y+1)$
$=\left(x^3-8\right)\left(x^3+1\right)$
$=\left(x^3-2^3\right)\left(x^3+1^3\right)$
$=(x-2)\left(x^2+2 x+4\right)(x+1)\left(x^2-x+1\right)$
$=(x-2)(x+1)\left(x^2+2 x+4\right)\left(x^2-x+1\right)$
View full question & answer→Question 262 Marks
Factorise:
$7\text{x}^2+2\sqrt{14}\text{x}+2$
Answer $7\text{x}^2+2\sqrt{14}\text{x}+2$
$=7\text{x}^2+\sqrt{2}\big(\sqrt{7}\text{x}\big)+\sqrt{2}\big(\sqrt{7}\text{x}\big)+2$
$=\sqrt{7}\text{x}\big(\sqrt{7}\text{x}+\sqrt{2}\big)+\sqrt{2}\big(\sqrt{7}\text{x}+\sqrt{2}\big)$
$=\big(\sqrt{7}\text{x}+\sqrt{2}\big)\big(\sqrt{7}\text{x}+\sqrt{2}\big)=\big(\sqrt{7}\text{x}+\sqrt{2}\big)^2$
View full question & answer→Question 272 Marks
Factorise: $\frac{\text{x}^3}{216}-8\text{y}^3$
Answer$\frac{\text{x}^3}{216}-8\text{y}^3$ $=\Big(\frac{\text{x}}{6}\Big)^3-(2\text{y})^3$ $=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\bigg[\Big(\frac{\text{x}}{6}\Big)^2+\Big(\frac{\text{x}}{6}\Big)(2\text{y})+(2\text{y})^2\bigg]$ $=\Big(\frac{\text{x}}{6}-2\text{y}\Big)\Big(\frac{\text{x}^2}{36}+\frac{\text{xy}}{3}+4\text{y}^2\Big)$
View full question & answer→Question 282 Marks
Factorise: $\sqrt{2}\text{x}^2+3\text{x}+\sqrt{2}$
Answer$\sqrt{2}\text{x}^2+3\text{x}+\sqrt{2}$ $=\sqrt{2}\text{x}^2+\text{x}+2\text{x}+\sqrt{2}$ $=\text{x}\big(\sqrt{2}\text{x}+1\big)+\sqrt{2}\big(\sqrt{2}\text{x}+1\big)$ $=\big(\sqrt{2}\text{x}+1\big)\big(\text{x}+\sqrt{2}\big)$
View full question & answer→Question 292 Marks
Factorise:
$9 x^2+18 x+8$
Answer$9 x^2+18 x+8$
$=9 x^2+12 x+6 x+8$
$=3 x(3 x+4)+2(3 x+4)$
$=(3 x+4)(3 x+2)$
View full question & answer→Question 302 Marks
Factorise:
$6 x^2-11 x-35$
Answer$6 x^2-11 x-35$
$=6 x^2-21 x+10 x-35$
$=3 x(2 x-7)+5(2 x-7)$
$=(2 x-7)(3 x+5)$
View full question & answer→Question 312 Marks
Factorise:
$15 x^2+2 x-8$
Answer$15 x^2+2 x-8$
$=15 x^2-10 x+12 x-8$
$=5 x(3 x-2)+4(3 x-2)$
$=(3 x-2)(5 x+4)$
View full question & answer→Question 322 Marks
Factorise:
$(a+b)^3-(a-b)^3$
AnswerWe know that, Since $a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
Therefore,
$(a+b)^3-(a-b)^3$
$=[a+b-(a-b)]\left[(a+b)^2+(a+b)(a-b)+(a-b)^2\right.$
$=(a+b-a+b)\left[a^2+b^2+2 a b+a^2-b^2+a^2+b^2-2 a b\right]$
$=2 b\left(3 a^2+b^2\right)$
View full question & answer→Question 332 Marks
Factorise:
$8(x+y)^3-27(x-y)^3$
Answer$8(x+y)^3-27(x-y)^3$
$=\left[2^3(x+y)^3\right]-\left[3^3(x-y)^3\right]$
$=[2(x+y)-3(x-y)]\left\{[2(x+y)]^2+2(x+y) 3(x-y)+[3(x-y)]^2\right\}$
$=(2 x+2 y-3 x+3 y)\left\{\left[4\left(x^2+y^2+2 x y\right)\right]+6\left(x^2-y^2\right)+\left[9\left(x^2+y^2-2 x y\right]\right\}\right.$
$=(-x+5 y)\left\{4 x^2+4 y^2+8 x y+6 x^2-6 y^2+9 x^2+9 y^2-18 x y\right\}$
$=(-x+5 y)\left(19 x^2+7 y^2-10 x y\right)$
View full question & answer→Question 342 Marks
Factorise: $(2 a+1)^3+(a-1)^3$
Answer$(2 a+1)^3+(a-1)^3$
$=(2 a+1+a-1)\left[(2 a+1)^2-(2 a+1)(a-1)+(a-1)^2\right]$
$=(3 a)\left[4 a^2+4 a+1-2 a^2+2 a-a+1+a^2-2 a+1\right]$
$=3 a\left(3 a^2+3 a+3\right)$
$=9 a\left(a^2+a+1\right)$
View full question & answer→Question 352 Marks
Factorise:
$a b\left(x^2+1\right)+x\left(a^2+b^2\right)$
Answer$a b\left(x^2+1\right)+x\left(a^2+b^2\right)$
$=a b x^2+a b+a^2 x+b^2 x$
$=a b x^2+a^2 x+a b+b^2 x$
$=a x(b x+a)+b(b x+a)$
$=(b x+a)(a x+b)$
View full question & answer→Question 362 Marks
Factorise: $\text{x}^2+\frac{1}{\text{x}^2}-2-3\text{x}+\frac{3}{\text{x}}$
Answer$\text{x}^2+\frac{1}{\text{x}^2}-2-3\text{x}+\frac{3}{\text{x}}$ $=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-3\Big(\text{x}-\frac{1}{\text{x}}\Big)$ $=\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}-3\Big)$
View full question & answer→Question 372 Marks
Factorise: $\text{x}^2-3\sqrt{5}\text{x}-20$
Answer$\text{x}^2-3\sqrt{5}\text{x}-20$ $=\text{x}^2-4\sqrt{5}\text{x}+\sqrt{5}\text{x}-20$ $=\text{x}(\text{x}-4\sqrt{5})+\sqrt{5}(\text{x}-4\sqrt{5})$ $=\text{x}(\text{x}-4\sqrt{5})(\text{x}+\sqrt{5})$
View full question & answer→Question 382 Marks
Factorise:
$1-27 a^3$
Answer$1-27 a^3$
$=(1)^3-(3 a)^3$
$=(1-3 a)\left[(1)^2+1 \times 3 a+(3 a)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=(1-3 a)\left(1+3 a+9 a^2\right)$
View full question & answer→Question 392 Marks
Expand:
$(2 a+5 b+7 c)^2$
Answer$(2 a+5 b+7 c)^2=[(2 a)+(-5 b)+(-7 c)]^2$
$=(2 a)^2+(-5 b)^2+(-7 c)^2+2(2 a)(-5 b)+2(-5 b)(-7 c)+2(2 a)(-7 c)$
$=4 a^2+25 b^2+49 c^2-20 a b+70 b c-28 a c$
View full question & answer→Question 402 Marks
Factorise: $2\text{x}^2-\text{x}+\frac{1}{8}$
Answer$2\text{x}^2-\text{x}+\frac{1}{8}$ $=2\text{x}^2-\frac{1}{2}\text{x}-\frac{1}{2}\text{x}+\frac{1}{8}$ $=\frac{\text{x}}{2}(4\text{x}-1)-\frac{1}{8}(4\text{x}-1)$ $=\Big(\frac{\text{x}}{2}-\frac{1}{8}\Big)(4\text{x}-1)$
View full question & answer→Question 412 Marks
Factorise:
$a b\left(x^2+y^2\right)-x y\left(a^2+b^2\right)$
Answer$a b\left(x^2+y^2\right)-x y\left(a^2+b^2\right)$
$=a b x^2+a b y^2-a^2 x y-b^2 x y$
$=a b x^2-a^2 x y+a b y^2-b^2 x y$
$=a x(b x-a y)+b y(a y-b x)$
$=(b x-a y)(a x-b y)$
View full question & answer→Question 422 Marks
Factorise:
$6-x-x^2$
Answer$6-x-x^2$
$=6+2 x-3 x-x^2$
$=2(3+x)-x(3+x)$
$=(3+x)(2-x)$
View full question & answer→Question 432 Marks
Factorise:
$x^2-32 x-105$
Answer$x^2-32 x-105$
$=x^2-35 x+3 x-105$
$=x(x-35)+3(x-35)$
$=(x-35)(x+3)$
View full question & answer→Question 442 Marks
Factorise: $\text{x}^2+6\sqrt{6}\text{x}+48$
Answer$\text{x}^2+6\sqrt{6}\text{x}+48$ $=\text{x}^2+4\sqrt{6}\text{x}+2\sqrt{6}\text{x}+48$ $=\text{x}(\text{x}+4\sqrt{6})+2\sqrt{6}(\text{x}+4\sqrt{6})$ $=(\text{x}+4\sqrt{6})(\text{x}+2\sqrt{6})$
View full question & answer→Question 452 Marks
Factorise: $\text{x}^2-2+\frac{1}{\text{x}^2}-\text{y}^2$
Answer$\text{x}^2-2+\frac{1}{\text{x}^2}-\text{y}^2$ $=\Big(\text{x}^2-2(\text{x}^2)\Big(\frac{1}{\text{x}^2}\Big)+\frac{1}{\text{x}^2}\Big)-\text{y}^2$ $=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-\text{y}^2$ $=\Big(\text{x}-\frac{1}{\text{x}}+\text{y}\Big)\Big(\text{x}-\frac{1}{\text{x}}-\text{y}\Big)$
View full question & answer→Question 462 Marks
Factorise:
$8(3 a-2 b)^2-10(3 a-2 b)$
Answer$8(3 a-2 b)^2-10(3 a-2 b)$
$=(3 a-2 b)[8(3 a-2 b)-10]$
$=(3 a-2 b) 2[4(3 a-2 b)-5]$
$=2(3 a-2 b)(12 a-8 b-5)$
View full question & answer→Question 472 Marks
Factorise:
$7 a^3+56 b^3$
Answer$7 a^3+56 b^3$
$=7\left(a^3+8 b^3\right)$
$=7\left[(a)^3+(2 b)^3\right]$
$=7(a+2 b)\left[a^2-a \times 2 b+(2 b)^2\right] \text { Since } a^3+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
$=7(a+2 b)\left(a^2-2 a b+4 b^2\right)$
View full question & answer→Question 482 Marks
Factorise:
$125 a^3+b^3+64 c^3-60 a b c$
Answer$125 a^3+b^3+64 c^3-60 a b c$
$=(5 a)^3+(b)^3+(4 c)^3-3 \times 5 a \times b \times 4 c$
$=(5 a+b+4 c)\left[(5 a)^2+(b)^2+(4 c)^2-5 a \times b-b \times 4 c-5 a \times 4 c\right]$
$=(5 a+b+4 c)\left(25 a^2+b^2+16 c^2-5 a b-4 b c-20 a c)\right.$
View full question & answer→Question 492 Marks
Expand: $\Big(1+\frac{2}{3}{\text{a}}\Big)^3$
Answer$\Big(1+\frac{2}{3}{\text{a}}\Big)^3$ $=\Big(\frac{2}{3}\text{a}\Big)^3+3\times\Big(\frac{2}{3}\text{a}\Big)^2\times1+3\text{a}\frac{2}{3}\text{a}\times(1)^2+(1)^3$ $=\frac{8}{27}\text{a}^3+\frac{4}{3}\text{a}^2+2\text{a}+1$
View full question & answer→Question 502 Marks
Factorise:
$3 a^3 b-243 a b^3$
Answer$3 a^3 b-243 a b^3$
$=3 a b\left(a^2-81 b^2\right)$
$=3 a b\left[(a)^2-(9 b)^2\right]$
$=3 a b(a+9 b)(a-9 b)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
View full question & answer→