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→Question 512 Marks
If $a, b, c$ are all nonzero and $a + b + c = 0$, prove that: $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3$
Answer$a+b+c=0 \Rightarrow a^3+b^3+c^3=3 a b c$ Thus, We have: $\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$ $=\frac{3\text{abc}}{\text{abc}}$ $=3$
View full question & answer→Question 522 Marks
Factorise: $2\text{x}^2+3\sqrt{3}\text{x}+3$
Answer$2\text{x}^2+3\sqrt{3}\text{x}+3$ $2\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+3$ $=2\text{x}\big(\text{x}+\sqrt{3}\big)+\sqrt{3}\big(\text{x}+\sqrt{3}\big)$ $=\big(\text{x}+\sqrt{3}\big)\big(2\text{x}+\sqrt{3}\big)$
View full question & answer→Question 532 Marks
Expand: $\Big(3\text{x}-\frac{5}{\text{x}}\Big)^3$
Answer $\Big(3\text{x}-\frac{5}{\text{x}}\Big)^3$
$=(3\text{x})^3-\Big(\frac{5}{\text{x}}\Big)^3-3(3\text{x})^2\Big(\frac{5}{\text{x}}\Big)+3(3\text{x})\Big(\frac{5}{\text{x}}\Big)^2$
$=27\text{x}^3-\frac{125}{\text{x}^3}-135\text{x}+\frac{225}{\text{x}}$
View full question & answer→Question 542 Marks
Evaluate:
$(107)^2$
Answer$(107)^2=(100+7)^2$
$=(100)^2+2 \times(100) \times(7)+(7)^2$
$=10000+1400+49$
$=11449$
View full question & answer→Question 552 Marks
Expand:
$(5 a-3 b)^3$
Answer$(5 a-3 b)^3$
$=(5 a)^3-(3 b)^3-3(5 a)^2(3 b)+3(5 a)(3 b)^2$
$=125 a^3-27 b^3-225 a^2 b+135 a b^2$
View full question & answer→Question 562 Marks
Factorise:
$x^2-32 x-105$
Answer$x^2-32 x-105$
$=x^2-35 x+3 x-105$
$=x(x-35)+3(x-35)$
$=x(x-35)(x+3)$
View full question & answer→Question 572 Marks
Factorise:
$6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
Answer $6\sqrt{3}\text{x}^2-47\text{x}+5\sqrt{3}$
$=6\sqrt{3}\text{x}^2-45\text{x}-2\text{x}+5\sqrt{3}$
$=3\sqrt{3}\text{x}\big(2\text{x}-5\sqrt{3}\big)-1\big(2\text{x}-5\sqrt{3}\big)$
$=\big(2\text{x}-5\sqrt{3}\big)\big(3\sqrt{3}\text{x}-1\big)$
View full question & answer→Question 582 Marks
Factorise:
$(a+b)^3-a-b$
Answer$(a+b)^3-a-b$
$=(a+b)^3-(a+b)$
$=(a+b)\left[(a+b)^2-1^2\right]\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(a+b)(a+b+1)(a+b-1)$
View full question & answer→Question 592 Marks
Factorise: $2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$
Answer$x =2\sqrt{2}\text{a}^3+16\sqrt{2}\text{b}^3+\text{c}^3-12\text{abc}$ $=\big(\sqrt{2}\text{a}\big)^3+\big(2\sqrt{2}\text{b}\big)^3+\text{c}^3-3\times\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)\times(\text{c})$ $=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\Big[\big(\sqrt{2}\text{a}\big)^2+(2\sqrt{2}\text{b})^2+\text{c}^2\\-\big(\sqrt{2}\text{a}\big)\times\big(2\sqrt{2}\text{b}\big)-\big(2\sqrt{2}\text{b}\big)\times(\text{c})-\big(\sqrt{2}\text{a}\big)\times(\text{c})\Big]$ $=\big(\sqrt{2}\text{a}+2\sqrt{2}\text{b}+\text{c}\big)\big(2\text{a}^2+8\text{b}^2+\text{c}^2-4\text{ab}-2\sqrt{2\text{bc}}-\sqrt{2}\text{ac}\big)$
View full question & answer→Question 602 Marks
Factorise: $\text{x}^2+\sqrt{2}\text{x}-24$
Answer$\text{x}^2+\sqrt{2}\text{x}-24$ $=\text{x}^2+4\sqrt{2}\text{x}-3\sqrt{2}\text{x}-24$ $=\text{x}(\text{x}+4\sqrt{2})-3\sqrt{2}(\text{x}+4\sqrt{2})$ $=(\text{x}-4\sqrt{2})(\text{x}-3\sqrt{2})$
View full question & answer→Question 612 Marks
Factorise:
$10 x^2-9 x-7$
Answer$10 x^2-9 x-7$
$=10 x^2+5 x-14 x-7$
$=5 x(2 x+1)-7(2 x+1)$
$=(2 x+1)(5 x-7)$
View full question & answer→Question 622 Marks
Factorise:
$x^2-21 x+90$
Answer$x^2-21 x+90$
$=x^2-6 x-15 x+90$
$=x(x-6)-15(x-6)$
$=(x-6)(x-15)$
View full question & answer→Question 632 Marks
Factorise:
$1029-3 x^3$
Answer$1029-3 x^3$
$=3\left(343-x^3\right)$
$=3\left[(7)^3-x^3\right]$
$=3\left[(7-x)\left(7^2+7 x+x^2\right)\right]$
$=3(7-x)\left(49+7 x+x^2\right)$
View full question & answer→Question 642 Marks
Factorise:
$x^2+11 x+30$
Answer$x^2+11 x+30$
$=x^2+6 x+5 x+30$
$=x(x+6)+5(x+6)$
$=(x+6)(x+5)$
View full question & answer→Question 652 Marks
Factorise:
$25 x^2-10 x+1-36 y^2$
Answer$25 x^2-10 x+1-36 y^2$
$=\left(25 x^2-10 x+1\right)-36 y^2$
$=\left[(5 x)^2-2(5 x)(1)+(1)^2\right]-(6 y)^2$
$=(5 x-1)^2-(6 y)^2$
$=(5 x-1-6 y)(5 x-1+6 y)$
View full question & answer→Question 662 Marks
Factorise: $\frac{3}{2}\text{x}^2+16\text{x}+10$
Answer$\frac{3}{2}\text{x}^2+16\text{x}+10$ $=\frac{3}{2}\text{x}^2+\text{x}+15\text{x}+10$ $=\frac{\text{x}}{2}(3\text{x}+2)+5(3\text{x}+2)$ $(3\text{x}+2)\Big(\frac{\text{x}}{2}+5\Big)$
View full question & answer→Question 672 Marks
Factorise:
$x-8 x y^3$
Answer$x-8 x y^3$
$=x\left(1-8 y^3\right)$
$=x\left[(1)^3-(2 y)^3\right]$
$=x(1-2 y)\left[(1)^2+1 \times 2 y+(2 y)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=x(1-2 y)\left(1+2 y+4 y^2\right)$
View full question & answer→Question 682 Marks
Factorise: $\text{x}^4+\frac{4}{\text{x}^4}$
Answer$\text{x}^4+\frac{4}{\text{x}^4}$ $=\text{x}^4+\frac{4}{\text{x}^4}+4-4$ $=\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)-2^2$ $=\bigg[\big(\text{x}^2\big)^2+\Big(\frac{2}{\text{x}^2}\Big)^2+2\big(\text{x}^2\big)\Big(\frac{2}{\text{x}^2}\Big)\bigg]-2^2$ $=\Big[\text{x}^2+\frac{2}{\text{x}^2}\Big]^2-2^2$ $=\Big(\text{x}^2+\frac{2}{\text{x}^2}+2\Big)\Big(\text{x}^2+\frac{2}{\text{x}^2}-2\Big)$
View full question & answer→Question 692 Marks
Factorise:
$x^6-729$
Answer$x^6-729$
$=\left(x^2\right)^3-(9)^3$
$=\left(x^2-9\right)\left[\left(x^2\right)^2+x^2 \times 9+(9)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=\left(x^2-9\right)\left(x^4+9 x^2+81\right)$
$=(x+3)(x-3)\left[\left(x^2+9\right)^2-(3 x)^2\right]$
$=(x+3)(x-3)\left(x^2+3 x+9\right)\left(x^2-3 x+9\right)$
View full question & answer→Question 702 Marks
Factorise:
$27 a^3-b^3+8 c^3-18 a b c$
Answer$27 a^3-b^3+8 c^3-18 a b c$
$=(3 a)^3+(-b)^3+(2 c)^3-3 \times(3 a) \times(-b) \times(2 c)$
$=[3 a+(-b)+2 c]\left[(3 a)^2+(-b)^2+(2 c)^2-3 a(-b) 2 c-3 a \times 2 c\right]$
$=(3 a-b+2 c)\left(9 a^2+b^2+4 c^2+3 a b+2 b c-6 a c\right)$
View full question & answer→Question 712 Marks
Factorise: $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$
Answer$1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ $=(1)^3+\Big(\frac{3}{5}\text{a}\Big)^3+3(1)^2\Big(\frac{3}{5}\text{a}\Big)+3(1)\Big(\frac{3}{5}\text{a}\Big)^2$ $=\Big(1+\frac{3}{5}\text{a}\Big)^3$ Hence, factorisation of $1+\frac{27}{125}\text{a}^3+\frac{9\text{a}}{5}+\frac{27\text{a}^2}{25}$ is $=\Big(1+\frac{3}{5}\text{a}\Big)^3$
View full question & answer→Question 722 Marks
Factorise: $21\text{x}^2-2\text{x}+\frac{1}{21}$
Answer$21\text{x}^2-2\text{x}+\frac{1}{21}$ $21\text{x}^2-\text{x}-\text{x}+\frac{1}{21}$ $=21\text{x}\Big(\text{x}-\frac{1}{21}\Big)-1\Big(\text{x}-\frac{1}{21}\Big)$ $=\Big(\text{x}-\frac{1}{21}\Big)(21\text{x}-1)$
View full question & answer→Question 732 Marks
Factorise:
$a^3+0.008$
Answer$a^3+0.008$
$=(a)^3+(0.2)^3$
$=(a+0.2)\left[(a)^2-a \times 0.2+(0.2)^2\right] \text { Since } a^3+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
$=(a+0.2)\left(a^2-0.2 a+0.04\right)$
View full question & answer→Question 742 Marks
Expand:
$(2 b-b+c)^2$
Answer$(2 b-b+c)^2=[(2 a)+(-b)+(c)]^2$
$=(2 a)^2+(-b)^2+(c)^2+2(2 a)(-b)+2(-b)(c)+4(a)(c)$
$=14 a^2+b^2+c^2-4 a b-2 b c+4 a c$
View full question & answer→Question 752 Marks
Factorise:
$\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
Answer $\sqrt{3}\text{x}^2-10\text{x}+8\sqrt{3}$
$=\sqrt{3}\text{x}^2+4\text{x}+6\text{x}+8\sqrt{3}$
$=\text{x}\big(\sqrt{3}\text{x}+4\big)+2\sqrt{3}\big(\sqrt{3}\text{x}+4\big)$
$=\big(\sqrt{3}\text{x}+4\big)\big(\text{x}+2\sqrt{3}\big)$
View full question & answer→Question 762 Marks
Factorise: $\text{x}^2+5\sqrt{5}\text{x}+30$
Answer$\text{x}^2+5\sqrt{5}\text{x}+30$ $=\text{x}^2+2\sqrt{5}\text{x}+3\sqrt{5}\text{x}+30$ $=\text{x}(\text{x}+2\sqrt{5})+3\sqrt{5}(\text{x}+2\sqrt{5})$ $=(\text{x}+2\sqrt{5})(\text{x}+3\sqrt{5})$
View full question & answer→Question 772 Marks
Factorise:
$125-8 x^3-27 y^3-90 x y$
Answer$125-8 x^3-27 y^3-90 x y$
$=5^3+(-2 x)^3+(-3 y)^3-3 \times 5 \times(-2 x) \times(-3 y)$
$=[5+(-2 x)+(-3 y)]\left[5^2+(-2 x)^2+(-3 y)^2-5 \times(-2 x)-(-2 x)(-3 y)-5 \times(-3 y)\right]$
$=(5-2 x-3 y)\left(25+4 x^2+9 y^2+10 x-6 x y+15 y\right)$
View full question & answer→Question 782 Marks
Factorise:
$4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
AnswerWe have:
$4 x^2+9 y^2+16 z^2+12 x y-24 y z-16 x z$
$=(2 x)^2+(3 y)^2+(-4 z)^2+2(2 x)(3 y)+2(3 y)(-4 z)+2(-4 z)(2 x)$
$=[(2 x)+(3 y)+(-4 z)]^2$
$=(2 x+3 y-4 z)^2$
View full question & answer→Question 792 Marks
Find the product.
$(x+y-z)\left(x^2+y^2+z^2-x y+y z+z x\right)$
Answer$(x+y-z)\left(x^2+y^2+z^2-x y+y z+z x\right)$
$=[x+y+(-z)]\left[x^2+y^2+(-z)^2-x y-y \times(-z)-[-z] \times x\right]$
$=x^3+y^3+(-z)^3-3 x \times y \times(-z)$
$=x^3+y^3+(-z)^3-3 x \times y \times(-z)$
$=x^3+y^3-z^3+3 x y z$
View full question & answer→Question 802 Marks
Factorise:
$x^2-(a+b) x+a b$
Answer$x^2-(a+b) x+a b$
$=x^2-a x-b x+a b$
$=x(x-a)-b(x-a)$
$=(x-a)(x-b)$
View full question & answer→Question 812 Marks
Factorise:
$x^2-4 x+3$
Answer$x^2-4 x+3$
$=x^2-3 x-x+3$
$=x(x-3)-1(x-3)$
$=(x-3)(x-1)$
View full question & answer→Question 822 Marks
Factorise: $\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$
Answer$\frac{2}{3}\text{x}^2-\frac{17}{3}\text{x}-28$ $=\frac{2}{3}\text{x}^2+\frac{7}{3}\text{x}-8\text{x}-28$ $=\frac{\text{x}}{2}(2\text{x}+7)-4(2\text{x}+7)$ $=\Big(\frac{\text{x}}{3}-4\Big)(2\text{x}+7)$
View full question & answer→Question 832 Marks
Factorise:
$125\text{a}^3+\frac{1}{8}$
Answer$125\text{a}^3+\frac{1}{8}$
We know that:
Since $a^2+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
Let us rewrite
$125\text{a}^3+\frac{1}{8}$
$=(5\text{a})^3+\Big(\frac{1}{2}\Big)^3$
$=\Big(5\text{a}+\frac{1}{2}\Big)\bigg[(5\text{a})^2-5\text{a}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\bigg]$
$=\Big(5\text{a}+\frac{1}{2}\Big)\Big(25\text{a}^2-\frac{5\text{a}}{2}+\frac{1}{4}\Big)$
View full question & answer→Question 842 Marks
Factorise:
$x^4-625$
Answer$x^4-625$
$=\left(x^2\right)^2-(25)^2$
$=\left(x^2+25\right)\left(x^2-25\right)$
$=\left(x^2+25\right)\left(x^2-5^2\right)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=\left(x^2+25\right)(x+5)(x-5)$
View full question & answer→Question 852 Marks
Factorise:
$9 a^2+3 a-8 b-64 b^2$
Answer$9 a^2+3 a-8 b-64 b^2$
$=9 a^2-64 b^2+3 a-8 b$
$=(3 a)^2-(8 b)^2+(3 a-8 b)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(3 a+8 b)(3 a-8 b)+(3 a-8 b)$
$=(3 a-8 b)(3 a+8 b+1)$
View full question & answer→Question 862 Marks
Factorise:
$9-a^2+2 a b-b^2$
Answer$9-a^2+2 a b-b^2$
$=9-\left(a^2-2 a b+b^2\right)$
$=3^2-(a-b)^2\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(3+a-b)(3-a+b)$
View full question & answer→Question 872 Marks
Expand:
$(-3 a+4 b-5 c)^2$
Answer$(-3 a+4 b-5 c)^2=[(-3 a)+(4 b)+(-5 c)]^2$
$=(-3 a)^2+(4 b)^2+(-5 c)^2+2(-3 a)(4 b)+2(4 b)(-5 c)+2(-3 a)(-5 c)$
$=9 a^2+16 b^2+25 c^2-24 a b-40 b c+30 a c$
View full question & answer→Question 882 Marks
Factorise:
$a^{12}-b^{12}$
Answer$a^{12}-b^{12}$
$=\left(a^6\right)^2-\left(b^6\right)^2$
$=\left(a^6-b^6\right)\left(a^6+b^6\right)$
$=\left[\left(a^3\right)^2-\left(b^3\right)^2\right]\left[\left(a^2\right)^3+\left(b^2\right)^3\right]$
$=\left(a^3-b^3\right)\left(a^3+b^3\right)\left[\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)\right]$
$=(a-b)\left(a^2+a b+b^2\right)(a+b)\left(a^2-a b+b^2\right)\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
$=(a-b)(a+b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)$
View full question & answer→Question 892 Marks
Evaluate:
$(99)^3$
Answer$(99)^3$
$=(100-1)^3$
$=(100)^3-(1)^3-3(100)^2 \times(1)+3(100)(1)^2$
$=1000000-1-30000+300$
$=1000300-30001$
$=970299$
View full question & answer→Question 902 Marks
Factorise:
$24 x^2-41 x+12$
Answer$24 x^2-41 x+12$
$=24 x^2-32 x-9 x+12$
$=8 x(3 x-4)-3(3 x-4)$
$=(3 x-4)(8 x-3)$
View full question & answer→Question 912 Marks
Factorise:
$27 x^3-y^3-z^3-9 x y z$
Answer$27 x^3-y^3-z^3-9 x y z$
$=(3 x)^3-y^3-z^3-3 \times(3 x) \times(-y) \times(-z)$
We know,
$a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$a=3 x, b=-y, c=-z$
$(3 x)^3-y^3-z^3-3 \times(3 x) \times(-y) \times(-z)$
$=(3 x-y-z)\left(9 x^2+y^2+z^2+3 x y-y z+3 x z\right)$
View full question & answer→Question 922 Marks
Evaluate:
$(103)^3$
Answer$(103)^3$
$=(100+3)^3$
$=(100)^3+(3)^3+3(100)^2 \times(3)+3(100)(3)^2$
$=1000000+27+90000+2700$
$=1092727$
View full question & answer→Question 932 Marks
Factorise:
$x^9-y^9$
Answer$x^9-y^9$
$=\left(x^3\right)^3-\left(y^3\right)^3$
$=\left[\left(x^3-y^3\right)\right]\left[\left(x^3\right)^2+x^3 y^3+\left(y^3\right)^2\right]$
$=\left[(x-y)\left(x^2+x y+y^2\right)\left(x^6+x^3 y^3+y^6\right)\right.$
View full question & answer→Question 942 Marks
Factorise:
$(a-b)^3+(b-c)^3+(c-a)^3$
Answer$(a-b)^3+(b-c)^3+(c-a)^3$
$\text { Putting }(a-b)=x,(b-c)=y \text { and }(c-a)=z$
$\text { We get: }(a-b)^3+(b-c)^3+(c-a)^3$
$=x^3+y^3+z^3[(x+y+z)=(a-b)+(b-c)+(c-a)=0]$
$=3 x y z\left[(x+y+z)=0 \Rightarrow x^3+y^3+z^3=3 x y z\right]$
$=3(a-b)(b-c)(c-a)$
View full question & answer→Question 952 Marks
Factorise:
$(a x+b y)^2+(b x-a y)^2$
Answer$(a x+b y)^2+(b x-a y)^2$
$=a^2 x^2+b^2 y^2+2 a b x y+b^2 x^2+a^2 y^2-2 a b x y$
$=a^2 x^2+b^2 y^2+b^2 x^2+a^2 y^2$
$=a^2 x^2+b^2 x^2+b^2 y^2+a^2 y^2$
$=x^2\left(a^2+b^2\right)+y^2\left(a^2+b^2\right)$
$=\left(a^2+b^2\right)\left(x^2+y^2\right)$
View full question & answer→Question 962 Marks
Factorise: $\text{x}^2+3\sqrt{3}\text{x}+6$
Answer$\text{x}^2+3\sqrt{3}\text{x}+6$ $=\text{x}^2+2\sqrt{3}\text{x}+\sqrt{3}\text{x}+6$ $=\text{x}(\text{x}+2\sqrt{3})+\sqrt{3}(\text{x}+2\sqrt{3})$ $=(\text{x}+2\sqrt{3})(\text{x}+\sqrt{3})$
View full question & answer→Question 972 Marks
Factorise:
$a^3-0.064$
Answer$a^3-0.064$
$=(a)^3-(0.4)^3$
$=(a-0.4)\left[(a)^2+a \times 0.4+(0.4)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=(a-0.4)\left(a^2+0.4 a+0.16\right)$
View full question & answer→Question 982 Marks
Factorise:
$81 x^4-y^4$
Answer$81 x^4-y^4$
$=\left(9 x^2\right)^2-\left(y^2\right)^2$
$=\left(9 x^2-y^2\right)\left(9 x^2+y^2\right)$
$=\left[(3 x)^2-y^2\right]\left(9 x^2+y^2\right)$
$=(3 x-y)(3 x+y)\left(9 x^2+y^2\right)$
View full question & answer→Question 992 Marks
Factorise:
$8 a^3+125 b^3-64 c^3+120 a b c$
Answer$8 a^3+125 b^3-64 c^3+120 a b c$
$=(2 a)^3+(5 b)^3+(-4 c)^3-3 \times(2 a) \times(5 b) \times(-4 c)$
$=(2 a+5 b-4 c)\left[(2 a)^2+(5 b)^2+(-4 c)^2-(2 a)(5 b)-(5 b)(-4 c)-(2 a) \times(-4 c)\right]$
$=(2 a+5 b-4 c)\left(4 a^2+25 b^2+16 c^2-10 a b+20 b c+8 a c\right)$
View full question & answer→Question 1002 Marks
Factorise:
$(x+2)^3+(x-2)^3$
Answer$(x+2)^3+(x-2)^3$
$=[(x+2)+(x-2)]\left[(x+2)^2-(x+2)(x-2)+(x-2)^2\right]$
$=(2 x)\left(x^2+4 x+4-x^2+4+x^2-4 x+4\right)$
$=2 x\left(x^2+12\right)$
View full question & answer→Question 1012 Marks
Factorise:
$8 a^3-b^3-4 a x+2 b x$
Answer$8 a^3-b^3-4 a x+2 b x$
$=8 a^3-b^3-2 x(2 a-b)$
$=(2 a)^3-(b)^3-2 x(2 a-b) \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=(2 a-b)\left[(2 a)^2+2 a \times b+(b)^2\right]-2 x(2 a-b)$
$=(2 a-b)\left(4 a^2+2 a b+b^2\right)-2 x(2 a-b)$
$=(2 a-b)\left(4 a^2+2 a b+b^2-2 x\right)$
View full question & answer→Question 1022 Marks
Factorise:
$a^2 x^2+\left(a x^2+1\right) x+a$
Answer$a^2 x^2+\left(a x^2+1\right) x+a$
$=a^2 x^2+a x^3+x+a$
$=a x^2(a+x)+1(x+a)$
$=\left(a x^2+1\right)(a+x)$
View full question & answer→Question 1032 Marks
Factorise:
$4 x^2-9 y^2-2 x-3 y$
Answer$4 x^2-9 y^2-2 x-3 y$
$=(2 x)^2-(3 y)^2-(2 x+3 y)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(2 x+3 y)(2 x-3 y)-(2 x+3 y)$
$=(2 x+3 y)(2 x-3 y-1)$
View full question & answer→Question 1042 Marks
Factorise:
$6 x^2-5 x-21$
Answer$6 x^2-5 x-21$
$=6 x^2+9 x-14 x-21$
$=3 x(2 x+3)-7(2 x+3)$
$=(3 x-7)(2 x+3)$
View full question & answer→Question 1052 Marks
Factorise:
$a-b-a^2+b^2$
Answer$a-b-a^2+b^2$
$=(a-b)-\left(a^2-b^2\right)$
$=(a-b)-(a-b)(a+b)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(a-b)(1-a-b)$
View full question & answer→Question 1062 Marks
Factorise:
$4 a^2-9 b^2-2 a-3 b$
Answer$4 a^2-9 b^2-2 a-3 b$
$=(2 a)^2-(3 b)^2-(2 a+3 b)$
$=(2 a-3 b)(2 a+3 b)-(2 a+3 b)$
$=(2 a+3 b)(2 a-3 b-1)$
View full question & answer→Question 1072 Marks
Factorise:
$x^2+y^2-z^2-2 x y$
Answer$x^2+y^2-z^2-2 x y$
$=\left(x^2+y^2-2 x y\right)-z^2$
$=(x-y)^2-z^2$
$=(x-y-z)(x-y+z)$
View full question & answer→Question 1082 Marks
Factorise:
$x^5+x^2$
Answer$x^5+x^2$
$=x^2\left(x^3+1\right)$
$=x^2(x+1)\left[(x)^2-x \times 1+(1)^2\right] \text { Since } a^3+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
$=x^2(x+1)\left(x^2-x+1\right)$
View full question & answer→Question 1092 Marks
Factorise:
$x^2+2 x y+y^2-a^2+2 a b-b^2$
Answer$x^2+2 x y+y^2-a^2+2 a b-b^2$
$=\left(x^2+2 x y+y^2\right)-\left(a^2-2 a b+b^2\right)$
$=(x+y)^2-(a-b)^2$
$=[(x+y)-(a-b)][(x+y)+(a-b)]$
$=(x+y-a+b)(x+y+a-b)$
View full question & answer→Question 1102 Marks
Factorise:
$a^2+a b(b+1)+b^3$
Answer$a^2+a b(b+1)+b^3$
$=a^2+a b^2+a b+b^3$
$=a^2+a b+a b^2+b^3$
$=a(a+b)+b^2(a+b)$
$=(a+b)\left(a+b^2\right)$
View full question & answer→Question 1112 Marks
Expand: $\Big(\frac{1}{2}\text{a}-\frac{1}{4}\text{b}+2\Big)^2$
Answer$\Big(\frac{1}{2}\text{a}-\frac{1}{4}\text{b}+2\Big)^2=\Big[\Big(\frac{\text{a}}{2}\Big)+\Big(-\frac{\text{b}}{4}\Big)+(2)\Big]^2$
$=\Big(\frac{\text{a}}{2}\Big)^2+\Big(-\frac{\text{b}}{4}\Big)^2+(2)^2$
$+2\Big(\frac{\text{a}}{2}\Big)\times\Big(\frac{-\text{b}}{4}\Big)(2)+2\Big(\frac{\text{a}}{2}\Big)(2)$
$=\frac{\text{a}^2}{4}+\frac{\text{b}^2}{16}+4-\frac{\text{ab}}{4}-\text{b}+2\text{a}$
View full question & answer→Question 1122 Marks
Factorise:
$3 a^7 b-81 a^4 b^4$
Answer$3 a^7 b-81 a^4 b^4$
$=3 a^4 b\left(a^3-27 b^3\right)$$=3 a^4 b\left[(a)^3-(3 b)^3\right]$
$=3 a^4 b(a-3 b)\left[(a)^2+a \times 3 b+(3 b)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=3 a^4 b(a-3 b)\left(a^2+3 a b+9 b^2\right)$
View full question & answer→Question 1132 Marks
Factorise:
$9 a^2+6 a+1-36 b^2$
Answer$9 a^2+6 a+1-36 b^2$
$=\left(9 a^2+6 a+1\right)-36 b^2$
$=\left[(3 a)^2+2(3 a)(1)+(1)^2\right]-(6 b)^2$
$=(3 a+1)^2-(6 b)^2$
$=(3 a+1-6 b)(3 a+1+6 b)$
View full question & answer→Question 1142 Marks
Factorise:
$8-27 b^3-343 c^3-126 b c$
Answer$8-27 b^3-343 c^3-126 b c$
$=(2)^3+(-3 b)^3+(-7 c)^3-3 \times(2) \times(-3 b) \times(-7 c)$
$=\left[2+(-3 b)+(-7 c)\left[(2)^2+(-3 b)^2+(-7 c)^2-(2)(-3 b)-(-3 b)(-7 c)-(2)(-7 c)\right]\right.$
$=(2-3 b-7 c)\left(4+9 b^2+49 c^2+6 b-21 b c+14 c\right)$
View full question & answer→Question 1152 Marks
Find the product.
$(x-2 y+3)\left(x^2+4 y^2+2 x y+6 y-3 x+9\right)$
Answer$(x-2 y+3)\left(x^2+4 y^2+2 x y+6 y-3 x+9\right)$
$=(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 \times(-2 y)-(-2 y) \times 3-3 x x\right]$
$=(x)^3+(-2 y)^3+(3)^3-3(x)(-2 y)(3)$
$=x^3-8 y^3+27+18 x y$
View full question & answer→Question 1162 Marks
Factorise:
$108 a^2-3(b-c)^2$
Answer$108 a^2-3(b-c)^2$
$=3\left[\left(36 a^2-(b-c)^2\right]\right.$
$=3\left[(6 a)^2-(b-c)^2\right]\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=3(6 a+b-c)(6 a-b+c)$
View full question & answer→Question 1172 Marks
Expand:$\Big(\frac{4}{5}\text{a}-2\Big)^3$
Answer$\Big(\frac{4}{5}\text{a}-2\Big)^3$ $\Big(\frac{4}{5}\text{a}\Big)^3-(2)^3-3\Big(\frac{4}{5}\text{a}\Big)^2(2)+3\Big(\frac{4}{5}\text{a}\Big)(2)^2$ $=\frac{64}{125}\text{a}^3-8-\frac{96}{25}\text{a}^2+\frac{48}{5}\text{a}$
View full question & answer→Question 1182 Marks
Factorise:
$x^3-3 x^2+3 x+7$
Answer$x^3-3 x^2+3 x+7$
$=x^3-3 x^2+3 x-1+8$
$=\left(x^3-3 x^2+3 x-1\right)+8$
$=(x-1)^3+2^3$
$=(x-1+2)\left[(x-1)^2-(x-1)(2)+2^2\right]$
$=(x+1)\left(x^2-2 x+1-2 x+2+4\right)$
$=(x+1)\left(x^2-4 x+7\right)$
View full question & answer→Question 1192 Marks
Factorise:
$x^2+19 x-150$
Answer$x^2+19 x-150$
$=x^2+25 x-6 x-150$
$=x(x+25)-6(x+25)$
$=(x+25)(x-6)$
View full question & answer→Question 1202 Marks
Factorise:
$(x+2)^3-(x-2)^3$
Answer$(x+2)^3-(x-2)^3$
$=[(x+2)-(x-2)]\left[(x+2)^2+(x+2)(x-2)+(x-2)^2\right]$
$=4\left(x^2+4 x+4+x^2-4+x^2-4 x+4\right)$
$=4\left(3 x^2+4\right)$
View full question & answer→Question 1212 Marks
Factorise:
$21 x^2+5 x-6$
Answer$21 x^2+5 x-6$
$=21 x^2+14 x-9 x-6$
$=7 x(3 x+2)-3(3 x+2)$
$=(3 x+2)(7 x-3)$
View full question & answer→Question 1222 Marks
Expand:
$(3 x+2)^3$
Answer$(3 x+2)^3$
$=(3 x)^3+3 \times(3 x)^2 x^2+3 \times 3 x \times(2)^2+(2)^3$
$=27 x^3+54 x^2+36 x+8$
View full question & answer→Question 1232 Marks
Factorise:
$x^2-22 x+120$
Answer$x^2-22 x+120$
$=x^2-10 x-12 x+120$
$=x(x-10)-12(x-10)$
$=(x-10)(x-12)$
View full question & answer→Question 1242 Marks
Factorise:
$x(x+y)^3-3 x^2 y(x+y)$
Answer$x(x+y)^3-3 x^2 y(x+y)$
$=x(x+y)\left[(x+y)^2-3 x y\right]$
$=x(x+y)\left(x^2+y^2+2 x y-3 x y\right)$
$=x(x+y)\left(x^2+y^2-x y\right)$
View full question & answer→Question 1252 Marks
Factorise:
$16 x^2+4 y^2+9 z^2-16 x y-12 y z+24 x z$
AnswerWe have:
$16 x^2+4 y^2+9 z^2-16 x y-12 y z+24 x z$
$=(4 x)^2+(-2 y)^2+(3 z)^2+2(4 x)(-2 y)+2(-2 y)(3 z)+2(3 z)(4 x)$
$=(4 x-2 y+3 z)^2\left[\text { using } a^2+b^2+c^2+2 a b+2 b c+2 c a=(a+b+c)^2\right]$
Hence, $16 x^2+4 y^2+9 z^2-16 x y-12 y z+24 x z=(4 x-2 y+3 z)^2$
View full question & answer→Question 1262 Marks
Factorise:
$9 x^2-3 x-20$
Answer$9 x^2-3 x-20$
$=9 x^2-15 x+12 x-20$
$=3 x(3 x-5)+4(3 x-5)$
$=(3 x-5)(3 x+4)$
View full question & answer→Question 1272 Marks
Factorise:
$1+2 a b-\left(a^2+b^2\right)$
Answer$1+2 a b-\left(a^2+b^2\right)$
$=1-\left(a^2+b^2-2 a b\right)$
$=(1)^2-(a-b)^2$
$=[1-(a-b)][1+(a-b)]$
$=(1-a+b)(1+a-b)$
View full question & answer→Question 1282 Marks
Factorise:
$216+27 b^3+8 c^3-108 b c$
Answer$216+27 b^3+8 c^3-108 b c$
$=(6)^3+(3 b)^3+(2 c)^3-3 \times 6 \times 3 b \times 2 c$
$=(6+3 b+2 c)\left[6^2+(3 b)^2+(2 c)^2-6 \times 3 b-3 b \times 2 c-2 c \times 6\right]$
$=(6+3 b+2 c)\left(36+9 b^2+4 c^2-18 b-6 b c-12 c\right)$
View full question & answer→Question 1292 Marks
Factorise:
$x^4 y^4-x y$
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\left\{(x y-1)\left[(x y)^2+(x y)(1)+(1)^2\right]\right\}$
$=x y(x y-1)\left(x^2 y^2+x y+1\right)$
View full question & answer→Question 1302 Marks
Factorise: $\text{x}^2-\sqrt{3}\text{x}-6$
Answer$\text{x}^2-\sqrt{3}\text{x}-6$
$=\text{x}^2-2\sqrt{3}\text{x}+\sqrt{3}\text{x}-6$
$=\text{x}(\text{x}-2\sqrt{3})+\sqrt{3}(\text{x}-2\sqrt{3})$
$=(\text{x}-2\sqrt{3})(\text{x}+\sqrt{3})$
View full question & answer→Question 1312 Marks
Factorise: $(a+2 b)^2+101(a+2 b)+100$
AnswerGiven equation: $(a+2 b)^2+101(a+2 b)+100$
Let $(a+2 b)=x$ Then,
we have $x^2+101 x+100=x^2+100 x+x+100$
$=x(x+100)+1(x+100)$
$=(x+100)(x+1)=(a+2 b+100)(a+2 b+1)$
View full question & answer→Question 1322 Marks
Factorise:
$1+b^3+8 c^3-6 b c$
Answer$1+b^3+8 c^3-6 b c$
$=(1)^3+(b)^3+(2 c)^3-3 \times 1 \times b \times 2 c$
$=(1+b+2 c)\left[1^2+b^2+(2 c)^2-1 \times b-b \times 2 c-1 \times 2 c\right]$
$=(1+b+2 c)\left(1^2+b^2+4 c^2-b-2 b c-2 c\right)$
View full question & answer→Question 1332 Marks
Factorise:
$16 x^4-1$
Answer$16 x^4-1$
$=\left(4 x^2\right)^2-(1)^2$
$=\left(4 x^2-1\right)\left(4 x^2+1\right)$
$=\left[(2 x)^2-(1)^2\right]\left(4 x^2+1\right)$
$=(2 x-1)(2 x+1)\left(4 x^2+1\right)$
View full question & answer→Question 1342 Marks
Factorise:
$3 x^2-14 x+8$
Answer$3 x^2-14 x+8$
$=3 x^2-12 x-2 x+8$
$=3 x(x-4)-2(x-4)$
$=(x-4)(3 x-2)$
View full question & answer→Question 1352 Marks
Factorise:
$2 x^2-7 x-15$
Answer$2 x^2-7 x-15$
$=2 x^2-10 x+3 x-15$
$=2 x(x-5)+3(x-5)$
$=(x-5)(2 x+3)$
View full question & answer→Question 1362 Marks
Factorise:
$16 x^4+54 x$
Answer$16 x^4+54 x$
$=2 x(8 \times 3+27)$
$=2 x\left[(2 x)^3+(3)^3\right] \text { Since } a^3+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
$=2 x(2 x+3)\left[(2 x)^2-2 x \times 3+3^2\right]$
$=2 x(2 x+3)\left(4 x^2-6 x+9\right)$
View full question & answer→Question 1372 Marks
Factorise:
$4 a^2-4 b^2+4 a+1$
Answer$4 a^2-4 b^2+4 a+1$
$=\left(4 a^2+4 a+1\right)-4 b^2$
$=\left[(2 a)^2+2 \times 2 a \times 1+(1)^2\right]-(2 b)^2$
$=(2 a+1)^2-(2 b)^2$
$=(2 a+1-2 b)(2 a+1+2 b)$
$=(2 a-2 b+1)(2 a+2 b+1)$
View full question & answer→Question 1382 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}+4\sqrt{3})(\text{x}-2\sqrt{3})$
View full question & answer→Question 1392 Marks
Factorise:
$5 x^2-16 x-21$
Answer$5 x^2-16 x-21$
$=5 x^2+5 x-21 x-21$
$=5 x(x+1)-21(x+1)$
$=(x+1)(5 x-21)$
View full question & answer→Question 1402 Marks
Factorise:Evaluate
$\left\{(999)^2-1\right\}$
Answer$\left\{(999)^2-1\right\}$
$=\left\{(999)^2-(1)^2\right\}$
$=\{(999-1)(999+1)\}$
$=998 \times 1000$
$=998000$
View full question & answer→Question 1412 Marks
Factorise: $\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}$
Answer$\text{x}^2+\frac{12}{35}\text{x}+\frac{1}{35}$
$=\text{x}^2+\frac{5\text{x}}{35}+\frac{\text{x}}{5}+\frac{1}{35}$
$=5\text{x}\Big(\frac{\text{x}}{5}+\frac{1}{35}\Big)+1\Big(\frac{\text{x}}{5}+\frac{1}{35}\Big)$
$=(5\text{x}+1)\Big(\frac{\text{x}}{5}+\frac{1}{35}\Big)$
View full question & answer→Question 1422 Marks
Expand:
$(a-2 b-3 c)^2$
Answer$(a-2 b-3 c)^2=[a+(-2 b)+(-3 c)]^2$
$=(a)^2+(-2 b)^2+(-3 c)^2+2(a)(-2 b)+2(-2 b)(-3 c)+2(a)(-3 c)$
$=a^2+4 b^2+9 c^2-4 a b+12 b c-6 a c$
View full question & answer→Question 1432 Marks
Factorise:
$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$
$=(a+b)^3-2^3$
$=[(a+b)-2]\left[(a+b)^2+(a+b) 2+2^2\right]$
$=(a+b-2)\left[(a+b)^2+2(a+b)+4\right]$
View full question & answer→Question 1442 Marks
Factorise:
$x^3-512$
Answer$x^3-512$
$=(x)^3-(8)^3$
$=(x-8)\left[(x)^2+x \times 8+(8)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
$=(x-8)\left(x^2+8 x+64\right)$
$=x^3+8 x^2+64 x-8 x^2-64 x-512$
$=x^3-512$
View full question & answer→Question 1452 Marks
Factorise:
$x^2-11 x-80$
Answer$x^2-11 x-80$
$=x^2-16 x+5 x-80$
$=x(x-16)+5(x-16)$
$=(x-16)(x+5)$
View full question & answer→Question 1462 Marks
Factorise:
$x^3-x^2+a x+x-a-1$
Answer$x^3-x^2+a x+x-a-1$
$=x^3-x^2+a x-a+x-1$
$=x^2(x-1)+a(x-1)+1(x-1)$
$=(x-1)\left(x^2+a+1\right)$
View full question & answer→Question 1472 Marks
Factorise:
$4(a+b)-6(a+b)^2$
Answer$4(a+b)-6(a+b)^2$
$=(a+b)[4-6(a+b)]$
$=2(a+b)(2-3 a-3 b)$
$=2(a+b)(2-3 a-3 b)$
View full question & answer→Question 1482 Marks
Factorise: $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}$
$=5\sqrt{5}\text{x}^2-15\text{x}-5\text{x}+3\sqrt{5}$
$=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)$
View full question & answer→Question 1492 Marks
Factorise:
$216\text{x}^3+\frac{1}{125}$
Answer$216\text{x}^3+\frac{1}{125}$
We know that:
Since $a^2+b^3=(a+b)\left(a^2-a \times b+b^2\right)$
Let us rewrite
$216\text{x}^3+\frac{1}{125}$
$=(6\text{x})^3+\Big(\frac{1}{5}\Big)^3$
$=\Big(6\text{x}+\frac{1}{5}\Big)\bigg[(6\text{x})^2-6\text{x}\times\frac{1}{5}+\Big(\frac{1}{5}\Big)^2\bigg]$
$=\Big(6\text{x}+\frac{1}{5}\Big)\Big(36\text{x}^2-\frac{6\text{x}}{5}+\frac{1}{25}\Big)$
View full question & answer→Question 1502 Marks
Factorise:
$x^2+20 x-69$
Answer$x^2+20 x-69$
$=x^2+23 x-3 x-69$
$=x(x+23)-3(x+23)$
$=(x+23)(x-3)$
View full question & answer→Question 1512 Marks
Factorise:
$15 x^2-x-28$
Answer$15 x^2-x-28$
$=15 x^2+20 x-21 x-28$
$=5 x(3 x+4)-7(3 x+4)$
$=(3 x+4)(5 x-7)$
View full question & answer→Question 1522 Marks
Factorise:
$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 a \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]$
$=(a+2 b+4 c)\left(a^2+4 b^2+16 c^2-2 a b-8 b c-4 c a\right)$
View full question & answer→Question 1532 Marks
Factorise:
$32 x^4-500 x$
Answer$32 x^4-500 x$
$=4 x\left(8 x^3-125\right)$
$=4 x\left[(2 x)^3-(5)^3\right]$
$=4 x\left[(2 x-5)\left[(2 x)^2+2 x \times 5+(5)^2\right] \text { Since } a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)\right.$
$=4 x(2 x-5)\left(4 x^2+10 x+25\right)$
View full question & answer→Question 1542 Marks
Factorise: $x^2-24 x-180$
Answer$x^2-24 x-180$
$=x^2-30 x+6 x-180$
$=x(x-30)+6(x-30)$
$=(x-30)(x+6)$
View full question & answer→Question 1552 Marks
Evaluate:
$(99)^2$
Answer$(99)^2=(100-1)^2$
$=[(100)+(-1)]^2$
$=(100)^2+2 \times(100) \times(-1)+(-1)^2$
$=10000-200+1$$=9801$
View full question & answer→Question 1562 Marks
Factorise:
$8\text{x}^3-\frac{1}{27\text{y}^3}$
AnswerWe know that:
Since $a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
Let us rewrite
$8\text{x}^3-\frac{1}{27\text{y}^3}$
$=(2\text{x})^3-\Big(\frac{1}{3\text{y}}\Big)^3$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\bigg[(2\text{x})^2+2\text{x}\times\frac{1}{3\text{y}}+\Big(\frac{1}{3\text{y}}\Big)^2\bigg]$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\Big(4\text{x}^2+\frac{2\text{x}}{3\text{y}}+\frac{1}{9\text{y}^2}\Big)$
View full question & answer→Question 1572 Marks
Factorise:
$x^2-26 x+133$
Answer$x^2-26 x+133$
$=x^2-19 x-7 x+133$
$=x(x-19)-7(x-19)$
$=(x-19)(x-7)$
View full question & answer→