Question 1012 Marks
Write the greatest common factor in each of the following terms.
$11 \mathrm{x}^2, 12 \mathrm{y}^2$
Answer$11 \mathrm{x}^2, 12 \mathrm{y}^2$
The $GCF$ of $11,12 \& 1$
Also, there is no common factor between $x^2$ and $y^2$
Hence, $GCF$ of $11 x^2$ and $12 y^2$ is $1$.
View full question & answer→Question 1022 Marks
Factorise the following, using the identity $\mathrm{a}^2+2 \mathrm{a} b+\mathrm{b}^2=(\mathrm{a}+\mathrm{b})^2$
$9 x^2+30 x+25$
Answer$9 x^2+30 x+25$
$ =(3 x)^2+2 \times 3 x \times 5+5^2$
$=(3 x+5)^2 $
$ =(3 x+5)(3 x+5)$
View full question & answer→Question 1032 Marks
Simplify:
$\left(s^2 \mathrm{t}+\mathrm{tq}^2\right)^2-(2 \mathrm{stq})^2$
Answer$\left(s^2 \mathrm{t}+\mathrm{tq}^2\right)^2-(2 \mathrm{stq})^2$
$ =\left(s^2 t\right)^2+\left(t q^2\right)^2+2 \times s^2 t \times t q^2-4 s^2 t^2 q^2 $
$=s^4 t^2+t^2 q^4+2 s^2 t^2 q^2-4 s^2 t^2 q^2 $
$ =s^4 t^2+t^2 q^4-2 s^2 t^2 q^2 $
View full question & answer→Question 1042 Marks
Factorise the following expressions.
$y^2+ 8zx - 2xy - 4yz$
Answer$y^2+ 8zx - 2xy - 4yz$
$= y^2- 2xy + 8zx - 4yz$
$= y(y - 2x) - 4z(y - 2x)$
$= (y - 2x)(y - 4z)$
View full question & answer→Question 1052 Marks
Factorise the following using the identity $a^2- b^2= (a + b)(a - b).$
$4x^2- 25y^2$
Answer$4x^2- 25y^2$
$(2x)^2- (5y)^2$
$= (2x - 5y)(2x + 5)$
View full question & answer→Question 1062 Marks
Expand the following, using suitable identities.$\Big(\frac{2}{3}\text{x}-\frac{3}{2}\text{y}\Big)^2$
Answer$\Big(\frac{2}{3}\text{x}-\frac{3}{2}\text{y}\Big)^2$$=\Big(\frac{2}{3}\text{x}\Big)^2+\Big(\frac{3}{2}\text{y}\Big)^2-2\times\frac{2}{3}\text{x}\times\frac{3}{2}\text{y}$
$=\frac{4}{9}\text{x}^2+\frac{9}{4}\text{y}^2-2\text{xy}$
View full question & answer→Question 1072 Marks
Multiply the following:
$-7st, -1, -13st^2$
Answer$-7st, -1, -13st^2$
$-7st × (-1) × (-13st^2)$
$= [-7 × (-1) × (-13)]st × (st^2)$
$= -91s^2t^3$
View full question & answer→Question 1082 Marks
If $p + q = 12$ and $pq = 22$, then find $p^2+ q^2$.
AnswerGiven, $p + q = 12$ and $pq = 22$
Since,
$(p+q)^2=p^2+q^2+2 p q $
$ (12)^2=p^2+q^2+2(22) $
$ p^2+q^2=(12)^2-44 $
$ p^2+q^2=144-44 $
$= 100$
View full question & answer→Question 1092 Marks
Using suitable identities, evaluate the following.
$(52)^2$
Answer$(52)^2$
$= (50 + 2)^2$
$= (50)^2+ (2)^2+ 2 × 50 × 2$
$= 2500 + 4 + 200$
$= 2704$
View full question & answer→Question 1102 Marks
Expand the following, using suitable identities.
$ (0.9 p-0.5 q)^2 $
Answer$ (0.9 p-0.5 q)^2 $
$ =(0.9 p)^2+(0.5 q)^2-2 \times 0.9 p \times 0.5 q $
$ =0.81 p^2+0.25 q^2-0.9 p q $
View full question & answer→Question 1112 Marks
Factorise the following. $x^2+ 9x + 20$
Answer$x^2+ 9x + 20$
$= x^2+ 5x + 4x + 5 × 4$
$= x(x + 5) + 4(x + 5)$
$= (x + 5)(x + 4)$
View full question & answer→Question 1122 Marks
Write the greatest common factor in each of the following terms.
$l^2m^2n,l^2mn^2, l^2mn^2$
Answer$l^2m^2n,l^2mn^2, l^2mn^2$
$l^2m^2n= l × l × m × m × n$
$l^2mn^2= l × m × m × n$
$l^2mn^2= l × l × m × n × n$
View full question & answer→Question 1132 Marks
Using suitable identities, evaluate the following. $101 \times 103$
Answer$101 × 103$
$= (100 + 1)(100 + 3)$
$= (100)^2+ (1 + 3)100 + 3 × 1$
$= 10000 + 400 + 3$
$= 10403$
View full question & answer→Question 1142 Marks
Multiply the following: $abc, (bc + ca)$
Answer$abc, (bc + ca)$
$abc \times (bc + ca)$
$= abc \times bc + abc \times ca$
$=a b^2 c^2+a^2 b c^2$
View full question & answer→Question 1152 Marks
Expand the following, using suitable identities.$\Big(\frac{2\text{x}}{3}-\frac{2}{3}\Big)\Big(\frac{2\text{x}}{3}+\frac{2\text{a}}{3}\Big)$
Answer$\Big(\frac{2\text{x}}{3}-\frac{2}{3}\Big)\Big(\frac{2\text{x}}{3}+\frac{2\text{a}}{3}\Big)$$=\Big(\frac{2\text{x}}{3}\Big)^2+\Big(-\frac{2}{3}+\frac{2\text{a}}{3}\Big)+\Big(-\frac{2}{3}+\frac{2\text{a}}{3}\Big)\frac{2\text{x}}{3}+\Big(-\frac{2}{3}\times\frac{2\text{a}}{3}\Big)$
$=\frac{4\text{x}^2}{9}+\frac{2\text{a}-2}{3}\times\frac{2}{3}\text{x}-\frac{4}{9}\text{a}$
$=\frac{4\text{x}^2}{9}+\frac{4}{9}(\text{a}-1)\text{x}-\frac{4}{9}\text{a}$
View full question & answer→Question 1162 Marks
Factorise $p^4+ q^4+ p^2q^2$.
AnswerWe have,
$ p^4+q^4+p^2 q^2 $
$ =p^4+q^4+2 p^2 q^2-2 p^2 q^2+p^2 q^2 $
$ =p^4+q^4+2 p^2 q^2-p^2 q^2 $
$ =\left[\left(p^2\right)^2+\left(q^2\right)^2+2 p^2 q^2\right]-p^2 q^2 $
$ =\left(p^2+q^2\right)^2-(p q)^2 $
$ =\left(p^2+q^2+p q\right)\left(p^2+q^2-p q\right) $
View full question & answer→Question 1172 Marks
Simplify:
$ (2.5 m+1.5 q)^2+(2.5 m-1.5 q)^2 $
Answer$ (2.5 m+1.5 q)^2+(2.5 m-1.5 q)^2 $
$ =(2.5 m)^2+(1.5 q)^2+2 \times 2.5 m \times 1.5 q+(2.5 m)^2+(1.5 q)^2-2 \times(2.5 m) \times(1.5 q) $
$ =6.25 m^2+2.25 q^2+6.25 m^2+2.25 q^2 $
$ =(6.25+6.25) m^2+(2.25+2.25) q^2 $
$ =12.5 m^2+4.5 q^2 $
View full question & answer→Question 1182 Marks
Factorise the following, using the identity $a^2+2 a b+b^2=(a+b)^2$
$16 x^2+40 x+25$
Answer$16 x^2+40 x+25$
$=(4 x)^2+2 \times 4 x \times 5+5^2$
$=(4 x+5)^2$
$=(4 x+5)(4 x+5)$
View full question & answer→Question 1192 Marks
Multiply the following:
$7pqr, (p - q + r)$
Answer$7pqr, (p - q + r)$
$7pqr \times (p - q + r)$
$= 7qr \times p - 7pqr \times q + 7qr \times r$
$=7 p^2 q r-7 p q^2 r+7 p q r^2$
View full question & answer→Question 1202 Marks
If $a + b = 25$ and $a^2+ b^2= 225$, then find $ab$.
AnswerGiven,
$a + b = 25$ and $a^2+ b^2= 225$
$(a + b)^2= a^2+ b^2+ 2ab (25)^2$
$= 225 + 2ab 2ab$
$= (25)^2- 225 2ab$
$= 625 - 225 2ab$
$= 400$
$2\text{ab}=\frac{400}{2}$
$= 200$
View full question & answer→Question 1212 Marks
Factorise the following using the identity $a^2- b^2= (a + b)(a - b).$
$ 28 a y^2-175 a x^2 $
Answer$ 28 a y^2-175 a x^2 $
$ =7 a\left(4 y^2-25 x^2\right) $
$ =7 a\left[(2 y)^2-(5 x)^2\right] $
$ =7 a(2 y-5 x)(2 y+5 x) $
View full question & answer→Question 1222 Marks
Using suitable identities, evaluate the following.
$(69.3)^2-(30.7)^2$
Answer$(69.3)^2-(30.7)^2$
$= (69.3 + 30.7)(69.3 - 30.7)$
$= 100 \times 38.6$
$= 3860$
View full question & answer→Question 1232 Marks
Add:
$ \left(5 x^2-3 x y+4 y^2-9\right)+\left(7 y^2+5 x y-2 x^2+13\right) $
Answer$ \left(5 x^2-3 x y+4 y^2-9\right)+\left(7 y^2+5 x y-2 x^2+13\right) $
$ =5 x^2-3 x y+4 y^2-9+7 y^2+5 x y-2 x^2+13 $
$ =\left(5 x^2-2 x^2\right)+(-3 x y+5 x y)+\left(4 y^2+7 y^2\right)+(-9+13)$
$ =3 x^2+2 x y+11 y^2+4 $
View full question & answer→Question 1242 Marks
Simplify:
$(1.5 p+1.2 q)^2-(1.5 p-1.2 q)^2$
Answer$(1.5 p+1.2 q)^2-(1.5 p-1.2 q)^2$
$= [(1.5p + 1.2q) + (1.5p - 1.2q)][(1.5p + 1.2q) - (1.5p − 1.2q)]$
$= [(1.5p + 1.5p) + (1.2q - 1.2q)][(1.5p - 1.5p) + (1.2q + 1.2q)]$
$= 3p × 2.4q$
$= 7.2pq$
View full question & answer→Question 1252 Marks
Factorise the following, using the identity $a^2-2 a b+b^2=(a-b)^2.$
$9\text{y}^2-4\text{xy}+\frac{\text{y}^2}{9}$
Answer$9\text{y}^2-4\text{xy}+\frac{\text{y}^2}{9}$$=\big(3\text{y}\big)^2-2\times3\text{y}\times\frac{2}{3}\text{x}+\Big(\frac{2}{3}\text{x}\Big)^2$
$=\Big(3\text{y}-\frac{2}{3}\text{x}\Big)^2$
$=\Big(3\text{y}-\frac{2}{3}\text{x}\Big)\Big(3\text{y}-\frac{2}{3}\text{x}\Big)$
View full question & answer→Question 1262 Marks
Factorise the following using the identity $a^2-b^2=(a+b)(a-b)$.
$\frac{\text{x}^3\text{y}}{9}-\frac{\text{xy}^3}{16}$
Answer$\frac{\text{x}^3\text{y}}{9}-\frac{\text{xy}^3}{16}$$=\text{xy}\Big(\frac{\text{x}^2}{9}-\frac{\text{y}^2}{16}\Big)$
$=\text{xy}\bigg[\Big(\frac{\text{x}}{3}\Big)^2-\Big(\frac{\text{y}}{4}\Big)^2\bigg]$
$=\text{xy}\Big(\frac{\text{x}}{3}+\frac{\text{y}}{4}\Big)\Big(\frac{\text{x}}{3}-\frac{\text{y}}{4}\Big)$
View full question & answer→Question 1272 Marks
Multiply the following:
$ -3 x^2 y,(5 y-x y) $
Answer$ -3 x^2 y,(5 y-x y) $
$ -3 x^2 y \times(5 y-x y) $
$ =-3 x^2 y \times 5 y+3 x^2 y \times x y $
$=-15 x^2 y^2+3 x^3 y^2 $
View full question & answer→Question 1282 Marks
The base of a parallelogram is ($2x + 3$ units) and the corresponding height is ($2x - 3$ units). Find the area of the parallelogram in terms of $x$. What will be the area of parallelogram of $x = 30$ units?
AnswerThe base & the corresponding height of a parallelogram are $(2? + 3)$ units & $(2x - 3)$units, respectively.
Area of a parallelogram = Base $\times $ Height
$= (2x + 3) × (2x - 3)$
$= (2x)^2- (3)^2$
$= (4x - 9)^2$ units
Now, if $x = 10.$
Then, the area of parallelogram $= 4 × (10)^2- 9$
$= 400 - 9$
$= 391$ sq. units
View full question & answer→Question 1292 Marks
Factorise the expressions and divide them as directed: $\left(x^3+x^2-132 x\right) \div x(x-11)$
Answer$\left(x^3+x^2-132 x\right) \div x(x-11)$
$=\frac{\text{x}^3+\text{x}^2-132\text{x}}{\text{x}(\text{x}-11)}$
$=\frac{\text{x}(\text{x}^2+\text{x}-132)}{\text{x}(\text{x}-11)}$
$=\frac{\text{x}^2+12\text{x}-11\text{x}-132}{\text{x}-11}$
$=\frac{\text{x}(\text{x}+12)-11(\text{x}+12)}{\text{x}-11}$
$=\frac{(\text{x}+12)(\text{x}-11))}{\text{x}-11}$
$=\text{x}+12$
View full question & answer→Question 1302 Marks
Expand the following, using suitable identities.$\Big(\frac{4\text{x}}{5}+\frac{\text{y}}{4}\Big)\Big(\frac{4\text{x}}{5}+\frac{3\text{y}}{4}\Big)$
Answer$\Big(\frac{4\text{x}}{5}+\frac{\text{y}}{4}\Big)\Big(\frac{4\text{x}}{5}+\frac{3\text{y}}{4}\Big)$$=\Big(\frac{4\text{x}}{5}\Big)^2+\Big(\frac{\text{y}}{4}+\frac{3\text{y}}{4}\Big)\frac{4\text{x}}{5}+\frac{\text{y}}{4}\times\frac{3\text{y}}{4}$
$=\frac{16}{25}\text{x}^2+\frac{4\text{xy}}{5}+\frac{3\text{y}^2}{16}$
View full question & answer→Question 1312 Marks
Multiply the following:
$ a, a^5, a^6 $
Answer$ a, a^5, a^6 $
$ a \times a^5 \times a^6=a^{1+5+6} $
$ =a^{12} $
View full question & answer→Question 1322 Marks
Expand the following, using suitable identities. $(2x - 5y)(2x - 5y)$
Answer$(2x - 5y)(2x - 5y)$
$ =(4 x)^2+(5 y)^2-2 \times 2 x \times 5 y $
$ =16 x^2+25 y^2-20 x y $
View full question & answer→Question 1332 Marks
Simplify:$\Big(\frac{3}{4}\text{x}-\frac{4}{3}\text{y}\Big)^2+2\text{xy}$
Answer$\Big(\frac{3}{4}\text{x}-\frac{4}{3}\text{y}\Big)^2+2\text{xy}$$=\Big(\frac{3}{4}\text{x}\Big)^2+\Big(\frac{4}{3}\text{y}\Big)^2-2\times\frac{3}{4}\text{x}\times\frac{4}{3}\text{y}+2\text{xy}$
$=\frac{9}{16}\text{x}^2+\frac{16}{9}\text{y}^2-2\text{xy}+2\text{xy}$
$=\frac{9}{16}\text{x}^2+\frac{16}{9}\text{y}^2$
View full question & answer→Question 1342 Marks
Simplify:$\big(\frac{7}{9}\text{a}+\frac{9}{7}\text{b}\Big)^2-\text{ab}$
Answer$\big(\frac{7}{9}\text{a}+\frac{9}{7}\text{b}\Big)^2-\text{ab}$$=\Big(\frac{7}{9}\text{a}\Big)^2+\Big(\frac{9}{7}\text{b}\Big)^2+2\times\frac{7}{9}\text{a}\times\frac{9}{7}\text{b}-\text{ab}$
$=\frac{49}{81}\text{a}^2+\frac{81}{49}\text{b}^2+2\text{ab}-\text{ab}$
$=\frac{49}{81}\text{a}^2+\text{ab}+\frac{81}{49}\text{b}^2$
View full question & answer→Question 1352 Marks
Expand the following, using suitable identities.
$ \left(x^2 y-x y^2\right)^2 $
Answer$ \left(x^2 y-x y^2\right)^2 $
$ =\left(x^2 y\right)^2+\left(x y^2\right)^2-2\left(x^2 y\right)^2\left(x y^2\right) $
$ =x^4 y^2+x^2 y^4-2 x^3 y^3 $
View full question & answer→Question 1362 Marks
Multiply the following: $(pq - 2r), (pq - 2r)$
Answer$(pq - 2r), (pq - 2r)$
$(pq - 2r)(pq - 2r)$
$= pq(pq - 2r) - 2r(pq - 2r)$
$ =p^2 q^2-2 p q r-2 r p a+4 r^2 $
$ =p^2 q^2-4 p q r+4 r^2 $
View full question & answer→Question 1372 Marks
Simplify:
$ (a b-c)^2+2 a b c $
Answer$ (a b-c)^2+2 a b c $
$ =(a b)^2+c^2-2 a b c+2 a b c $
$ =a^2 b^2+c^2 $
View full question & answer→Question 1382 Marks
If $\text{x}-\frac{1}{\text{x}}=7$ then find the value of $\text{x}^2+\frac{1}{\text{x}^2}$.
AnswerGiven,$\text{x}-\frac{1}{\text{x}}=7$
Since,$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2.\text{x}.\frac{1}{\text{x}}$
$7^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\text{x}^2+\frac{1}{\text{x}^2}=49+2$
$\text{x}^2+\frac{1}{\text{x}^2}=51$
View full question & answer→Question 1392 Marks
Write the greatest common factor in each of the following terms.
$ 3 x^3 y^2 z,-6 x y^3 z^2, 12 x^2 y z^3 $
Answer$ 3 x^3 y^2 z,-6 x y^3 z^2, 12 x^2 y z^3 $
$ 3 ?^3 ?^2 ?=3 \times x \times x \times x \times y \times y \times z $
$ -6 x y^3 z^2=-3 \times 2 \times x \times y \times y \times y \times z \times z $
$ 12 x^2 y z^3=3 \times 4 \times x \times x \times y \times z \times z \times z G C F=3 x y z $
View full question & answer→Question 1402 Marks
Using suitable identities, evaluate the following.
$(98)^2$
Answer$ (98)^2$
$ =(100-2)^2 $
$ =(100)^2+(2)^2-2 \times 100 \times 2 $
$ =10000+4-400 $
$ =9604$
View full question & answer→Question 1412 Marks
Expand the following, using suitable identities. $(2x + 9)(2x - 7)$
Answer$(2x + 9)(2x - 7)$
$= (2x + 9)[2x + (-7)]$
$= (2x)^2+ [9 + (-7)] 2x + 9 × (-7)$
$= 4x^2+ 4x - 6$
View full question & answer→Question 1422 Marks
Factorise the following expressions.
$ a^3 x-x^4+a^2 x^2-a x^3 $
Answer$ a^3 x-x^4+a^2 x^2-a x^3 $
$ =x\left(a^3-x^2+a^2 x-a x^2\right) $
$ =x\left(a^3+a^2 x-x^3-ax^2\right) $
$ =x\left[a^2(a+x)-x^2(x+?)\right] $
$ =x\left[(x+a)\left(a^2-x^2\right)\right] $
$ =x\left(a^2-x^2\right)(a+x) $
View full question & answer→Question 1432 Marks
Factorise the following expressions.
$ a x^3-b x^2+c x $
Answer$ a x^3-b x^2+c x $
$ =x\left(a x^2-b x+c\right) $
View full question & answer→Question 1442 Marks
Multiply the following:
$ 15 x y^2, 17 y z^2 $
Answer$ 15 x y^2, 17 y z^2 $
$ 15 x y^2 \times 17 y z^2 $
$ =(15 \times 17) xy^2 \times y z^2 $
$=255 x y^3 z^3 $
View full question & answer→Question 1452 Marks
Factorise the following using the identity $a^2- b^2= (a + b)(a - b).$
$ y^4-625 $
Answer$ y^4-625 $
$ =\left(y^2\right)^2-(25)^2 $
$ =\left(y^2+25\right)\left(y^2-25\right) $
$ =\left(y^2+25\right)\left(y^2-5^2\right) $
$ =\left(y^2+25\right)(y+5)(y-5) $
View full question & answer→Question 1462 Marks
Factorise the following expressions. $-xy - ay$
Answer$-xy - ay = -y(x + a)$
View full question & answer→Question 1472 Marks
Subtract: $7p(3q + 7p)$ from $8p(2p - 7q)$
AnswerThe required difference is given by
$8p(2p - 7q) - 7p(3q + 7p)$
$ =16 p^2-56 p q-21 p q-49 p^2 $
$ =\left(16 p 2-49 p^2\right)+(-56 p q-21 p q) $
$ =-33 p^2-77 p q $
View full question & answer→Question 1482 Marks
Simplify:
$ (4.5 a+1.5 b)^2+(4.5 b+1.5 a)^2$
Answer$ (4.5 a+1.5 b)^2+(4.5 b+1.5 a)^2$
$ =(4.5 a)^2+(1.5 b)^2+2 \times 4.5 a \times 1.5 b+(4.5 b)^2+(1.5 a)^2+2 \times 4.5 b \times 1.5 a $
$ =20.25 a^2+2.25 b^2+13.5 a b+20.25 b^2+2.25 a^2+13.5 a b $
$ =40.5 a^2+4.5 b^2+27 a b $
View full question & answer→Question 1492 Marks
The sum of first n natural numbers is given by the expression $\frac{\text{n}^2}{2}+\frac{\text{n}}{2}$. Factorise this expression.
Answer The sum of first n natural numbers $=\frac{\text{n}^2}{2}+\frac{\text{n}}{2}$
Factorisation of given expression $=\frac{1}{2}(\text{n}^2+\text{n})=\frac{1}{2}\text{n}(\text{n}+1)$ View full question & answer→Question 1502 Marks
Factorise the following, using the identity $a^2+ 2ab + b^2= (a + b)^2$
$x^2+ 6x + 9$
Answer$x^2+ 6x + 9$
$= x^2+ 2 × 3 × x + 3^2$
$= (x + 3)^2$
$= (x + 3)(x + 3)$
View full question & answer→