Question 12 Marks
$(79)^2-(69)^2$
AnswerHere, we will use the identity $(a - b) (a + b) = a^2- b^2$
Let us consider the following expression:
$(79)^2-(69)^2$
$= (79 + 69)(79 - 69)$
$= 148 × 10$
$= 1480$
View full question & answer→Question 22 Marks
Add the following algebric expressions: $\frac{2}{3}\text{a,}\frac{3}{5}\text{a,}-\frac{6}{5}$
AnswerTo add the like terms, we proceed as follows: $\frac{2}{3}\text{a}+\frac{3}{5}\text{a}+\big(-\frac{6}{5}\text{a}\big)$ $=\frac{2}{3}\text{a}+\frac{3}{5}\text{a}-\frac{6}{5}\text{a}$ $=\big(\frac{2}{3}+\frac{3}{5}-\frac{6}{5}\big)\text{a}$ (Distributive Law) $=\frac{1}{15}\text{a}$
View full question & answer→Question 32 Marks
Find the following products: $ \left(x^2+4\right)\left(x^2+9\right) $
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$ \left(x^2+4\right)\left(x^2+9\right) $
$ =\left(x^2\right)^2+(4+9)\left(x^2\right)+4 \times 9 $
$ =x^4+13 x^2+36 $
View full question & answer→Question 42 Marks
Find the following products: $\big(\text{x}+\frac{4}{3}\big)\big(\text{x}+\frac{3}{4}\big)$
AnswerHere, we will use the identity$(x+a)(x+b)=x^2+(a+b) x+a b$.
$\big(\text{x}+\frac{4}{3}\big)\big(\text{x}+\frac{3}{4}\big)$
$=\text{x}^2+\big(\frac{4}{3}+\frac{3}{4}\big)\text{x}+\frac{4}{3}×\frac{3}{4}$
$=\text{x}^2+\frac{25}{12}\text{x}+1$
View full question & answer→Question 52 Marks
Find the following products: $(x - 3)( x - 2)$
AnswerHere, we will use the identity $(x - a)(x - b) = x^2- (a + b)x + ab.$
$(x - 3)( x - 2)$
$=x^2-(3+2) x+3 \times 2 $
$ =x^2-5 x+6$
View full question & answer→Question 62 Marks
Multiply: $ \left(2 x^2-1\right)\left(4 x^3+5 x^2\right) $
AnswerTo multiply, we will use distributive law as follows:
$ \left(2 x^2-1\right)\left(4 x^3+5 x^2\right) $
$ =2 x^2\left(4 x^3+5 x^2\right)-1\left(4 x^3+5 x^2\right)$
$ =8 x^5+10 x^4-4 x^3-5 x^2 $
Thus, the answer is $ =8 x^5+10 x^4-4 x^3-5 x^2 $.
View full question & answer→Question 72 Marks
Find product: $\big(\frac{4}{3}\text{pq}^2\big)×\big(\frac{−1}{4}\text{p}^2\text{r}\big)×\big(16\text{p}^2\text{q}^2\text{r}^2\big)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$ We have:
$\big(\frac{4}{3}\text{pq}^2\big)×\big(\frac{−1}{4}\text{p}^2\text{r}\big)×\big(16\text{p}^2\text{q}^2\text{r}^2\big)$
$=\big[\frac{4}{3}×\big(\frac{−1}{4}\big)×16]×\big(\text{p}×\text{p}^2×\text{p}^2\big)$
$×\big(\text{q}^2×\text{q}^2\big)×\big(\text{r}×\text{r}^2\big)$
$=\big[\frac{4}{3}×\big(\frac{−1}{4}\big)×16]×\big(\text{p}^{1+2+2}\big)×\big(\text{q}^{2+2}\big)×\big(\text{r}^{1+2}\big)$
$=−\frac{16}{3}\text{p}^5\text{q}^4\text{r}^3$
Thus, the answer is $=−\frac{16}{3}\text{p}^5\text{q}^4\text{r}^3$
View full question & answer→Question 82 Marks
Simplify the following using the identities: $178 × 178 - 22 × 22$
Answer$178 × 178 - 22 × 22$
Let us consider the following expression:
Using the identity $(a + b) (a - b) = a^2 - b^2$
$178 × 178 - 22 × 22$
$= 178^2- 22^2$
$= (178 + 22)(178 - 22)$
$= 200 × 156$
$= 31200$
Thus, the answer is $31200.$
View full question & answer→Question 92 Marks
Find the following products: $(3x + 5)(3x + 11)$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$(3x + 5)(3x + 11)$
$= (3x)^2+ (5 + 11)(3x) + 5 × 11$
$= 9x^2+ 48x + 55$
View full question & answer→Question 102 Marks
Find following product: $ \left(-4 x^2\right) \times\left(-6 x y^2\right) \times\left(-3 y z^2\right) $
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m× a^n= a^{m+n}$.
We have:
$ \left(-4 x^2\right) \times\left(-6 x y^2\right) \times\left(-3 y z^2\right) $
$ =[(-4) \times(-6) \times(-3)] \times\left(x^2 \times x\right) \times\left(y^2 \times y\right) \times z^2 $
$ {[(-4) \times(-6) \times(-3)] \times\left(x^2+1\right) \times\left(y^2+1\right) \times z^2} $
$ =-72 x^3 y^3 z^2 $
View full question & answer→Question 112 Marks
Simplify: $ \left(m^2-n^2 m\right)^2+2 m^3 n^2 $
AnswerTo simplify, we will proceed as follows:
$ \left(m^2-n^2 m\right)^2+2 m^3 n^2 $
$ =(m 2)^2+(n 2 m)^2\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =m^4+n^4 m^2 $
View full question & answer→Question 122 Marks
Evaluate the following: $34 \times 36$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$34 × 36$
$= (30 + 4) (30 + 6)$
$= 30^2+ (4 + 6)30 + 4 × 6$
$= 900 + 300 + 24$
$= 1224$
View full question & answer→Question 132 Marks
Find the following products: $\big(\text{x}+\frac{1}{5}\big)\big(\text{x} + 5)$
AnswerHere, we will use the identity $(x+a)(x+b)=x^2+(a+b) x+a b$.
$\big(\text{x}+\frac{1}{5}\big)\big(\text{x} + 5)$
$=\text{x}^2+\big(\frac{1}{5}+5)\text{x}+\frac{1}{5}×5$
$=\text{x}^2+\frac{26}{5}\text{x}+1$
View full question & answer→Question 142 Marks
Evaluate the following: $994 \times 1006$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$994 × 1006$
$= (1000 - 6) × (1000 + 6)$
$= 1000^2+ (6 - 6) × 1000 - 6 × 6$
$= 1000000 - 36$
$= 999964$
View full question & answer→Question 152 Marks
Find the following products: $(z^2+ 2) (z^2- 3)$
AnswerHere, we will use the identity $(x + a)(x - b) = x^2+ (a - b)x - ab.$
$\left(z^2+2\right)\left(z^2-3\right) $
$ =\left(z^2\right) 2+(2-3)\left(z^2\right)-2 \times 3 $
$ =z^4-z^2-6 $
View full question & answer→Question 162 Marks
Show that: $(4 p q+3 q)^2-(4 p q-3 q)^2=48 p q^2$
Answer$(4 p q+3 q)^2-(4 p q-3 q)^2=48 p q^2$
LHS $= (4pq + 3q)^2- (4pq - 3q)^2$
$= 4(4pq) (3q) [∵(a + b)^2- (a + b)^2= 4ab]$
$= 48pq^2$
= RHS
View full question & answer→Question 172 Marks
Find product: $(2.3xy) \times (0.1x) \times (0.16)$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.
We have:
$(2.3xy) × (0.1x) × (0.16)$
$= (2.3 × 0.1 × 0.16) × (x × x) × y$
$= (2.3 × 0.1 × 0.16) × (x^{1+1}) × y$
$= 0.0368x^2y$
Thus, the answer is $0.0368x^2y$.
View full question & answer→Question 182 Marks
Multiply:
$ \left(2 x y+3 y^2\right)\left(3 y^2-2\right) $
AnswerTo multiply, we will use distributive law as follows:
$ \left(2 x y+3 y^2\right)\left(3 y^2-2\right) $
$ =2 x y\left(3 y^2-2\right)+3 y^2\left(3 y^2-2\right) $
$ =6 x y^3-4 x y+9 y^4-6 y^2 $
$ =9 y^4+6 x y^3-6 y^2-4 x y $
Thus, the answer is $ 9 y^4+6 x y^3-6 y^2-4 x y $
View full question & answer→Question 192 Marks
Using the formula for squaring a binomial, evaluate the following:
$ (999)^2$
AnswerHere, we will use the identity $(a + b)^2= a^2+ 2ab + b^2$.
$ (999)^2 $
$ =(1000-1)^2 $
$ =(1000)^2-2 \times 1000 \times 1+1^2 $
$ =1000000-2000+1 $
$ =998001 $
View full question & answer→Question 202 Marks
Find the following products: $(2x^2- 3) (2x^2+ 5)$
AnswerHere, we will use the identity $(x - a)(x + b) = x^2+ (b + a)x - ab.$
$ \left(2 x^2-3\right)\left(2 x^2+5\right) $
$ =\left(2 x^2\right)^2+(5-3)\left(2 x^2\right)-3 \times 5 $
$ =4 x^4+4 x^2-15 $
View full question & answer→Question 212 Marks
Add the following algebric expressions:
$3 a^2 b-4 a^2 b, 9 a^2 b$
View full question & answer→Question 222 Marks
Multiply:
$ \left(2 x^2 y^2-5 x y^2\right)\left(x^2-y^2\right)$
AnswerTo multiply, we will use distributive law as follows:
$ \left(2 x^2 y^2-5 x y^2\right)\left(x^2-y^2\right)$
$ =2 x^2 y^2\left(x^2-y^2\right)-5 x y^2\left(x^2-y^2\right) $
$ =2 x^4 y^2-2 x^2 y^4-5 x^3 y^2+5 x y^4 $
Thus, the answer is $ 2 x^4 y^2-2 x^2 y^4-5 x^3 y^2+5 x y^4 $.
View full question & answer→Question 232 Marks
$(467)^2- (33)^2$
AnswerHere, we will use the identity $(a - b) (a + b) = a^2- b^2$
Let us consider the following expression:
$(467)^2- (33)^2$
$= (467 + 33)(467 - 33)$
$= 500 × 434$
$= 217000$
View full question & answer→Question 242 Marks
Simplify:
$ (4 m-8 n)^2+(7 m+8 n)^2 $
AnswerTo simplify, we will proceed as follows:
$ (4 m-8 n)^2+(7 m+8 n)^2 $
$ =2(7 m)^2+2(8 n)^2\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =98 m^2+128 n^2 $
View full question & answer→Question 252 Marks
Simplify the following using the identities: $1.73 \times 1.73 - 0.27 \times 0.27$
AnswerLet us consider the following expression:
$1.73 × 1.73 - 0.27 × 0.27$
Using the identity $(a + b) (a - b) = a^2- b^2$
we get:
$1.73 × 1.73 - 0.27 × 0.27 = 1.732 - 0.272 = (1.73 + 0.27)(1.73 - 0.27) = 2 × 1.46 = 2.92$
Thus, the answer is $2.92.$
View full question & answer→Question 262 Marks
Evaluate the following: $103 \times 96$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$103 × 96$
$= (100 + 3) (100 - 4)$
$= 100^2+ (3 - 4)100 - 3 × 4$
$= 10000 - 100 - 12$
$= 9888$
View full question & answer→Question 272 Marks
Find the following products: $\big(\text{z}+\frac{3}{4}\big)\big(\text{z}+\frac{4}{3}\big)$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$\big(\text{z}+\frac{3}{4}\big)\big(\text{z}+\frac{4}{3}\big)$
$=\text{z}^2+\big(\frac{3}{4}+\frac{4}{3})\text{z}+\frac{3}{4}×\frac{4}{3}$
$=\text{z}^2+\frac{25}{12}\text{x}+1$
View full question & answer→Question 282 Marks
Find the following products: $(y^2- 4)(y^2- 3)$
AnswerHere, we will use the identity $(x - a)(x - b) = x^2- (a + b)x + ab.$
$ \left(y^2-4\right)\left(y^2-3\right) $
$ =(y 2)^2-(4+3)\left(y^2\right)+4 \times 3 $
$ =y^4-7 y^2+12 $
View full question & answer→Question 292 Marks
Find product: $\big(\frac{7}{9}\text{ab}^2\big)×\big(\frac{15}{7}\text{ac}^2\text{b}\big)×\big(\frac{−3}{5}\text{a}^2\text{c})$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$
We have:
$\big(\frac{7}{9}\text{ab}^2\big)×\big(\frac{15}{7}\text{ac}^2\text{b}\big)×\big(\frac{−3}{5}\text{a}^2\text{c})$
$=\big[\big(\frac{7}{9}\big)×\big(\frac{15}{7}\big)×\big(−\frac{3}{5}\big)\big]×(\text{a}×\text{a}×\text{a}^2)\\×(\text{b}^2×\text{b})×(\text{c}^2×\text{c})$
$=\big[\big(\frac{7}{9}\big)×\big(\frac{15}{7}\big)×\big(−\frac{3}{5}\big)\big]×\big(\text{a}^{1+1+2}\big)\\×\big(\text{b}^{2+1}\big)×\big(\text{c}^{2+1}\big)$
$=−\text{a}^4\text{b}^3\text{c}^3$ Thus, the answer is $=−\text{a}^4\text{b}^3\text{c}^3$
View full question & answer→Question 302 Marks
Subtract: $2a - b$ from $3a - 5b$
Answer$(3a - 5b) - (2a - b) $
$= (3a - 5b) - 2a + b$
$= 3a - 5b - 2a + b$
$= 3a - 2a - 5b + b$
$= a - 4b$
View full question & answer→Question 312 Marks
Simplify: $ a^2 b\left(a-b^2\right)+a b^2\left(4 a b-2 a^2\right)-a^3 b(1-2 b) $
AnswerTo simplify, we will use distributive law as follows:
$ a^2 b\left(a-b^2\right)+a b^2\left(4 a b-2 a^2\right)-a^3 b(1-2 b) $
$ =a^3 b-a^2 b^3+4 a^2 b^3-2 a^3 b^2-a^3 b+2 a^3 b^2 $
$ =a^3 b-a^3 b-a^2 b^3+4 a^2 b^3-2 a^3 b^2+2 a^3 b^2 $
$ =3 a^2 b^3 $
View full question & answer→Question 322 Marks
Find following product:
$ (-5 a) \times\left(-10 a^2\right) \times\left(-2 a^3\right) $
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m× a^n= a^{m+n}$. We have:
$ (-5 a) \times\left(-10 a^2\right) \times\left(-2 a^3\right) $
$ =[(-5) \times(-10) \times(-2)] \times\left(a \times a^2 \times a^3\right) $
$ =[(-5) \times(-10) \times(-2)] \times\left(a^{1+2+3}\right) $
$ =-100 a^6 $
View full question & answer→Question 332 Marks
Multiply:
$ \left(x^2+y^2\right) b y(3 a+2 b) $
AnswerTo multiply, we will use distributive law as follows:
$ \left(x^2+y^2\right) b y(3 a+2 b) $
$ =x^2(3 a+2 b)+y^2(3 a+2 b) $
$ =3 a x^2+2 b x^2+3 a y^2+2 b y^2 $
Thus, the answer is $3 a x^2+2 b x^2+3 a y^2+2 b y^2 $.
View full question & answer→Question 342 Marks
Simplify:
$ a^2(2 a-1)+3 a+a^3-8 $
AnswerTo simplify, we will use distributive law as follows:
$ a^2(2 a-1)+3 a+a^3-8 $
$ =2 a^3-a^2+3 a+a^3-8 $
$ =2 a^3+a^3-a^2+3 a-8 $
$ =3 a^3-a^2+3 a-8 $
View full question & answer→Question 352 Marks
Show that: $\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
Answer$\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$\text{LHS}=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}$
$=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\times \frac{4\text{m}}{3}\times \frac{3\text{n}}{4}$
$=\big(\frac{4\text{m}}{3}\big)^2+\big(\frac{3\text{n}}{4}\big)^2 $
$\big[\because (\text{a}−\text{b}\big)^2+2\text{ab}=\text{a}^2+\text{b}^2\big]$
$=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$=\text{RHS}$ Because $LHS$ is equal to $RHS,$ the given equation is verified.
View full question & answer→Question 362 Marks
Show that:
$ (3 x+7)^2-84 x=(3 x-7)^2 $
AnswerTo simplify, we will proceed as follows:
$ (3 x+7)^2-84 x=(3 x-7)^2 $
$ \text { LHS }=(3 x+7)^2-84 x $
$ =(3 x+7)^2-4 \times 3 x \times 7 $
$ =(3 x-7)^2\left[\because(a+b)(a-b)=a^2-b^2\right] $
$= RHS$
Because $LHS$ is equal to $RHS,$ the given equation is verified.
View full question & answer→Question 372 Marks
Subtract:
$2 x^3-4 x^2+3 x+5$ from $4 x^3+x^2+x+6$
Answer$ \left(4 x^3+x^2+x+6\right)-\left(2 x^3-4 x^2+3 x+5\right)$
$ =4 x^3+x^2+x+6-2 x^3+4 x^2-3 x-5 $
$ =4 x^3-2 x^3+x^2+4 x^2+x-3 x+6-5 \text { (Collecting like terms) } $
$ =2 x^3+5 x^2-2 x+1 $
View full question & answer→Question 382 Marks
Find product: $\big(\frac{4}{3}\text{u}^2\text{vw}\big)×\big(−5\text{uvw}^2)×\big(\frac{1}{3}\text{v}^2\text{wu})$
AnswerTo multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.
We have:
$\big(\frac{4}{3}\text{u}^2\text{vw}\big)×\big(−5\text{uvw}^2)×\big(\frac{1}{3}\text{v}^2\text{wu})$
$=\big[\frac{4}{3}×(−5)×\frac{1}{3}\big]×\big(\text{u}^2×\text{u}×\text{u}\big)$
$×\big(\text{v}×\text{v}×\text{v}^2)×\big(\text{w}×\text{w}^2×\text{w}\big)$
$=\big[\frac{4}{3}×(−5)×\frac{1}{3}\big]×\big(\text{u}^{2+1+1}\big)$
$×\big(\text{v}^{1+1+2}\big)×\big(\text{w}^{1+2+1}\big)$
$=−\frac{20}{9}\text{u}^4\text{v}^4\text{w}^4$
Thus, the answer is $=−\frac{20}{9}\text{u}^4\text{v}^4\text{w}^4$
View full question & answer→Question 392 Marks
Show that:
$(9 a-5 b)^2+180 a b=(9 a+5 b)^2$
Answer$(9 a-5 b)^2+180 a b=(9 a+5 b)^2$
$LHS= (9a - 5b)^2+ 180ab$
$= (9a + 5b)^2$ $ [∵(a + b)(a - b) = a^2- b^2]$
$= RHS$
Because $LHS$ is equal to $RHS,$ the given equation is verified.
View full question & answer→Question 402 Marks
Find the product of the following binomials: $(2x + y)(2x + y)$
AnswerWe will use the identify $(a + b)^2= a^2+ 2ab + b^2$ in the given expression to find the product.
$ (2 x+y)(2 x+y) $
$ =(2 x+y)^2 $
$ =(2 x)^2+2(2 x)(y)+y^2 $
$ =4 x^2+4 x y+y^2 $
View full question & answer→Question 412 Marks
Multiply: $\big(3\text{x}^2\text{y} − 5\text{xy}^2\big) \ \text{by} \ \big(\frac{1}{5}\text{x}^2 + \frac{1}{3}\text{y}^2\big)$
AnswerTo multiply, we will use distributive law as follows: $\big(3\text{x}^2\text{y} − 5\text{xy}^2\big) \ \text{by} \ \big(\frac{1}{5}\text{x}^2 + \frac{1}{3}\text{y}^2\big)$ $=\frac{1}{5}\text{x}^2\big(3\text{x}^2\text{y}−5\text{xy}^2\big)+\frac{1}{3}\text{y}^2\big(3\text{x}^2\text{y}−5\text{xy}^2\big)$ $=\frac{3}{5}\text{x}^4\text{y}−\text{x}^3\text{y}^2+\text{x}^2\text{y}^3−\frac{5}{3}\text{xy}^4$ Thus, the answer is $=\frac{3}{5}\text{x}^4\text{y}−\text{x}^3\text{y}^2+\text{x}^2\text{y}^3−\frac{5}{3}\text{xy}^4.$
View full question & answer→Question 422 Marks
Evaluate the following: $102 \times 106$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$102 × 106$
$= (100 + 2)(100 + 6)$
$= 100^2+ (2 + 6)100 + 2 × 6$
$= 10000 + 800 + 12$
$= 10812$
View full question & answer→Question 432 Marks
Find the following products:$ \left(y^2+12\right)\left(y^2+6\right) $
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$ \left(y^2+12\right)\left(y^2+6\right) $
$ =\left(y^2\right)^2+(12+6)\left(y^2\right)+12 \times 6 $
$ =y^4+18 y^2+72 $
View full question & answer→Question 442 Marks
Simplify: $a(b - c) - b(c - a) - c(a - b)$
AnswerTo simplify, we will use distributive law as follows:
$a(b - c)-b(c - a)-c(a - b)$
$= ab - ac - bc + ba - ca + cb$
$= ab + ba - ac - ca - bc + cb$
$= 2ab - 2ac$
$= 0$
View full question & answer→Question 452 Marks
Multiply: $\big(−\frac{\text{a}}{7}+\frac{\text{a}^2}{9}) \ \text{by}\big(\frac{\text{b}}{2}−\frac{\text{b}^2}{3}\big)$
AnswerTo multiply, we will use distributive law as follows:$\big(−\frac{\text{a}}{7}+\frac{\text{a}^2}{9}) \ \text{by}\big(\frac{\text{b}}{2}−\frac{\text{b}^2}{3}\big)$
$=\big(−\frac{\text{a}}{7}\big)\big(\frac{\text{b}}{2}−\frac{\text{b}^2}{3}\big)+\big(\frac{\text{a}^2}{9}\big)\big(\frac{\text{b}}{2}−\frac{\text{b}^2}{3}\big)$
$=(−\frac{\text{ab}}{14}+\frac{ab^2}{21}\big)+\big(\frac{\text{a}^2\text{b}}{18}−\frac{\text{a}^2\text{b}^2}{27}\big)$
$=−\frac{\text{ab}}{14}+\frac{\text{ab}^2}{21}+\frac{\text{a}^2\text{b}}{18}−\frac{\text{a}^2\text{b}^2}{27}$
Thus, the answer is $−\frac{\text{ab}}{14}+\frac{\text{ab}^2}{21}+\frac{\text{a}^2\text{b}}{18}−\frac{\text{a}^2\text{b}^2}{27}.$
View full question & answer→Question 462 Marks
Show that: $(a - b)(a + b) + (b - c)(b + c) + (c - a)( c + a) = 0$
Answer$(a - b)(a + b) + (b - c)(b + c) + (c - a)( c + a) = 0$
$LHS = (a - b)(a + b) + (b - c)(b + c) + (c - a)( c + a)$
$=a^2-b^2+b^2-c^2+c 2-a^2$
$= 0$
$= RHS$
Because $LHS$ is equal to $RHS$, the given equation is verified.
View full question & answer→Question 472 Marks
Find the following products: $ \big(\text{p}^2 + 16\big) \big(\text{p}^2−\frac{1}{4}\big)$
AnswerHere, we will use the identity $(x+a)(x-b)=x^2+(a-b) x-a b$.
$ \big(\text{p}^2 + 16\big) \big(\text{p}^2−\frac{1}{4}\big)$
$=\big(\text{p}^2\big)2+\big(16−\frac{1}{4}\big)\big(\text{p}^2\big)−16×\frac{1}{4}$
$=\text{p}^4+\frac{63}{4}\text{p}^2−4$
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Find the following products: $(x - 11)(x + 4)$
AnswerHere, we will use the identity $(x - a)(x + b) = x^2+ (b - a)x - ab.$
$(x - 11)(x + 4)$
$= x^2+ (4 - 11)x - 11 × 4$
$= x^2- 7x - 44$
View full question & answer→Question 492 Marks
Evaluate the following: $53 \times 55$
AnswerHere, we will use the identity $(x + a)(x + b) = x^2+ (a + b)x + ab.$
$53 × 55$
$= (50 + 3) (50 + 5)$
$= 50^2+ (3 + 5) 50 + 3 × 5$
$= 2500 + 400 + 15$
$= 2915$
View full question & answer→Question 502 Marks
Multiply: $ (a-1) b y\left(0.1 a^2+3\right) $
AnswerTo multiply, we will use distributive law as follows:
$ (a-1) b y\left(0.1 a^2+3\right) $
$ =0.1 a^2(a-1)+3(a-1) $
$ =0.1 a^3-0.1 a^2+3 a-3 $
Thus, the answer is $0.1 a^3-0.1 a^2+3 a-3 $.
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