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:
$(79)^2-(69)^2$
$=(79+69)(79-69)$
$=148 \times 10$
$=1480$
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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$\left(x^2+4\right)\left(x^2+9\right)$
$=\left(x^2\right)^2+(4+9)\left(x^2\right)+4 \times 9$
$=x^4+13 x^2+36$
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 $(x-a)(x-b)=x^2-(a+b) x+a b$.
$(x-3)(x-2)$
$=x^2-(3+2) x+3 \times 2$
$=x^2-5 x+6$
View full question & answer→Question 62 Marks
Multiply:
$\left(2 x^2-1\right) \text { by }\left(4 x^3+5 x^2\right)$
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:
$178 \times 178-22 \times 22$
Answer$178 \times 178-22 \times 22$
Let us consider the following expression: Using the identity $(a+b)(a-b)=a^2-b^2$
$178 \times 178-22 \times 22$
$=178^2-22^2$
$=(178+22)(178-22)$
$=200 \times 156$
$=31200$
Thus, the answer is 31200 .
View full question & answer→Question 92 Marks
Find the following products:
$(3 x+5)(3 x+11)$
AnswerHere, we will use the identity $(x+a)(x+b)=x^2+(a+b) x+a b$.
$(3 x+5)(3 x+11)$
$=(3 x)^2+(5+11)(3 x)+5 \times 11$
$=9 x^2+48 x+55$
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, $a^{ m } \times a ^{ n }= a ^{ m + n }$.
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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$34 \times 36$
$=(30+4)(30+6)$
$=30^2+(4+6) 30+4 \times 6$
$=900+300+24$
$=1224$
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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$994 \times 1006$
$=(1000-6) \times(1000+6)$
$=1000^2+(6-6) \times 1000-6 \times 6$
$=1000000-36$
$=999964$
View full question & answer→Question 152 Marks
Find the following products:
$\left(z^2+2\right)\left(z^2-3\right)$
AnswerHere, we will use the identity $(x+a)(x-b)=x^2+(a-b) x-a b$.
$\left(z^2+2\right)\left(z^2-3\right)$
$=\left(z^2\right) 2+(2-3)\left(z^2\right)-2 \times 3$
$=z^4-z^2-6$
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$(4 p q+3 q)^2-(4 p q-3 q)^2=48 p q^2$
$\text { LHS }=(4 p q+3 q)^2-(4 p q-3 q)^2$
$=4(4 p q)(3 q)\left[\because(a+b)^2-(a+b)^2=4 a b\right]$
$=48 p q^2$
$=\text { RHS }$
View full question & answer→Question 172 Marks
Find product:
$(2.3 x y) \times(0.1 x) \times(0.16)$
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:
$(2.3 x y) \times(0.1 x) \times(0.16)$
$=(2.3 \times 0.1 \times 0.16) \times(x \times x) \times y$
$=(2.3 \times 0.1 \times 0.16) \times\left(x^{1+1}\right) \times y$
$=0.0368 x^2 y$
Thus, the answer is $0.0368 x^2 y$.
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 $(a+b)^2=a^2+2 a b+b^2$.
$(999)^2$
$=(1000-1)^2$
$=(1000)^2-2 \times 1000 \times 1+1^2$
$=1000000-2000+1$
$=998001$
View full question & answer→Question 202 Marks
Find the following products:
$\left(2 x^2-3\right)\left(2 x^2+5\right)$
AnswerHere, we will use the identity $(x-a)(x+b)=x^2+(b+a) x-a b$.
$\left(2 x^2-3\right)\left(2 x^2+5\right)$
$=\left(2 x^2\right)^2+(5-3)\left(2 x^2\right)-3 \times 5$
$=4 x^4+4 x^2-15$
View full question & answer→Question 212 Marks
Add the following algebric expressions:
$3 a^2 b-4 a^2 b, 9 a^2 b$
AnswerTo add the like terms, we proceed as follows:
$3 a^2 b+\left(-4 a^2 b\right)+9 a^2 b$
$=3 a^2 b-4 a^2 b+9 a^2 b \text { (Distribute Law) }$
$=(3-4+9) a^2 b$
$=8 a^2 b$
View full question & answer→Question 222 Marks
Multiply:
$\left(2 x^2 y^2-5 x y^2\right) \text { by }\left(x^2-y^2\right)$
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:
$(467)^2-(33)^2$
$=(467+33)(467-33)$
$=500 \times 434$
$=217000$
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:
$1.73 \times 1.73-0.27 \times 0.27$
Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$1.73 \times 1.73-0.27 \times 0.27=1.732-0.272=(1.73+0.27)(1.73-0.27)=2 \times 1.46=2.92$
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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$103 \times 96$
$=(100+3)(100-4)$
$=100^2+(3-4) 100-3 \times 4$
$=10000-100-12$
$=9888$
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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$\big(\text{z}+\frac{3}{4}\big)\big(\text{z}+\frac{4}{3}\big)$
$=\text{z}^2+\big(\frac{3}{4}+\frac{4}{3})\text{z}+\frac{3}{4}×\frac{4}{3}$
$=\text{z}^2+\frac{25}{12}\text{x}+1$
View full question & answer→Question 282 Marks
Find the following products:
$\left(y^2-4\right)\left(y^2-3\right)$
AnswerHere, we will use the identity $(x-a)(x-b)=x^2-(a+b) x+a b$.
$\left(y^2-4\right)\left(y^2-3\right)$
$=(y 2)^2-(4+3)\left(y^2\right)+4 \times 3$
$=y^4-7 y^2+12$
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, $a^m \times a^n=a^{m+n}$.
We have:
$(-5 a) \times\left(-10 a^2\right) \times\left(-2 a^3\right)$
$=[(-5) \times(-10) \times(-2)] \times\left(a \times a^2 \times a^3\right)$
$=[(-5) \times(-10) \times(-2)] \times\left(a^{1+2+3}\right)$
$=-100 a^6$
Thus, the answer is $-100 a^6$
View full question & answer→Question 332 Marks
Multiply:
$\left(x^2+y^2\right) \text { by }(3 a+2 b)$
AnswerTo multiply, we will use distributive law as foll
$\left(x^2+y^2\right) b y(3 a+2 b)$
$=x^2(3 a+2 b)+y^2(3 a+2 b)$
$=3 a x^2+2 b x^2+3 a y^2+2 b y^2$
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$\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$\text{LHS}=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}$
$=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2×\frac{4\text{m}}{3}×\frac{3\text{n}}{4}$
$=\big(\frac{4\text{m}}{3}\big)^2+\big(\frac{3\text{n}}{4}\big)^2 $ $\big[∵ (\text{a}−\text{b}\big)^2+2\text{ab}=\text{a}^2+\text{b}^2\big]$
$=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$=\text{RHS}$
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:
$(3 x+7)^2-84 x=(3 x-7)^2$
$\text { LHS }=(3 x+7)^2-84 x$
$=(3 x+7)^2-4 \times 3 x \times 7$
$=(3 x-7)^2\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=\text { RHS }$
Because LHS is equal to RHS, the given equation is verified.
View full question & answer→Question 372 Marks
Subtract:
$2 x^3-4 x^2+3 x+5 \text { from } 4 x^3+x^2+x+6$
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$(9 a-5 b)^2+180 a b=(9 a+5 b)^2$
$\text { LHS }=(9 a-5 b)^2+180 a b$
$=(9 a+5 b)^2\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=\text { RHS }$
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:
$(2 x+y)(2 x+y)$
AnswerWe will use the identify $(a+b)^2=a^2+2 a b+b^2$ 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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$102 \times 106$
$=(100+2)(100+6)$
$=100^2+(2+6) 100+2 \times 6$
$=10000+800+12$
$=10812$
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 $(x+a)(x+b)=x^2+(a+b) x+a b$.
$\left(y^2+12\right)\left(y^2+6\right)$
$=\left(y^2\right)^2+(12+6)\left(y^2\right)+12 \times 6$
$=y^4+18 y^2+72$
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$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$
$\text { LHS }=(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)$
$=a^2-b^2+b^2-c^2+c 2-a^2$
$=0$
$=\text { RHS }$
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 $(x+a)(x-b)=x^2+(a-b) x-a b$.
$\left(p^2+16\right)\left(p^2-\frac{1}{4}\right)$
$=\left(p^2\right) 2+\left(16-\frac{1}{4}\right)\left(p^2\right)-16 \times \frac{1}{4}$
$=p^4+\frac{63}{4} p^2-4$
View full question & answer→Question 482 Marks
Find the following products:
$(x-11)(x+4)$
AnswerHere, we will use the identity $(x-a)(x+b)=x^2+(b-a) x-a b$.
$(x-11)(x+4)$
$=x^2+(4-11) x-11 \times 4$
$=x^2-7 x-44$
View full question & answer→Question 492 Marks
Evaluate the following:
$53 \times 55$
AnswerHere, we will use the identity $(x+a)(x+b)=x^2+(a+b) x+a b$.
$53 \times 55$
$=(50+3)(50+5)$
$=50^2+(3+5) 50+3 \times 5$
$=2500+400+15$
$=2915$
View full question & answer→Question 502 Marks
Multiply:
$(a-1) by $\left(0.1 a^2+3\right)$
AnswerTo multiply, we will use distributive law as follows:
$(a-1) \text { by }\left(0.1 a^2+3\right)$
$=0.1 a^2(a-1)+3(a-1)$
$=0.1 a^3-0.1 a^2+3 a-3$
Thus, the answer is $0.1 a^3-0.1 a^2+3 a-3$
View full question & answer→Question 512 Marks
Find the following products:
$(x+4)(x+7)$
AnswerHere, we will use the identity $(x+a)(x+b)=x^2+(a+b) x+a b$.
$(x+4)(x+7)$
$=x^2+(4+7) x+4 \times 7$
$=x^2+11 x+28$
View full question & answer→Question 522 Marks
Simplify:
$a^2 b\left(a^3-a+1\right)-a b\left(a^4-2 a^2+2 a\right)-b\left(a^3-a^2-1\right)$
AnswerTo simplify, we will use distributive law as follows:
$a^2 b\left(a^3-a+1\right)-a b\left(a^4-2 a^2+2 a\right)-b\left(a^3-a^2-1\right)$
$=a^5 b-a^3 b+a^2 b-a^5 b+2 a^3 b-2 a^2 b-a^3 b+a^2 b+b$
$=a^5 b-a^5 b-a^3 b+2 a^3 b-a^3 b+a^2 b-2 a^2 b+a^2 b+b$
$=b$
View full question & answer→Question 532 Marks
Using the formula for squaring a binomial, evaluate the following:
$(1001)^2$
AnswerHere, we will use the identity $(a+b)^2=a^2+2 a b+b^2$.
$(1001)^2$
$=(1000+1)^2$
$=(1000)^2+2 \times 1000 \times 1+1^2$
$=1000000+2000+1$
$=1002001$
View full question & answer→Question 542 Marks
Simplify the following using the formula $(a+b)^2=a^2+2 a b+b^2$ :
$(82)^2-(18)^2$
AnswerHere, we will use the identity $(a-b)(a+b)=a^2-b^2$
Let us consider the following expression:
$(82)^2-(18)^2$
$=(82+18)(82-18)$
$=100 \times 64$
$=6400$
View full question & answer→Question 552 Marks
Evaluate the following:
$109 \times 107$
AnswerHere, we will use the identity $(x+a)(x+b)=x^2+(a+b) x+a b$.
$109 \times 107$
$=(100+9)(100+7)$
$=100^2+(9+7) 100+9 \times 7$
$=10000+1600+63$
$=11663$
View full question & answer→Question 562 Marks
Find the following products:
$(x+7)(x-5)$
AnswerHere, we will use the identity $(x+a)(x-b)=x^2+(a-b) x-a b \cdot(x+7)(x-5)$
$=x^2+(7-5) x-7 \times 5$
$=x^2+2 x-35$
View full question & answer→Question 572 Marks
Using the formula for squaring a binomial, evaluate the following: $(102)^2$
AnswerHere, we will use the identity $(a+b)^2=a^2+2 a b+b^2$.
$(102)^2$
$=(100+2)^2$
$=(100)^2+2 \times 100 \times 2+2^2$
$=10000+400+4$
$=10404$
View full question & answer→Question 582 Marks
Find the following products:
$\big(\text{y}^2+\frac{5}{7}\big)\big(\text{y}^2−\frac{14}{5}\big)$
AnswerHere, we will use the identity $(x+a)(x-b)=x^2+(a-b) x-a b$.
$\left(y^2+\frac{5}{7}\right)\left(y^2-\frac{14}{5}\right)$
$=\left(y^2\right)^2+\left(\frac{5}{7}-\frac{14}{5}\right)\left(y^2\right)-\frac{5}{7} \times \frac{14}{5}$
$=y^4-\frac{73}{35} y^2-2$
View full question & answer→Question 592 Marks
Using the formula for squaring a binomial, evaluate the following:
$(703)^2$
AnswerHere, we will use the identity $(a+b)^2=a^2+2 a b+b^2$.
$(703)^2$
$=(700+3)^2$
$=(700)^2+2 \times 700 \times 3+3^2$
$=490000+4200+9$
$=494209$
View full question & answer→Question 602 Marks
Evaluate the following:
$35 \times 37$
AnswerHere, we will use the identity $(x+a)(x+b)=x^2+(a+b) x+a b$.
$35 \times 37$
$=(30+5)(30+7)$
$=30^2+(5+7) 30+5 \times 7$
$=900+360+35$
$=1295$
View full question & answer→Question 612 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
= 0
View full question & answer→Question 622 Marks
Using the formula for squaring a binomial, evaluate the following:
$(99)^2$
AnswerHere, we will use the identity $(a+b)^2=a^2+2 a b+b^2$.
$(99)^2$
$=(100-1)^2$
$=(100)^2-2 \times 100 \times 1+1^2$
$=10000-200+1$
$=9801$
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