Question 14 Marks
Multiply the monomial by the binomial and find the value of $x=-1, y=0.25$ and $z=0.05$ :
$15 y^2(2-3 x)$
AnswerTo find the product, we will use distributive law as follows:
$15 y^2(2-3 x)$
$=15 y^2 \times 2-15 y^2 \times 3 x$
$=30 y^2-45 x y^2$
Substituting $x=-1$ and $y=0.25$ in the result, we get:
$30 y^2-45 x y^2$
$=30(0.25)^2-45(-1)(0.25)^2$
$=30 \times 0.0625-[45 \times(-1) \times 0.0625]$
$=30 \times 0.0625-[45 \times(-1) \times 0.0625]$
$=1.875-(-2.8125)$
$=1.875+2.8125$
$=4.6875$
View full question & answer→Question 24 Marks
Find the value of $\left(5 x^6\right) \times\left(-1.5 x^2 y^3\right) \times\left(-12 x y^2\right)$ when $x=1, y=0.5$.
AnswerFirst multiply the expressions and then substitute the values for the variables.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^{ m } \times a ^{ n }$ $=a^{m+n}$.
We have:
$\left(5 x^6\right) \times\left(-1.5 x^2 y^3\right) \times\left(-12 x y^2\right) \\
=[5 \times(-1.5) \times(-12)] \times\left(x^6 \times x^2 \times x\right) \times\left(y^3 \times y^2\right) \\
=[5 \times(-1.5) \times(-12)] \times\left(x^{6+2+1}\right) \times\left(y^{3+2}\right) \\
=90 x^9 y^5 \\
\therefore\left(5 x^6\right) \times\left(-1.5 x^2 y^3\right) \times\left(-12 x y^2\right)=90 x^9 y^5$
Substituting $x=1$ and $y=0.5$ in the result, we get:
$=90 x^9 y^5 \\
=90(1)^9(0.5)^5 \\
=90 \times 1 \times 0.03125 \\
=2.8125$
Thus, the answer is 2.8125 .
View full question & answer→Question 34 Marks
If $\text{x}+\frac{1}{\text{x}}=20,$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
AnswerLet us consider the following equation:
$\text{x}+\frac{1}{\text{x}}=20,$
Squaring both sides, we get:
$\big(\text{x}+\frac{1}{\text{x}}\big)^2=(20)^2=400$ $\Big[\big(\text{a}+\text{b}\big)^2=\text{a}^2 +\text{b}^2 +2\text{ab}\Big]$
$\big(\text{x}+\frac{1}{\text{x}}\big)^2=400$
$⇒\text{x}^2+2×\text{x}×\frac{1}{\text{x}}+\big(\frac{1}{\text{x}}\big)^2=400 $
$⇒\text{x}^2+2+\frac{1}{\text{x}^2}=400$
$⇒\text{x}^2+\frac{1}{\text{x}^2}=398$ (Subtracting 2 from both sides)
Thus, the answer is 398.
View full question & answer→Question 44 Marks
Find product:
Express product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (-5x)
Question 54 Marks
Evaluate $\left(-8 x^2 y^5\right) \times(-20 x y)$ for $x=2.5$ and $y=1$
Answer
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a ^{ m } \times a ^{ n }$ $=a^{m+n}$.
We have:
$\left(-8 x^2 y^6\right) \times(-20 x y)$
$=[(-8) \times(-20)] \times\left(x^2 \times x\right) \times\left(y^6 \times y\right)$
$=[(-8) \times(-20)] \times\left(x^2+1\right) \times\left(y^6+1\right)$
$=160 x^3 y^7$
$\left(-8 x^2 y^6\right) \times(-20 x y)=160 x^3 y^7$
Substituting $x=2.5$ and $y=1$ in the result, we get:
$=160 x^3 y^7$
$=160(2.5)^3(1)^7$
$=160 \times 15.625$
$=2500$
Thus, the answer is 2500 .
View full question & answer→Question 64 Marks
Find the product $-3 y\left(x y+y^2\right)$ and find its value for $x=4$ and $y=5$.
AnswerTo find the product, we will use distributive law as follows:- $-3 y\left(x y+y^2\right)$
$=-3 y \times x y+(-3 y) \times y^2$
$=-3 x y^{1+1}-3 y^{1+2}$
$=-3 x y^2-3 y^3$
Substituting $x =4$ and $y =5$ in the result, we get:
$-3 x y^2-3 y^3$
$=-3(4)(5)^2-3(5)^3$
$=-3(4)(25)-3(125)$
$=-300-375$
$=-675$
Thus, the product is $\left(-3 x y^2-y^3-3 x y^2-3 y^3\right)$, and its value for $x=4$ and $y=5$ is $(-675)$.
View full question & answer→Question 74 Marks
Find the product:
$-\frac{8}{27}\text{xyz}\big(\frac{3}{2}\text{xyz}^2-\frac{9}{4}\text{xy}^2\text{z}^2\big)$
AnswerTo find the product, we will use distributive law as follows:$-\frac{8}{27}\text{xyz}\big(\frac{3}{2}\text{xyz}^2-\frac{9}{4}\text{xy}^2\text{z}^2\big)$
$=\Big\{\big(−\frac{8}{27}\text{xyz}\big)\big(\frac{3}{2}\text{xyz}^2\big)\Big\}−\Big\{\big(−\frac{8}{27}\text{xyz}\big)\big(\frac{9}{4}\text{xy}^2\text{z}^{3}\big)\Big\}$
$=\Big\{\big(−\frac{8}{27}\times\frac{3}{2}\big)\big(\text{x}×\text{x}\big)×\big(\text{y}×\text{y}\big)×\big(\text{z}×\text{z}^2\big)\Big\}\\−\Big\{\big(−\frac{8}{27}×\frac{9}{4}\big)\big(\text{x}×\text{x}\big)×\big(\text{y}×\text{y}^2)×\big(\text{z}×\text{z}^3\big) \Big\}$
$=\Big\{\big(−\frac{8}{27}\times\frac{3}{2}\big)\big(\text{x}^{1+1}\text{y}^{1+1}\text{z}^{1+2}\big)\Big\}\\−\Big\{\big(−\frac{8}{27}×\frac{9}{4}\big)\big(\text{x}^{1+1}\text{y}^{1+2}\text{z}^{1+3}\big)\Big\}$
$=−\frac{4}{9}\text{x}^2\text{y}^2\text{z}^3+\frac{2}{3}\text{x}^2\text{y}^3\text{z}^4$
Thus, the answer is $−\frac{4}{9}\text{x}^2\text{y}^2\text{z}^3+\frac{2}{3}\text{x}^2\text{y}^3\text{z}^4$
View full question & answer→Question 84 Marks
Take away:
$\frac{6}{5}\text{x}^{2}-\frac{4}{5}\text{x}^3+\frac{5}{6}+\frac{3}{2}\text{x}$ from $\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}$
AnswerThe difference is given by:
$\big(\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}\big)\\-\big(\frac{6}{5}\text{x}^{2}-\frac{4}{5}\text{x}^3+\frac{5}{6}+\frac{3}{2}\text{x}\big)$
$\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}-\frac{6}{5}\text{x}^{2}\\+\frac{4}{5}\text{x}^3-\frac{5}{6}-\frac{3}{2}\text{x}$
$=\frac{\text{x}^3}{3}+\frac{4}{5}\text{x}^3-\frac{5}{2}\text{x}^2-\frac{6}{5}\text{x}+\frac{3}{5}\text{x}\\-\frac{3}{2}\text{x}+\frac{1}{4}-\frac{5}{6}$
$=\big(\frac{5+12}{15}\big)\text{x}^3+\big(\frac{−25−12}{10}\big)\text{x}^2\\+\big(\frac{6−15}{10}\big)\text{x}+\big(\frac{6−20}{24}\big)$ (Collecting like terms)
$=\frac{17}{15}\text{x}^3-\frac{37}{10}\text{x}^2-\frac{9}{10}\text{x}-\frac{7}{12}$ (Combining like terms)
View full question & answer→Question 94 Marks
Find the following product and verify the result for $x=-1, y=-2$ :
$(3 x-5 y)(x+y)$
AnswerTo multiply, we will use distributive law as follows:
$(3 x-5 y)(x+y)$
$=3 x(x+y)-5 y(x+y)$
$=3 x^2+3 x y-5 x y-5 y^2=3 x^2$
$-2 x y-5 y^2$
$=\therefore$ $(3 x-5 y)(x+y)$
$=3 x^2-2 x y-5 y^2$
Now,
we put $x=-1$ and $y=-2$ on both sides to verify the result.
$LHS =(3 x-5 y)(x+y)$
$=\{3(-1)-5(-2)\}-1+(-2)\}$
$=(-3+10)(-3)$
$=(7)(-3)=-21$
$\text { RHS }=3 x^2-2 x y-5 y^2$
$=3(-1)^2-2(-1)(-2)-5(-2)^2$
$=3 \times 1-4-5 \times 4$
$=3-4-20$
$=-21$
Because LHS is equal to RHS, the result is verified.
View full question & answer→Question 104 Marks
Express product as a monomials and verify the result in each case for $x=1$ :
$\left(5 x^4\right)^{\circ}\left(x^2\right)^3 \times(2 x)^2$
AnswerWe have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{ m } \times$ $a^n=a^{m+n}$ and $\left(a^m\right)^n=a^{m n}$.
We have:
$\left(5 x^4\right) \times\left(x^2\right)^3 \times(2 x)^2$
$=\left(5 x^4\right) \times\left(x^6\right) \times\left(2^2 \times x^2\right)$
$=\left(5 \times 2^2\right) \times\left(x^4 \times x^6 \times x^2\right)$
$=\left(5 \times 2^2\right) \times\left(x^{4+6+2}\right)$
$=20 x^{12}$
$\therefore\left(5 x^4\right) \times\left(x^2\right)^3 \times(2 x)^2=20 x^{12}$
$\text { LHS }=\left(5 x^4\right) \times\left(x^2\right)^3 \times(2 x)^2$
$=\left(5 \times 1^4\right) \times\left(1^2\right)^3 \times(2 \times 1)^2$
$=(5 \times 1) \times\left(1^6\right) \times(2)^2$
$=5 \times 1 \times 4$
$=20$
Put $x=1$ in RHS, we get:
$\text { RHS }=20 x^{12}$
$=20 \times(1)^{12}$
$=20 \times 1$
$=20$
$\because$ LHS $=$ RHS for $x =1$, therefore, the result is correct.
Thus, the answer is $20 x^{12}$.
View full question & answer→Question 114 Marks
Subtract the sum of $31-4 m-7 n^2$ and $21+3 m-4 n^2$ from the sum of $91+2 m-3 n^2$ and $-31+m+4 n^2$.
AnswerWe have to subtract the sum of $\left(31-4 m-7 n^2\right)$ and $\left(2 l+3 m-4 n^2\right)$
from the sum of $\left(91+2 m-3 n^2\right)$ and $\left(-31+m+4 n^2\right)\left\{\left(91+2 m-3 n^2\right)+\left(-31+m+4 n^2\right)\right\}-\left\{\left(31-4 m-7 n^2\right)+(21+3 m-\right.$ $4 n^2$ )
$=\left(9 \mid-31+2 m+m-3 n^2+4 n^2\right)-\left(3 \mid+2 l-4 m+3 m-7 n^2-4 n^2\right)$
$=\left(6 l+3 m+n^2\right)-\left(5 I-m-11 n^2\right)$ (Combining like terms inside the parentheses)
$=6 l+3 m+n^2-5 l+m+11 n^2=6 l-51+3 m+m+n^2+11 n^2$ (Collecting like terms)
$=1+4 m+12 n ^2$ (Combining like terms)
Thus, the required solution is $I+4 m+12 n^2$.
View full question & answer→Question 124 Marks
Simplify:
$(2.5 p-1.5 q)^2-(1.5 p-2.5 q)^2$
AnswerTo simplify, we will proceed as follows:
$(2.5 p-1.5 q)^2-(1.5 p-2.5 q)^2$
$=(2.5 p)^2+(1.5 q)^2-2(2.5 p)(1.5 q)-\left[(1.5 p)^2+(2.5 q)^2-2(1.5 p)(2.5 q)\right]$
$=(2.5 p)^2-(1.5 p)^2+(1.5 q)^2-(2.5 q)^2$
$=[(2.5 p+1.5 p)(2.5 p-1.5 p)]+[(1.5 q+2.5)(1.5 q-2.5 q)]$
$=4 p \times p+4 q \times(-q)$
$=4 p^2-4 q^2$
$=4\left(p^2-4 q^2\right)$
View full question & answer→Question 134 Marks
Evaluate following when x = 2, y = −1.
$$$\big(\frac{3}{5}\text{x}^2\text{y}\big)×\big(−\frac{15}{4}\text{xy}^2\big)×\big(\frac{7}{9}\text{x}^2\text{y}^2)$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a ^{ m } \times a ^{ n }= a ^{ m + n }$.
We have:$\big(\frac{3}{5}\text{x}^2\text{y}\big)×\big(−\frac{15}{4}\text{xy}^2\big)×\big(\frac{7}{9}\text{x}^2\text{y}^2)$
$=\big[\frac{3}{5}×\big(−\frac{15}{4}\big)×\frac{7}{9}\big]×\big(\text{x}^2×\text{x}×\text{x}^2\big)\\×\big(\text{y}×\text{y}^2×\text{y}^2)$
$=\big[\frac{3}{5}×\big(−\frac{15}{4}\big)×\frac{7}{9}\big]×\big(\text{x}^{2+1+2}\big)\\×\big(\text{y}^{1+2+2}\big) $
$=−\frac{7}{4}\text{x}^5\text{y}^5$
$\therefore\big(\frac{3}{5}\text{x}^2\text{y}\big)×\big(−\frac{15}{4}\text{xy}^2\big)×\big(\frac{7}{9}\text{x}^2\text{y}^2)=−\frac{7}{4}\text{x}^5\text{y}^5$
Substituting x = 2 and y = -1 in the result, we get:$−\frac{7}{4}\text{x}^5\text{y}^5$
$=−\frac{7}{4}(2)^5(−1)^5$
$=\big(−\frac{7}{4}\big)×32×(−1)$
$=56$
Thus, the answer is 56.
View full question & answer→Question 144 Marks
Evaluate $\left(2.3 a ^5 b^2\right) \times\left(1.2 a ^2 b^2\right)$ when $a =1$ and $b =0.5$.
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m \times a^n$
$=a^{m+n} .\left(2.3 a^5 b^2\right) \times\left(1.2 a^2 b^2\right)$
$=(2.3 \times 1.2) \times\left(a^5 \times a^2\right) \times\left(b^2 \times b^2\right)$
$=(2.3 \times 1.2) \times\left(a^5+2\right) \times\left(b^2+2\right)$
$=2.76 a^7 b^4$
$\left(2.3 a^5 b^2\right) \times\left(1.2 a^2 b^2\right)=2.76 a^7 b^4$
Substituting $a=1$ and $b=0.5$ in the result, we get: $2.76 a^7 b^4$
$=2.76(1)^7(0.5)^4$
$=2.76 \times 1 \times 0.0625$
$=0.1725$
Thus, the answer is $=0.1725$
View full question & answer→Question 154 Marks
Express product as a monomials and verify the result in each case for x = 1:
$\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)$
AnswerWe have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a ^{ m \times}$ $a^n=a^{m+n}$.
$\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)$
$=\big[4×(−3)×\frac{4}{5}\big]×\big(\text{x}^2×\text{x}×\text{x}^3\big)$
$=\big[4×(−3)×\frac{4}{5}\big]×\big(\text{x}^{2+1+3}\big)$
$=−\frac{48}{5}\text{x}^6$
$\therefore\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)=−\frac{48}{5}\text{x}^6$
Substituting x = 1 in LHS, we get:
$\text{LHS}=\big(4\text{x}^2\big) × (-3\text{x}) × \big(\frac{4}{5}\text{x}^3\big)$
$=\big(4×1^2\big)×(-3×1)×\big(\frac{4}{5}×1^3)$
$=4×(−3)×\frac{4}{5}$
$=−\frac{48}{5}$
Putting x = 1 in RHS, we get:
$\text{RHS}=−\frac{48}{5}\text{x}^6$
$=−\frac{48}{5}×1^6$
$=−\frac{48}{5}$
$\because$ LHS = RHS for x = 1, therefore, the result is correct
Thus, the answer is $−\frac{48}{5}\text{x}^6.$
View full question & answer→Question 164 Marks
Simplify:
$(2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right)$
AnswerTo simplify, we will proceed as follows:
To simplify, we will proceed as follows:
$(2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right)$
$=\left((2 x)^2-1^2\right)\left(4 x^2+1\right)\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=\left(4 x^2-1\right)\left(4 x^2+1\right)\left(16 x^4+1\right)$
$=\left\{\left(4 x^2\right)^2-\left(1^2\right)^2\right\}\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=\left(16 x^4-1\right)\left(16 x^4+1\right)$
$=\left(16 x^4\right)^2-1^2\left[\because(a+b)(a-b)=a^2-b^2\right]$
$=256 x^8-1$
View full question & answer→Question 174 Marks
Simplify:
$(5 x+3)(x-1)(3 x-2)$
AnswerTo simplify, we will proceed as follows:
$(5 x+3)(x-1)(3 x-2)$
$=[(5 x+3)(x-1)](3 x-2)$
$=[5 x(x-1)+3(x-1)](3 x-2)$
$=\left[5 x^2-5 x+3 x-3\right](3 x-2)(\text { Distributive law) }$
$=\left[5 x^2-2 x-3\right](3 x-2)$
$=3 x\left(5 x^2-2 x-3\right)-2\left(5 x^2-2 x-3\right)$
$=15 x^3-6 x^2-9 x-\left[10 x^2-4 x-6\right]$
$=15 x^3-6 x^2-9 x-10 x^2+4 x+6$
$=15 x^3-6 x^2-10 x^2-9 x+4 x+6 \text { (Rearranging) }$
$=15 x^3-16 x^2-5 x+6 \text { (Combining like terms) }$
Thus, the answer is $15 x^3-16 x^2-5 x+6$.
View full question & answer→Question 184 Marks
Multiply the monomial by the binomial and find the value of $x=-1, y=0.25$ and $z=0.05$ :
$x z\left(x^2+y^2\right)$
AnswerTo find the product, we will use distributive law as follows:xz $\left(x^2+y^2\right)$
$=x z \times x^2+x z \times y^2$
$=x^3 z+x y^2 z$
Substituting $x=-1, y=0.25$ and $z=0.05$ in the result, we get:
$x^3 z+x y^2 z$
$=(-1)^3(0.05)+(-1)(0.25)^2(0.05)$
$=(-1)(0.05)+(-1)(0.0625)(0.05)$
$=-0.05-0.003125$
$=-0.053125$
View full question & answer→Question 194 Marks
Simplify:
$\left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right)$
AnswerTo simplify, we will proceed as follows:
$\left(2 x^2+3 x-5\right)\left(3 x^2-5 x+4\right)$
$=2 x^2\left(3 x^2-5 x+4\right)+3 x\left(3 x^2-5 x+4\right)-5\left(3 x^2-5 x+4\right) \text { (Distributive law) }$
$=6 x^4-10 x^3+8 x^2+9 x^3-15 x^2+12 x-15 x^2+25 x-20$
$=6 x^4-10 x^3+9 x^3+8 x^2-15 x^2-15 x^2+12 x+25 x-20 \text { (Rearranging) }$
$=6 x^4-x 3-22 x^2+36 x-20$
Thus, the answer is $6 x^4-x 3-22 x^2+36 x-20$.
View full question & answer→Question 204 Marks
Simplify:
$\left(x^2-3 x+2\right)(5 x-2)-\left(3 x^2+4 x-5\right)(2 x-1)$
AnswerTo simplify, we will proceed as follows: $\left(x^2-3 x+2\right)(5 x-2)-\left(3 x^2+4 x-5\right)(2 x-1)$
$=\left[\left(x^2-3 x+2\right)(5 x-2)\right]-\left[\left(3 x^2+4 x-5\right)(2 x-1)\right]$
$=\left[5 x\left(x^2-3 x+2\right)-2\left(x^2-3 x+2\right)\right]-\left[2 x\left(3 x^2+4 x-5\right)-1 \times\left(3 x^2+4 x-5\right)\right] \text { (Distributive law) }$
$=\left[5 x^3-15 x^2+10 x-\left(2 x^2-6 x+4\right)\right]-\left[6 x^3+8 x^2-10 x-3 x^2-4 x+5\right]$
$=5 x^3-15 x^2+10 x-2 x^2+6 x-4-6 x^3-8 x^2+10 x+3 x^2+4 x-5$
$=5 x^3-6 x^3-15 x^2-2 x^2-8 x^2+3 x^2+10 x+6 x+10 x+4 x-5-4 \text { (Rearranging) }$
$=-x 3-22 x^2+30 x-9 \text { (Combining like terms) }$
Thus, the answer is $-x 3-22 x^2+30 x-9$.
View full question & answer→Question 214 Marks
Express product as a monomials and verify the result in each case for $x =1$ :
$\left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5)$
AnswerWe have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^m \times$ $a^n=a^{m+n}$ and $\left(a^m\right)^n=a^{m n}$.
We have:
$\left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5)$
$=\left(x^6\right) \times(2 x) \times(-4 x) \times 5$
$=[2 \times(-4) \times 5] \times\left(x^6 \times x \times x\right)$
$=[2 \times(-4) \times 5] \times\left(x^{6+1+1}\right)$
$=-40 x^8$
$\therefore\left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5)=-40 x^8$
$\text { LHS }=\left(x^2\right)^3 \times(2 x) \times(-4 x) \times(5)$
$=\left(1^2\right)^3 \times(2 \times 1) \times(-4 \times 1) \times 5$
$=1^6 \times 2 \times(-4) \times 5$
$=1 \times 2 \times(-4) \times 5$
$=-40$
Putting $x=1$ in RHS, we get:
$\text { RHS }=-40 x^8$
$=-40(1)^8$
$=-40 \times 1$
$=-40$
$\because$ LHS $=$ RHS for $x=1$; therefore, the result is correct
Thus, the answer is $-40 x^8$.
View full question & answer→Question 224 Marks
Simplify:
$(5-x)(6-5 x)(2-x)$
AnswerTo simplify, we will proceed as follows:( 5 - x $)(6-5 x)(2-x)$
$=[(5-x)(6-5 x)](2-x)$
$=[5(6-5 x)-x(6-5 x)](2-x)(\text { Distributive law) }$
$=\left(30-25 x-6 x+5 x^2\right)(2-x)$
$=\left(30-31 x+5 x^2(2-x)\right.$
$=2\left(30-31 x+5 x^2\right)-x\left(30-31 x+5 x^2\right)$
$=60-62 x+10 x^2-30 x+31 x^2-5 x^3$
$=60-62 x+10 x^2-30 x+31 x^2-5 x^3$
$=60-62 x-30 x+10 x^2+31 x^2-5 x^3 \text { (Rearranging) }$
$=60-92 x+41 x^2-5 x^3 \text { Combining like terms) }$
Thus, the answer is $60-92 x+41 x^2-5 x^3$.
View full question & answer→Question 234 Marks
Simplify:
$(3 x-2)(2 x-3)+(5 x-3)(x+1)$
AnswerTo simplify, we will proceed as follows:
$(3 x-2)(2 x-3)+(5 x-3)(x+1)$
$=[(3 x-2)(2 x-3)]+[(5 x-3)(x+1)]$
$=[3 x(2 x-3)-2(2 x-3)]+[5 x(x+1)-3(x+1)] \text { (Distributive law) }$
$=6 x^2-9 x-4 x+6+5 x^2+5 x-3 x-3$
$=6 x^2+5 x^2-9 x-4 x+5 x-3 x-3+6$
$=11 x^2-11 x+3 \text { (Combining like terms) }$
Thus, the answer is $11 x^2-11 x+3$.
View full question & answer→Question 244 Marks
Simplify:
$(3 x+2 y)(4 x+3 y)-(2 x-y)(7 x-3 y)$
AnswerTo simplify, we will proceed as follows:
$(3 x+2 y)(4 x+3 y)-(2 x-y)(7 x-3 y)$
$=[(3 x+2 y)(4 x+3 y)]-[(2 x-y)(7 x-3 y)]$
$=[3 x(4 x+3 y)+2 y(4 x+3 y)]-[2 x(7 x-3 y)-y(7 x-3 y)] \text { (Distributive law) }$
$=12 x^2+9 x y+8 x y+6 y^2-\left[14 x^2-6 x y-7 x y+3 y^2\right]$
$=12 x^2+9 x y+8 x y+6 y^2-14 x^2+6 x y+7 x y-3 y^2$
$=12 x^2-14 x^2+9 x y+8 x y+6 x y+7 x y+6 y^2-3 y^2$
$=-2 x^2+30 x y+3 y^2 \text { (Rearranging) }$
Thus, the answer is $-2 x^2+30 x y+3 y^2$.
View full question & answer→Question 254 Marks
Evaluate $\left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right)$ when $x=1$ and $y=0.5$
AnswerTo multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a ^{ m } \times a ^{ n }$
$=a^{m+n}$.
$\left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right)$
$=(3.2 \times 2.1) \times\left(x^6 \times x^2\right) \times\left(y^3 \times y^2\right)$
$=(3.2 \times 2.1) \times\left(x^{6+2}\right) \times\left(y^{3+2}\right)$
$=6.72 x^8 y^5$
Substituting $x=1$ and $y =0.5$ in the result,
we get:
$6.72 x^8 y^5$
$=6.72(1)^8(0.5)^5$
$=6.72 \times 1 \times 0.03125$
$=0.21$
Thus, the answer is 0.21 .
View full question & answer→Question 264 Marks
Find the product $24 x ^2(1-2 x )$ and evaluate its value for $x =3$.
Answer
To find the product, we will use distributive law as follows: $24 x^2(1-2 x)$
$=24 x^2 \times 1-24 x^2 \times 2 x$
$=24 x^2-48 x^{1+2}$
$=24 x^2-48 x^3$
Substituting $x=3$ in the result, we get:
$24 x^2-48 x^3$
$=24(3)^2-48(3)^3$
$=24 \times 9-48 \times 27$
$=216-1296$
$=-1080$
Thus, the product is $\left(24 x^2-48 x^3\right)$ and its value for $x=3$ is -1080 .
View full question & answer→Question 274 Marks
Multiply the monomial by the binomial and find the value of $x=-1, y=0.25$ and $z=0.05$ :
$-3 x\left(y^2+z^2\right)$
AnswerTo find the product, we will use distributive law as follows:
$-3 x\left(y^2+z^2\right)$
$=-3 x \times y^2+(-3 x) \times z^2$
$=-3 x y^2-3 x z^2$
Substituting $x=-1, y=0.25$ and $z=0.05$ in the result.
we get:
$-3 x y^2-3 x z^2$
$=-3(-1)(0.25)^2-3(-1)(0.05)^2$
$=-3(-1)(0.0625)-3(-1)(0.0025)$
$=01875+0.0075$
$=0.195$
View full question & answer→Question 284 Marks
Simplify the following using the identities:
$\frac{8.63×8.63−1.37\times1.37}{0.726}$
AnswerLet us consider the following expression:$\frac{8.63×8.63−1.37\times1.37}{0.726}=\frac{8.63^2−1.37^2}{0.726}$
Using the identity $(a+b)(a-b)=a^2-b^2$ $\frac{8.63×8.63−1.37\times1.37}{0.726}=\frac{8.63^2−1.37^2}{0.726}\\=\frac{(8.63+1.37)(8.63−1.37)}{0.726}$
$\frac{8.63×8.63−1.37\times1.37}{0.726}=\frac{(8.63+1.37)(8.63−1.37)}{0.726}$
$\frac{8.63×8.63−1.37\times1.37}{0.726}=\frac{10×7.26}{0.726}$
$\frac{8.63×8.63−1.37×1.37}{0.726}=100$
Thus, the answer is 100.
View full question & answer→Question 294 Marks
Simplify:
$(5 x-3)(x+2)-(2 x+5)(4 x-3)$
AnswerTo simplify, we will proceed as follows: $(5 x-3)(x+2)-(2 x+5)(4 x-3)$
$=[(5 x-3)(x+2)]-[(2 x+5)(4 x-3)]$
$=[5 x(x+2)-3(x+2)]-[2 x(4 x-3)+5(4 x-3)] \text { (Distributive law) }$
$=5 x^2+10 x-3 x-6-8 x^2+6 x-20 x+15$
$=5 x^2-8 x^2+10 x-3 x+6 x-20 x-6+15 \text { (Rearranging) }$
$=-3 x^2-7 x+9$
View full question & answer→Question 304 Marks
Simplify:
$x^2(x-y) y^2(x+2 y)$
AnswerTo simplify, we will proceed as follows: $x^2(x-y) y^2(x+2 y)$
$=\left[x^2(x-y)\right]\left[y^2(x+2 y)\right]$
$=\left(x^3-x^2 y\right)\left(x y^2+2 y^3\right)$
$=x^3\left(x y^2+2 y^3\right)-x^2 y\left(x y^2+2 y^3\right)$
$=x^4 y^2+2 x^3 y^3-\left[x^3 y^3+2 x^2 y^4\right]$
$=x^4 y^2+2 x^3 y^3-x^3 y^3-2 x^2 y^4$
$=x^4 y^2+x^3 y^3-2 x^2 y^4$
Thus, the answer is $x^4 y^2+x^3 y^3-2 x^2 y^4$.
View full question & answer→