Algebraic Expressions

Terms related to algebraic expression Constant : A quantity which has a fixed value, i.e., whose value does not change is called a constant. E.g., $8,-7,0,6\frac{7}{8}$. etc. are all constants.

Variable: A symbol which can be assigned different numerical values is called a variable. In algebra, the variables are denoted by the letters in the English alphabet, viz.; $\mathrm{x},\mathrm{y},\mathrm{z},\dots .\mathrm{a},\mathrm{b},\mathrm{c}$ etc. E.g., The perimeter $P$ of a square of a side $S$ is given by the formula, $P=4\times S$, here 4 is a constant while $P$ and $S$ are variables.

1.0Algebraic expression

A combination of constants and variables connected by signs of fundamental operations $(+,-,\times$ and $÷$ ) is called an algebraic expression. E.g., $5-3x+4x^{2}y$ is an algebraic expression consisting of three terms, namely $5,-3x$ and $4x^{2}y$. Expression can also be obtained by multiplying variable by variable. $x\times x=x^2$ and $x\times x\times x=x^3$ Factors: Each term of an algebraic expression consists of a product of constants and variables. A constant factor is called a numerical factor, while a variable factor is known as literal factor. E.g., In the expression $3x^2-4xy+7y^2;3x^2,-4xy,7y^2$ are all terms of the above expression. Each term is further made of factors, i.e., $3x^2$ can be written as $3\times x\times x,-4xy$ as $-4\times x\times y$ and $7y^2$ as $7\times y\times y$. Here $3,x,x$ are factors of $3x^2$, $-4,x,y$ are factors of $-4xy$ and $7,y,y$ are factors of $7y^2$. Thus, an expression, its terms and factors of the terms can be represented by a tree diagram to make it easily comprehensible to you.

Factors are numbers that can't be further factorized $>1$ is not taken as separate factor.

Coefficients: In a term, coefficient is either a numerical factor, an algebraic factor or the product of two or more factors.

E.g., In the algebraic expression $10\mathrm{xy},10$ is the coefficient of $\mathrm{xy},10\mathrm{x}$ is the coefficient of y and 10 y is the coefficient of $x$. The numerical part is called the numerical coefficient and literal part or variable or variable part is called the literal coefficient. When the numerical coefficient is not given. It is always understood to be 1 . E.g.,

Always consider the "sign" while taking the coefficient. E.g., In $-7x^2$ coefficient of $y^2$ is $-7x$ and numerical coefficient is -7 not ' 7 '.

Like and unlike terms

Like terms:

• Like terms or similar terms are terms which have the same literal factor. They may differ in their numerical coefficient. E.g., (i) $9x^2,5x^2$ and $-2x^2$ are like terms. (ii) $-6x^{2}y$ and $4y{x}^2$ are like terms.

Unlike terms :

• Unlike terms are terms which have different literal factors. E.g., $7x^2$ and $9y^2$, the literal factors are $x^2$ in $7x^2$ and $y^2$ in $9y^2$, which are different.

2.0Various types of algebraic expressions

Monomial: An algebraic expression which contains only one term, is called a monomial. Thus, $5\mathrm{x},2\mathrm{xy},-3a^{2}b,-7$, etc, are all monomials.

Binomial : An algebraic expression containing two unlike terms is called a binomial. Thus, $(2a+3b),(8-3x),\left(x^2-4x{y}^2\right)$ etc, are all binomials.

Trinomial : An algebraic expression containing three terms is called a trinomial. Thus, $(a+2b+5c),(x+2y-3z),\left(x^3-y^3-z^3\right)$, etc, are all trinomials.

Quadrinomial : An algebraic expression containing four terms is called a quadrinomial. Thus, $(x+y+z-5),\left(x^3+y^3+z^3+3xyz\right)$, etc, are all quadrinomials.

Polynomial : An expression containing one or more terms is called a polynomial. An algebraic expression of the form $a+bx+c{x}^2+d{x}^3+\dots$. , where $a,b,c,d$ are constants and $x$ is a variable is called a polynomial in x . The degree of the polynomial is the greatest power of the variable present in the polynomial. In any polynomial, the power of variable is a non-negative integer. E.g., $9x^4-\frac{9}{2}{x}^2-\frac{7}{5}x-2$ is a polynomial in $x$ of degree 4 . If the polynomial is in two or more variables, then the sum of the powers of the variables in each term is taken and the greatest sum is the degree of the polynomial. Note : Every polynomial is an expression, but every expression need not be a polynomial.

3.0Operations on Algebraic Expressions

Addition of algebraic expressions

While adding algebraic expressions, we collect the like terms and add them. The sum of several like terms is another like term whose coefficient is the sum of the coefficients of those like terms.

4.0Subtraction of algebraic expressions

The difference of two like terms is a like term whose coefficient is the difference of the numerical coefficients of the two like terms.

Rule for subtraction of algebraic expressions: Change the sign of each term of the expression to be subtracted and then add.

Note: At the time of adding or subtracting the expression, arrange the expressions so that the like terms are grouped together or arrange the expression in rows so that like terms appear in columns and then add or subtract.

If the operation of subtraction is performed by writing in column form, then the sign of each term to be subtracted must be changed and then the terms are added or subtracted.

• While adding or subtracting algebraic expressions, like terms will be added or subtracted to like terms only.

5.0Multiplication of algebraic expressions

Before taking up the product of algebraic expressions, let us look at two simple rules. (i) The product of two factors with like signs is positive, and the product of two factors with unlike signs is negative. (ii) If $a$ is any variable and $m,n$ are positive integers then $a^m\times a^n=\left(a^{m+n}\right)$

Thus, $x^3.x^5=x^{3+5}=x^8$ and $x^{7+1}=x^8$, etc.

Multiplication of monomials - Rules:

(i) The coefficient of the product of two monomials is equal to the product of their coefficients.

(ii) The variable part in the product of two monomials is equal to the product of the variables in the given monomials. These rules may be extended for the product of three or more monomials.

Volume of cuboid with length ( $\ell$ ), breadth ($b$) and height ( $h$ ) is $lbh$ cubic units. If a trinomial is a perfect square, then it is the square of the binomial.

Multiplication of a monomial and a binomial

Let $\mathrm{p},\mathrm{q}$ and r be three monomials. Then, by distributive law of multiplication over addition, we have $\mathrm{p}\times (\mathrm{q}+\mathrm{r})=(\mathrm{p}\times \mathrm{q})+(\mathrm{p}\times \mathrm{r})$

• $\mathrm{x}\times \mathrm{x}={\mathrm{x}}^2$
• $x+x=2x$

Multiplication of two Binomials

Suppose $(a+b)$ and $(c+d)$ are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below: $(\mathrm{a}+\mathrm{b})\times (\mathrm{c}+\mathrm{d})=\mathrm{a}\times (\mathrm{c}+\mathrm{d})+\mathrm{b}\times (\mathrm{c}+\mathrm{d})=(\mathrm{a}\times \mathrm{c}+\mathrm{a}\times \mathrm{d})+(\mathrm{b}\times \mathrm{c}+\mathrm{b}\times \mathrm{d})=\mathrm{ac}+\mathrm{ad}+\mathrm{bc}+\mathrm{bd}$. This method is known as the horizontal method.

6.0Division

7.0Division of Monomials

To divide one algebraic term by another, the power or exponent rule for division is used. (i) The power of all the factors in the denominator is subtracted from the power of like factors in the numerator. (ii) The numerical coefficient of the numerator and denominator is divided by their HCF.

If either, or both terms are fractions, then the division sign is changed to multiplication and the second term (fraction) is inverted. Then division is then carried out as explained in (i) above or simplify by reducing to the lowest terms.

• Division by 0 is not defined. $\frac{x}{0}=$ not defined

8.0Value of an expression

The value of an expression can be found by substituting the given value of literal in the expression. Eg. value of $x^2+2x+1$ when $x=5$ is $5^2+2\times 5+1=25+10+1=36$.

9.0Formulas and rules using algebraic expressions

Perimeter formulas

(i) Perimeter of an equilateral triangle $=3\ell$, where $\ell$ is the length of the side of the equilateral triangle. (ii) Perimeter of a square $=4\ell$, where $\ell$ is the length of the side of the square. (iii) Perimeter of a regular pentagon $=5\ell$, where $\ell$ is the length of the side of the pentagon.

Area formulas

(i) Area of a square $={\ell }^2$, where $\ell$ is the side of the square. (ii) Area of a rectangle $=\ell \times \mathrm{b}$, where $\ell$ and b are respectively the length and the breadth of the rectangle. (iii) Area of triangle $=\frac{b\times h}{2}$, where $b$ is the base of the triangle and $h$ is the height of the triangle.

10.0Rules for number patterns

(i) If a natural number is denoted by n , its successor is $(\mathrm{n}+1)$. (ii) If a natural number is denoted by $n,2n$ is an even number and $(2n+1)$ is an odd number.

Make similar pattern with basic figures as shown:

(The number of line segments required to make the figure is given to the right. Also, the expression for the number of line segments required to make $n$ letters is also given).

11.0Numerical Ability

Q. Draw a tree diagram of the expression $4x^2+5y-8z+1$. Answer:

Q. Write down the numerical as well as the literal coefficient of each of the following expression. (a) $7x^2{y}^{2}z$ (b) $-28x^{2}yz{q}^2$ Solution:

Q. Identify the like terms in each of the following: (a) $9x^2,8xy,-2xy,\frac{5}{2}xy$ (b) $8m^2{n}^2,-m{p}^2,-m^2{n}^2,n{m}^2$ Solution: (a) Like terms: $8xy,-2xy,\frac{5}{2}xy$ (b) Like terms: $8{\mathrm{m}}^2{\mathrm{n}}^2,-{\mathrm{m}}^2{\mathrm{n}}^2$

Q. Find the coefficient of $x$ in the following: (a) $x$ (b) $-3x{y}^2$ (c) $4p^2\mathbf{x}$ Solution: (a) 1 (b) $-3y^2$ (c) $4p^2$

Q. Which of the following is a polynomial? (a) $7{\mathbf{a}}^2-3ab+5b^2+9$ (b) $\frac{9}{b^2}+a+ab$ Solution: (a) $7a^2-3ab+5b^2+9$ is a polynomial in a and $b$. The sum of the powers of variables of each term is $2,2,2$ respectively. (b) $\frac{9}{b^2}+a+ab$ is not a polynomial in $a$ and $b$. The sum of the powers of variables of each term is $-2,1,2$ respectively.

Q. State the degree of the polynomial: ${\mathbf{a}}^2{\mathbf{b}}^2-\mathbf{ab}+3{\mathbf{ab}}^2$. Solution: $a^2{b}^2-ab+3a{b}^2$ is a polynomial in $a$ and $b$. The sum of the powers of variables of each term is $4,2,3$ respectively. $\therefore$ The degree is highest power of variable of a term in the expression. Here, highest power is 4 . So, it is a polynomial of degree 4 .

Q. Classify the following polynomials as monomial, binomial or trinomial? $3\mathrm{x}+4\mathrm{y}+\mathrm{z},3{\mathrm{x}}^2-2{\mathrm{x}}^2,\mathrm{x}+3\mathrm{y}-2\mathrm{x},4{\mathrm{p}}^2-2{\mathrm{z}}^2-7{\mathrm{z}}^2-{\mathrm{p}}^2$ Solution: $3x+4y+z$ has 3 terms, so it is a trinomial. $3x^2-2x^2=x^2$ has 1 term, so it is a monomial. $x+3y-2x=3y-x$ has 2 terms, so it is a binomial. $4p^2-2z^2-7z^2-p^2=3p^2-9z^2$ has 2 terms, so it is a binomial.

Q. Add: $5{\mathrm{x}}^2-7\mathrm{x}+3,-8{\mathrm{x}}^2+2\mathrm{x}-5$ and $7{\mathrm{x}}^2-\mathrm{x}-2$ Solution: Required sum $=\left(5x^2-7x+3\right)+\left(-8x^2+2x-5\right)+\left(7x^2-x-2\right)$ $=5{\mathrm{x}}^2-8{\mathrm{x}}^2+7{\mathrm{x}}^2-7\mathrm{x}+2\mathrm{x}-\mathrm{x}+3-5-2$ [collecting like terms] $=(5-8+7)x^2+(-7+2-1)x+(3-5-2)$quad[ adding like terms] $=4x^2-6x-4$

Q. Add : $\left(3x^2-\frac{1}{5}x+\frac{7}{3}\right)+\left(-\frac{1}{4}{x}^2+\frac{1}{3}x-\frac{1}{6}\right)+\left(-2x^2-\frac{1}{2}x+5\right)$ Solution: Required sum $=\left(3x^2-\frac{1}{5}x+\frac{7}{3}\right)+\left(-\frac{1}{4}{x}^2+\frac{1}{3}x-\frac{1}{6}\right)+\left(-2x^2-\frac{1}{2}x+5\right)$ $=3x^2-\frac{1}{4}{\mathrm{x}}^2-2{\mathrm{x}}^2-\frac{1}{5}\mathrm{x}+\frac{1}{3}\mathrm{x}-\frac{1}{2}\mathrm{x}+\frac{7}{3}-\frac{1}{6}+5$ [collecting like terms] $=\left(3-\frac{1}{4}-2\right){\mathrm{x}}^2+\left(-\frac{1}{5}+\frac{1}{3}-\frac{1}{2}\right)\mathrm{x}+\left(\frac{7}{3}-\frac{1}{6}+5\right)$ [adding like terms] $=\left(\frac{12-1-8}{4}\right)x^2+\left(\frac{-6+10-15}{30}\right)x+\left(\frac{14-1+30}{6}\right)$ $=\frac{3}{4}{x}^2-\frac{11}{30}x+\frac{43}{6}$

Q. Subtract $\left(2x^2-5x+7\right)$ from $\left(3x^2+4x-6\right)$ Solution: We have, $\left(3x^2+4x-6\right)-\left(2x^2-5x+7\right)$ $=3x^2+4x-6-2x^2+5x-7$ $=(3-2)x^2+(4+5)x+(-6-7)$ $={\mathrm{x}}^2+9\mathrm{x}-13$

Q. Take away $\left(\frac{8}{5}{x}^2-\frac{2}{3}{x}^3+\frac{3}{2}x-1\right)$ from $\left(\frac{x^3}{5}-\frac{3}{2}{x}^2+\frac{2}{3}x+\frac{1}{4}\right)$ Solution: We have, $\left(\frac{x^3}{5}-\frac{3}{2}{x}^2+\frac{2}{3}x+\frac{1}{4}\right)-\left(\frac{8}{5}{\mathrm{x}}^2-\frac{2}{3}{\mathrm{x}}^3+\frac{3}{2}\mathrm{x}-1\right)$ $=\frac{x^3}{5}-\frac{3}{2}{x}^2+\frac{2}{3}x+\frac{1}{4}-\frac{8}{5}{x}^2+\frac{2}{3}{x}^3-\frac{3}{2}x+1$ $=\left(\frac{1}{5}+\frac{2}{3}\right)x^3+\left(-\frac{3}{2}-\frac{8}{5}\right)x^2+\left(\frac{2}{3}-\frac{3}{2}\right)x+\left(\frac{1}{4}+1\right)$ $=\frac{(3+10)}{15}{x}^3+\left(\frac{-15-16}{10}\right)x^2+\frac{(4-9)}{6}x+\frac{(1+4)}{4}$ $=\frac{13}{15}{x}^3-\frac{31}{10}{x}^2-\frac{5}{6}x+\frac{5}{4}$

Q. $-8a{b}^{2}c,3a^{2}b$ and $-\frac{1}{6}$ Solution: $\left(-8a{b}^{2}c\right)\times \left(3a^{2}b\right)\times \left(-\frac{1}{6}\right)$

$$\begin{gathered} =\left(-8\times 3\times \frac{-1}{6}\right)\times \left(a\times a^2\times b^2\times b\times c\right) \\ =4a^{(1+2)}\times b^{(2+1)}\times c=4a^3{b}^{3}c \end{gathered}$$

Q. Multiply $\frac{5}{8}{a}^3{b}^2,12a^{2}b$ and $6c$ Solution: $\left(\frac{5}{8}{a}^3{b}^2\right)\times \left(12a^{2}b\right)\times (6c)$

$$\begin{gathered} =\left(\frac{5}{8}\times 12\times 6\right)\times \left({\mathrm{a}}^3\times {\mathrm{a}}^2\times {\mathrm{b}}^2\times \mathrm{b}\times \mathrm{c}\right)=45\times {\mathrm{a}}^{(3+2)}\times {\mathrm{b}}^{(2+1)}\times \mathrm{c} \\ =45{\mathrm{a}}^5{\mathrm{b}}^3\mathrm{c}. \end{gathered}$$

Q. Multiply $4,\frac{5}{12}x$ and $-8x^{2}y$ Solution: $4\times \left(\frac{5}{12}x\right)\times \left(-8x^{2}y\right)$

$$\begin{gathered} =\left[4\times \frac{5}{12}\times (-8)\right]\times \left(\mathrm{x}\times {\mathrm{x}}^2\times \mathrm{y}\right) \\ =\frac{-40}{3}\times {\mathrm{x}}^{(1+2)}\times \mathrm{y}=\frac{-40}{3}{\mathrm{x}}^3\mathrm{y} \end{gathered}$$

Q. Multiply $-a,a^{2}bc$ and $\frac{-2}{5}{a}^2{c}^2$ Solution: $(-a)\times \left(a^{2}bc\right)\times \left(\frac{-2}{5}a{b}^2{c}^2\right)$

$$\begin{gathered} =\left(-1\times \frac{-2}{5}\right)\times a\times a^2\times a\times b\times b^2\times c\times c^2 \\ =\frac{2}{5}\times a^{(1+2+1)}\times b^{(1+2)}\times c^{(1+2)}=\frac{2}{5}{a}^4{b}^3{c}^3. \end{gathered}$$

Q. Multiply $\left(-5x^{2}y\right),\left(\frac{-2}{3}x{y}^{2}z\right),\left(\frac{8}{15}xy{z}^2\right)$ and $\left(\frac{-1}{4}z\right)$ Verify the result when $x=1,y=2$ and $z=3$. Solution: We have $\left(-5x^{2}y\right)\times \left(\frac{-2}{3}x{y}^{2}z\right)\times \left(\frac{8}{15}xy{z}^2\right)\times \left(\frac{-1}{4}z\right)$ $=\left(-5\times \frac{-2}{3}\times \frac{8}{15}\times \frac{-1}{4}\right)\times \left({\mathrm{x}}^2\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times {\mathrm{y}}^2\times \mathrm{y}\times \mathrm{z}\times {\mathrm{z}}^2\times \mathrm{z}\right)$ $=\frac{-4}{9}\times {\mathrm{x}}^{(2+1+1)}\times {\mathrm{y}}^{(1+2+1)}\times {\mathrm{z}}^{(1+2+1)}=\frac{-4}{9}{\mathrm{x}}^4{\mathrm{y}}^4{\mathrm{z}}^4$.

• Verify: $\left(-5x^{2}y\right)=-5\times (1{)}^2\times (2)=-10$ $\left(\frac{-2}{3}x{y}^{2}z\right)=\frac{-2}{3}\times (1)\times (2{)}^2\times 3=-8$ $\left(\frac{8}{15}{\mathrm{xyz}}^2\right)=\frac{8}{15}\times 1\times 2\times 3^2=\frac{8}{15}\times 2\times 9=\frac{48}{5}$ $\left(-\frac{1}{4}z\right)=-\frac{1}{4}\times 3=\frac{-3}{4}$ $\left(-5x^{2}y\right)\times \left(\frac{-2}{3}x{y}^{2}z\right)\times \left(\frac{8}{15}xy{z}^2\right)\times \left(-\frac{1}{4}z\right)$ $(-10)\times (-8)\times \left(\frac{48}{5}\right)\times \left(-\frac{3}{4}\right)=-576$ $\frac{-4}{9}{x}^4{y}^4{z}^4=\frac{-4}{9}(1{)}^4\times (2{)}^4\times (3{)}^4=\frac{-4}{9}\times 16\times 81=-576$ Hence $\left(-5x^{2}y\right)\times \left(\frac{-2}{3}x{y}^{2}z\right)\times \left(\frac{8}{15}xy{z}^2\right)\times \left(-\frac{1}{4}z\right)=\left(-\frac{4}{9}\right)x^4{y}^4{z}^4$

Q. Multiply : $\frac{9}{2}{x}^{2}y$ by $(x+2y)$. Solution: We have $\frac{9}{2}{x}^{2}y\times (x+2y)=\left(\frac{9}{2}{x}^{2}y\times x\right)+\left(\frac{9}{2}{x}^{2}y\times 2y\right)$ [by distributive law] $=\left(\frac{9}{2}\times x^2\times x\times y\right)+\left(\frac{9}{2}\times 2\times x^2\times y\times y\right)$ $=\left\{\frac{9}{2}{x}^{(2+1)}\times y\right\}+\left\{9\times x^2\times y^{(1+1)}\right\}$ $=\frac{9}{2}{x}^{3}y+9x^2{y}^2$.

Q. Multiply: $\left(3x-\frac{4}{5}{y}^{2}x\right)$ by $\frac{1}{2}xy$. Solution: We have $\left(3x-\frac{4}{5}{y}^{2}x\right)\times \frac{1}{2}xy=\left(3x\times \frac{1}{2}xy\right)-\left(\frac{4}{5}{y}^{2}x\times \frac{1}{2}xy\right)$ $=\left(3\times \frac{1}{2}\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\right)-\left(\frac{4}{5}\times \frac{1}{2}\times {\mathrm{y}}^2\times \mathrm{y}\times \mathrm{x}\times \mathrm{x}\right)$ $=\left\{\frac{3}{2}\times x^{(1+1)}\times y\right\}-\left\{\frac{2}{5}\times y^{(2+1)}\times x^{(1+1)}\right\}=\left(\frac{3}{2}{x}^{2}y\right)-\left(\frac{2}{5}{y}^3{x}^2\right)$. column method we have $3\mathrm{x}-\frac{4}{5}{\mathrm{y}}^2\mathrm{x}$ $\times \frac{1}{2}xy$ $\frac{3}{2}{x}^{2}y-\frac{2}{5}{y}^3{x}^2$

Q. Multiply $\left(\frac{1}{5}x-\frac{1}{4}y\right)$ and $\left(5x^2-4y^2\right)$. Solution: We have: Using horizontal method, we have $\left(\frac{1}{5}x-\frac{1}{4}y\right)\times \left(5x^2-4y^2\right)=\frac{1}{5}x\left(5x^2-4y^2\right)-\frac{1}{4}y\left(5x^2-4y^2\right)$ $=\left(\frac{1}{5}x\times 5x^2\right)-\left(\frac{1}{5}x\times 4y^2\right)-\left(\frac{1}{4}y\times 5x^2\right)+\left(\frac{1}{4}y\times 4y^2\right)$ $=x^3-\frac{4}{5}x{y}^2-\frac{5}{4}{x}^{2}y+y^3$

Q. Multiply $(5x^2-6x+9)$ by $(2x-3)$. Solution: By column method, we have

$\begin{array}{r}5x^2-6x+9\\ \times 2x-3\end{array}$$10x^3-12x^2+18x$

multiplication by 2 x

$-15x^2+18x-27$

multiplication by - 3 Add: $10x^3-27x^2+36x-27$ [multiplication by $(2x-3)$] $\therefore \left(5x^2-6x+9\right)\times (2x-3)=10x^3-27x^2+36x-27$.

Q. $4a÷a$ Solution: $4\mathrm{a}÷\mathrm{a}=\frac{4\mathrm{a}}{\mathrm{a}}=4{\mathrm{a}}^{1-1}=4{\mathrm{a}}^0=4$

Q. $4a^{2}b÷ab$ Solution: $4{\mathrm{a}}^2\mathrm{b}÷\mathrm{ab}=\frac{4{\mathrm{a}}^2\mathrm{b}}{\mathrm{ab}}=4{\mathrm{a}}^{2-1}{\mathrm{b}}^{1-1}=4\mathrm{a}$

Q. $9a^4{b}^2÷3a^{2}b$ Solution: $9a^4{b}^2÷3a^{2}b=\frac{9a^4{b}^2}{3a^{2}b}=3a^{4-2}{b}^{2-1}=3a^{2}b$

Q. If $p=-2$, find the value of: (i) $4p+8$ (ii) $-3p^2+5p+7$ (iii) $-2p^3-3p^2+4p+7$ Solution: (i) Value of $(4\mathrm{p}+8)$, when $\mathrm{p}=-2$ is $4(-2)+8=-8+8=0$ (ii) Value of $\left(-3p^2+5p+7\right)$ when $p=-2$ is $-3(-2{)}^2+5(-2)+7$ $=-3\times 4-10+7=-12-10+7=-15$ (iii) Value of $\left(-2p^3-3p^2+4p+7\right)$ when $\mathrm{p}=-2$ is $-2(-2{)}^3-3(-2{)}^2+4(-2)+7$ $=16-12-8+7=3$

Q. If $a=1,b=-1$, find the value of: (i) $a^2+b^2$ $(a-b{)}^2=a^2+b^2-2ab$ (ii) $a^2+ab+b^2$ (iii) ${\mathbf{a}}^2-b^2$ Solution: (i) Value of $a^2+b^2$, when $a=1,b=-1$ is $(1{)}^2+(-1{)}^2=1+1=2$ (ii) Value of $a^2+ab+b^2$, when $a=1,b=-1$ is $(1{)}^2+(1)(-1)+(-1{)}^2=1-1+1=1$ (iii) Value of ${\mathrm{a}}^2-{\mathrm{b}}^2$, when $\mathrm{a}=1,\mathrm{b}=-1$ is $(1{)}^2-(-1{)}^2=1-1=0$

Q. Find the value of ' $a$ ' that will make the expression $3x^2-8x+$ a equal to 5 , at $x=2$. Solution: Value of $3{\mathrm{x}}^2-8\mathrm{x}+\mathrm{a}$, when $\mathrm{x}=2$ is $3(2{)}^2-8(2)+\mathrm{a}=12-16+\mathrm{a}=\mathrm{a}-4$ Now, $\mathrm{a}-4=5$ $\Rightarrow \mathrm{a}=9$

Q. Find the value of $m$ if the expression $x^4-5x^3+mx-4$ equals -1 when $x=-1$. Solution: The value of $\left(x^4-5x^3+mx-4\right)$ is -1 , when $x=-1$ $\therefore (-1{)}^4-5(-1{)}^3+m(-1)-4=-1$ $\Rightarrow 1+5-\mathrm{m}-4=-1$ $\Rightarrow 2-\mathrm{m}=-1$ $\Rightarrow -\mathrm{m}=-1-2=-3$ $\Rightarrow \mathrm{m}=3$

Q. The side of a square is $5cm$. Find its perimeter and area. Solution: Side of square $(\ell )=5\mathrm{cm}$ Perimeter $=4(\ell )=4\times 5=20\mathrm{cm}$ Area $={\ell }^2=5\times 5=25{\mathrm{cm}}^2$

Q. If the perimeter of square is 36 cm . Find its area. Solution: Perimeter of square $=36\mathrm{cm}$ Side $(\ell )=\frac{36}{4}\mathrm{cm}=9\mathrm{cm}$ Area of square $={\ell }^2$ $=(9{)}^2=81{\mathrm{cm}}^2$

Q. If $W=XP+2r$, then $X$ is equal to Solution: $\mathrm{XP}=\mathrm{W}-2\mathrm{r}$ $X=\frac{W-2r}{P}$