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Chapter 10

Algebraic Expressions and Identities — Exercise 10.1

Class - 8 ML Aggarwal Understanding ICSE Mathematics



Exercise 10.1

Question 1(i)

Identify monomials, binomials and trinomials from the following algebraic expression:

5p×q×r25p × q × r^2

Answer

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

5p×q×r25p × q × r^2 is a product of factors, so it has only one term.

Hence, 5p×q×r2\bm{5p × q × r^2} is a monomial.

Question 1(ii)

Identify monomials, binomials and trinomials from the following algebraic expression:

3x2×y÷2z3x^2 \times y \div 2z

Answer

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

3x2×y÷2z3x^2 \times y \div 2z involves only multiplication and division, so it has only one term.

Hence, 3x2×y÷2z\bm{3x^2 \times y \div 2z} is a monomial.

Question 1(iii)

Identify monomials, binomials and trinomials from the following algebraic expression:

−3+7x2-3 + 7x^2

Answer

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

−3+7x2-3 + 7x^2 has two terms, namely -3 and 7x27x^2.

Hence, −3+7x2\bm{-3 + 7x^2} is a binomial.

Question 1(iv)

Identify monomials, binomials and trinomials from the following algebraic expression:

5a2−3b2+c2\dfrac{5a^2 - 3b^2 + c}{2}

Answer

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

5a2−3b2+c2=5a22−3b22+c2\dfrac{5a^2 - 3b^2 + c}{2} = \dfrac{5a^2}{2} - \dfrac{3b^2}{2} + \dfrac{c}{2}, which has three terms.

Hence, 5a2−3b2+c2\bm{\dfrac{5a^2 - 3b^2 + c}{2}} is a trinomial.

Question 1(v)

Identify monomials, binomials and trinomials from the following algebraic expression:

7x5−3xy7x^5 - \dfrac{3x}{y}

Answer

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

7x5−3xy7x^5 - \dfrac{3x}{y} has two terms, namely 7x57x^5 and −3xy-\dfrac{3x}{y}.

Hence, 7x5−3xy\bm{7x^5 - \dfrac{3x}{y}} is a binomial.

Question 1(vi)

Identify monomials, binomials and trinomials from the following algebraic expression:

5p÷3q−3p2×q25p \div 3q - 3p^2 \times q^2

Answer

Multiplication and division do not separate the terms of an algebraic expression. Only the parts separated by '+' or '-' sign are counted as terms.

5p÷3q−3p2×q25p \div 3q - 3p^2 \times q^2 has two terms, namely 5p÷3q5p \div 3q and 3p2×q23p^2 \times q^2.

Hence, 5p÷3q−3p2×q2\bm{5p \div 3q - 3p^2 \times q^2} is a binomial.

Question 2

Identify which of the following expressions are polynomials. If so write their degrees:

(i) 25x4−3x2+5x−1\dfrac{2}{5}x^4 - \sqrt{3}x^2 + 5x - 1

(ii) 7x3−3x2+57x^3 - \dfrac{3}{x^2} + \sqrt{5}

(iii) 4a3b2−3ab4+5ab+234a^3b^2 - 3ab^4 + 5ab + \dfrac{2}{3}

(iv) 2x2y−3xy+5y3+32x^2y - \dfrac{3}{xy} + 5y^3 + \sqrt{3}

Answer

An algebraic expression is a polynomial if the powers of the variables in each term are non-negative integers. The degree is the greatest sum of the powers of the variables in a term.

(i) 25x4−3x2+5x−1\dfrac{2}{5}x^4 - \sqrt{3}x^2 + 5x - 1

The powers of xx in the terms are 4, 2, 1 and 0, which are all non-negative integers.

So, it is a polynomial and the greatest power is 4.

Hence, it is a polynomial of degree 4.

(ii) 7x3−3x2+57x^3 - \dfrac{3}{x^2} + \sqrt{5}

Here 3x2=3x−2\dfrac{3}{x^2} = 3x^{-2}, in which the power of xx is -2, a negative integer.

Hence, it is not a polynomial.

(iii) 4a3b2−3ab4+5ab+234a^3b^2 - 3ab^4 + 5ab + \dfrac{2}{3}

The powers of aa and bb in each term are non-negative integers, so it is a polynomial.

The sums of the powers of the terms are 3 + 2 = 5, 1 + 4 = 5, 1 + 1 = 2 and 0.

The greatest sum is 5.

Hence, it is a polynomial of degree 5.

(iv) 2x2y−3xy+5y3+32x^2y - \dfrac{3}{xy} + 5y^3 + \sqrt{3}

Here 3xy=3x−1y−1\dfrac{3}{xy} = 3x^{-1}y^{-1}, in which the powers of xx and yy are negative integers.

Hence, it is not a polynomial.

Question 3(i)

Add the following expressions:

ab−bc,bc−ca,ca−abab - bc, bc - ca, ca - ab

Answer

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

ab−bc+bc−ca+−ab+ca0\begin{array}{rrrr} & ab & - bc & \\ & & + bc & - ca \\ + & - ab & & + ca \\ \hline & 0 & & \\ \end{array}

Hence, the required sum = 0.

Question 3(ii)

Add the following expressions:

5p2q2+4pq+7,3+9pq−2p2q25p^2q^2 + 4pq + 7, 3 + 9pq - 2p^2q^2

Answer

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

5p2q2+4pq+7+−2p2q2+9pq+33p2q2+13pq+10\begin{array}{rrrr} & 5p^2q^2 & + 4pq & + 7 \\ + & - 2p^2q^2 & + 9pq & + 3 \\ \hline & 3p^2q^2 & + 13pq & + 10 \\ \end{array}

Hence, the required sum = 3p2q2+13pq+10\bm{3p^2q^2 + 13pq + 10}.

Question 3(iii)

Add the following expressions:

l2+m2+n2,lm+mn,mn+nl,nl+lml^2 + m^2 + n^2, lm + mn, mn + nl, nl + lm

Answer

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

l2+m2+n2lm+mn+mn+nl++lm+nll2+m2+n2+2lm+2mn+2nl\begin{array}{rrrrrrr} & l^2 & + m^2 & + n^2 & & & \\ & & & & lm & + mn & \\ & & & & & + mn & + nl \\ + & & & & + lm & & + nl \\ \hline & l^2 & + m^2 & + n^2 & + 2lm & + 2mn & + 2nl \\ \end{array}

The sum can also be written as l2+m2+n2+2(lm+mn+nl)l^2 + m^2 + n^2 + 2(lm + mn + nl).

Hence, the required sum = l2+m2+n2+2(lm+mn+nl)\bm{l^2 + m^2 + n^2 + 2(lm + mn + nl)}.

Question 3(iv)

Add the following expressions:

4x3−7x2+9,3x2−5x+4,7x3−11x+1,6x2−13x4x^3 - 7x^2 + 9, 3x^2 - 5x + 4, 7x^3 - 11x + 1, 6x^2 - 13x

Answer

Write each expression in a separate row, keeping the like terms one below the other, and then add them.

4x3−7x2+9+3x2−5x+47x3−11x+1++6x2−13x11x3+2x2−29x+14\begin{array}{rrrrr} & 4x^3 & - 7x^2 & & + 9 \\ & & + 3x^2 & - 5x & + 4 \\ & 7x^3 & & - 11x & + 1 \\ + & & + 6x^2 & - 13x & \\ \hline & 11x^3 & + 2x^2 & - 29x & + 14 \\ \end{array}

Hence, the required sum = 11x3+2x2−29x+14\bm{11x^3 + 2x^2 - 29x + 14}.

Question 4(i)

Subtract:

8a+3ab−2b+78a + 3ab - 2b + 7 from 14a−5ab+7b−514a - 5ab + 7b - 5

Answer

Subtracting,

14a−5ab+7b−58a+3ab−2b+7−−+−6a−8ab+9b−12\begin{array}{rrrrr} & 14a & - 5ab & + 7b & - 5 \\ & 8a & + 3ab & - 2b & + 7 \\ &- &- &+ &- \\ \hline & 6a & - 8ab & + 9b & - 12 \\ \end{array}

Hence, the required difference = 6a−8ab+9b−12\bm{6a - 8ab + 9b - 12}.

Question 4(ii)

Subtract:

8xy+4yz+5zx8xy + 4yz + 5zx from 12xy−3yz−4zx+5xyz12xy - 3yz - 4zx + 5xyz

Answer

Subtracting,

12xy−3yz−4zx+5xyz8xy+4yz+5zx−−−4xy−7yz−9zx+5xyz\begin{array}{rrrrr} & 12xy & - 3yz & - 4zx & + 5xyz \\ & 8xy & + 4yz & + 5zx & \\ &- &- &- & \\ \hline & 4xy & - 7yz & - 9zx & + 5xyz \\ \end{array}

Hence, the required difference = 4xy−7yz−9zx+5xyz\bm{4xy - 7yz - 9zx + 5xyz}.

Question 4(iii)

Subtract:

4p2q−3pq+5pq2−8p+7q−104p^2q - 3pq + 5pq^2 - 8p + 7q - 10 from 18−3p−11q+5pq−2pq2+5p2q18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q

Answer

Subtracting,

5p2q−2pq2+5pq−3p−11q+184p2q+5pq2−3pq−8p+7q−10−−++−+p2q−7pq2+8pq+5p−18q+28\begin{array}{rrrrrrr} & 5p^2q & - 2pq^2 & + 5pq & - 3p & - 11q & + 18 \\ & 4p^2q & + 5pq^2 & - 3pq & - 8p & + 7q & - 10 \\ &- &- &+ &+ &- &+ \\ \hline & p^2q & - 7pq^2 & + 8pq & + 5p & - 18q & + 28 \\ \end{array}

Hence, the required difference = p2q+8pq−7pq2+5p−18q+28\bm{p^2q + 8pq - 7pq^2 + 5p - 18q + 28}.

Question 5

Subtract the sum of 3x2+5xy+7y2+33x^2 + 5xy + 7y^2 + 3 and 2x2−4xy−3y2+72x^2 - 4xy - 3y^2 + 7 from 9x2−8xy+11y29x^2 - 8xy + 11y^2

Answer

First we find the sum of the two given expressions by the column method.

3x2+5xy+7y2+3+2x2−4xy−3y2+75x2+xy+4y2+10\begin{array}{rrrrr} & 3x^2 & + 5xy & + 7y^2 & + 3 \\ + & 2x^2 & - 4xy & - 3y^2 & + 7 \\ \hline & 5x^2 & + xy & + 4y^2 & + 10 \\ \end{array}

Now we subtract this sum from 9x2−8xy+11y29x^2 - 8xy + 11y^2.

9x2−8xy+11y25x2+xy+4y2+10−−−−4x2−9xy+7y2−10\begin{array}{rrrrr} & 9x^2 & - 8xy & + 11y^2 & \\ & 5x^2 & + xy & + 4y^2 & + 10 \\ &- &- &- &- \\ \hline & 4x^2 & - 9xy & + 7y^2 & - 10 \\ \end{array}

Hence, the required difference = 4x2−9xy+7y2−10\bm{4x^2 - 9xy + 7y^2 - 10}.

Question 6

What must be subtracted from 3a2−5ab−2b2−33a^2 - 5ab - 2b^2 - 3 to get 5a2−7ab−3b2+3a5a^2 - 7ab - 3b^2 + 3a?

Answer

Let the required expression be A.

According to the question,

⇒ (3a2−5ab−2b2−3)−A=5a2−7ab−3b2+3a(3a^2 - 5ab - 2b^2 - 3) - A = 5a^2 - 7ab - 3b^2 + 3a

⇒ A=(3a2−5ab−2b2−3)−(5a2−7ab−3b2+3a)A = (3a^2 - 5ab - 2b^2 - 3) - (5a^2 - 7ab - 3b^2 + 3a)

Subtracting,

3a2−5ab−2b2−35a2−7ab−3b2+3a−++−−2a2+2ab+b2−3a−3\begin{array}{rrrrrr} & 3a^2 & - 5ab & - 2b^2 & & - 3 \\ & 5a^2 & - 7ab & - 3b^2 & + 3a & \\ &- &+ &+ &- & \\ \hline & - 2a^2 & + 2ab & + b^2 & - 3a & - 3 \\ \end{array}

Hence, −2a2+2ab+b2−3a−3\bm{-2a^2 + 2ab + b^2 - 3a - 3} must be subtracted.

Question 7

The perimeter of a triangle is 7p2−5p+117p^2 - 5p + 11 and two of its sides are p2+2p−1p^2 + 2p - 1 and 3p2−6p+33p^2 - 6p + 3. Find the third side of the triangle.

Answer

As the perimeter of a triangle is the sum of its three sides,

Third side = Perimeter - Sum of the other two sides

First we find the sum of the two given sides by the column method.

p2+2p−1+3p2−6p+34p2−4p+2\begin{array}{rrrr} & p^2 & + 2p & - 1 \\ + & 3p^2 & - 6p & + 3 \\ \hline & 4p^2 & - 4p & + 2 \\ \end{array}

Now we subtract this sum from the perimeter 7p2−5p+117p^2 - 5p + 11.

7p2−5p+114p2−4p+2−+−3p2−p+9\begin{array}{rrrr} & 7p^2 & - 5p & + 11 \\ & 4p^2 & - 4p & + 2 \\ &- &+ &- \\ \hline & 3p^2 & - p & + 9 \\ \end{array}

Hence, the third side of the triangle = 3p2−p+9\bm{3p^2 - p + 9}.

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