WBBSE Solutions For Class 7 Maths Algebra Chapter 4 Polynomials Exercise 4 Solved Example Problems

Algebra Chapter 4 Polynomials Exercise 4 Solved Example Problems

Polynomials Introduction

In this chapter, our aim is to study an algebraic expression, to learn what is meant by the coefficient and variable in an algebraic expression, to discuss the term and finally to apply four basic operations namely addition, subtraction, multiplication and division on a polynomial.

Here one thing should be noted with care addition and subtraction in algebra is permissible only between like terms.

On the other hand, the permissibility of multiplication and division is not restricted between like terms but can also be performed between, unlike terms.

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Expression

By 20 metres, 500 gms and 4 hours the magnitudes of length, mass and time are expressed respectively.

Here it should be noted that the first part in each case is a number while the second part is unit.

Such things, which express some quantity is called an expression. In the expression 20 metres, the numerical value of length is 20 and its unit is metre.

WBBSE Solutions For Class 7 Maths Algebra Chapter 4 Polynomials Exercise 4 Solved Example Problems

Co-efficient and variable

If x denotes a fixed length then 8x also denotes another fixed length (which is 8 times the length denoted by x); so 8x is an expression.

Similarly, if y denotes a fixed mass then 10y also denotes another mass; so 10y is an expression.

Of the algebraic expressions 8x, 10y the first parts are numbers while the second parts are unknown. Thus, if “any number” is denoted by the alphabetic symbols x, y, z or a, b, c then their values are not fixed.

Therefore, as the values of the alphabetic symbols may vary, they are called variables. Again, if a number appears as a multiple of a variable then the number is called co-efficient. For example, in 5a the co-efficient of the variable a is 5.

It is not necessary that a co-efficient will always be a pure number. For example, in. the expression 9 abx if the variable be x then its co-efficient will be 9 ab. In this case a and b will denote fixed numbers, not variables. So, 9 ab will represent a number.

Again, if in the expression 9abx, ab is a variable then the co-efficient of ab will be 9x. In that case, the expression should be written as 9xab.

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WBBSE Class 7 Geography Multiple Choice Questions WBBSE Solutions For Class 7 Maths

 

If an expression is indicated by a single alphabetic symbol then its co-efficient is 1. For example, in the expression x, the coefficient of x is 1. In the expression, a the co-efficient of a is 1. So, it is clear that, any expression will have a coefficient.

Term

When two or more expressions are related with each other by sign of addition or subtraction then a polynomial is formed and each expression of that polynomial is called a term.

If there is one term in an expression, then it is called a monomial. For example, 5x is a monomial.

If there are two or three terms then they are called binomial and trinomial respectively. For example, 5a + 6b is a binomial and la – 126 + 8c is a trinomial.

If there are many terms in an expression then it is called a polynomial.

For example, a + 5b 8c + 9d + 7e + 8f – 11x is a polynomial.

Like and unlike terms

The terms in which the numerical coefficients are different but the algebraic symbols are the same are called like terms. The terms which are not like are called unlike terms.

For example, 2a and 5a are like terms, 2xy and 12xy are like terms but xy and 9x are unlike terms, 55 and 8c are unlike terms.

The operations of addition and subtraction may be performed between the like terms only, but not between the unlike terms.

We know that, the sum of 5 mangoes and 6 mangoes is (5 + 6) mangoes or 11 mangoes.

Likewise, the sum of 5x and 6x is (5x + 6x) or, 11x.

Again, the sum of 10 mangoes and 18 apples is, 10 mangoes +18 apples.

Likewise, the sum of 10x and 18y = 10x + 18y; the summation between 10x and I8y is not permissible as they are unlike terms.

WBBSE Class 7 Polynomial Examples

Algebraic sum

We have mentioned earlier that summation is permissible between like terms only. The summation of like terms having same signs is similar to the process of addition in arithmetic.

For example, the sum of -5x, 7x and 8x = 5x + 7x + 8x = 20x.

Again, the sum of – 3x, -4x and -8x = – 3x – 4x – 8x = – 15x.

When the terms to be added are like but of opposite signs, their summation is not similar to the process of arithmetic.

For example, the sum of – 5x, 7x and – 4x = – 5x + 7x – 4x – 2x – 4x = – 2x.

Here, subtracting 5x from 7x we get 2x.

After subtracting 4x from 2x we get – 2x. But in arithmetic it is not possible to subtract 4x from 2x as in arithmetic we say that, the greater number cannot be subtracted from a smaller number.

Algebra Chapter 4 Polynomials Exercise 4 Some Examples Of Summation

 Example 1. Find the sum of : \(4 x^2,-3 x^2, 7 x^2\)

Solution :

Given

\(4 x^2,-3 x^2, 7 x^2\)

The required sum = \(4 x^2-3 x^2+7 x^2=\left(4 x^2+7 x^2\right)-3 x^2=11 x^2-3 x^2-8 x^2\)

∴ \(8 x^2\)

\(4 x^2,-3 x^2, 7 x^2=8 x^2\)

Example 2. Find the sum of  2xy, 5xy, 9xy and – 7xy.

Solution :

Given:

2xy, 5xy, 9xy And – 7xy

Here the required sum = – 2xy + 5xy + 9xy – 7xy

= (5xy + 9xy) – (2xy+ 7xy)

= 14 xy -9xy = 5xy

2xy, 5xy, 9xy and – 7xy = 5xy

Solved Problems for Class 7 Polynomials

Example 3. Find the sum of \(\frac{2}{5}\) ab, – \(\frac{3}{5}\) ab, \(\frac{4}{5}\) ab, and – \(\frac{1}{5}\),

Solution:

Here the required sum

= \(\frac{2}{5}\) ab, – \(\frac{3}{5}\) ab, \(\frac{4}{5}\) ab, and – \(\frac{1}{5}\),

= (\(\frac{2}{5}\) ab + \(\frac{4}{5}\) ab) – (\(\frac{3}{5}\) ab + \(\frac{1}{5}\)) ab

= \(\frac{6}{5}\) ab- \(\frac{4}{5}\) ab = \(\frac{2}{5}\) ab

= \(\frac{2}{5}\) ab

Example 4. Find the sum of the following expressions: 5x – 8y + z, – 2x + 7y — 5z, 3x + 5y + 3z.

Solution:

Given:

5x – 8y + z, – 2x + 7y — 5z, 3x + 5y + 3z

\(\begin{array}{r}
5 x-8 y+z \\
-2 x+7 y-5 z \\
3 x+5 y+3 z \\
\hline 6 x+4 y-z \\
\hline
\end{array}\)

∴ The required sum = 6x + 4y – z.

5x – 8y + z, – 2x + 7y – 5z, 3x + 5y + 3z. = 6x + 4y – z.

Example 5. Find the sum of the following expressions: \(p^2-p q+q^2, 2 p^2-3 q^2, 3 p q+8 q^2\)

Solution:

Given:

\(p^2-p q+q^2, 2 p^2-3 q^2, 3 p q+8 q^2\) \(\begin{aligned}
& p^2-p q+q^2 \\
& 2 p^2-3 q^2 \\
& 3 p q+8 q^2 \\
& \hline 3 p^2+2 p q+6 q^2 \\
& \hline
\end{aligned}\)

∴The required sum = \(3 p^2+2 p q+6 q^2\)

\(p^2-p q+q^2, 2 p^2-3 q^2, 3 p q+8 q^2=3 p^2+2 p q+6 q^2\)

Example 6. Simplify \(-5 a^2+2 b^2+2 a b+3 a^2-6 b^2-5 a b+3 a\).

Solution:

The given expression

= \(-5 a^2+2 b^2+2 a b+3 a^2-6 b^2-5 a b+3 a\)

= \(\left(5 a^2+3 a^2\right)+\left(2 b^2-6 b^2\right)+(2 a b-5 a b)+3 a\)

∴ \(8 a^2-4 b^2-3 a b+3 a\)

\(5 a^2+2 b^2+2 a b+3 a^2-6 b^2-5 a b+3 a=8 a^2-4 b^2-3 a b+3 a\)

Class 7 Maths Polynomials Exercise Solutions

Example 7. Find the sum of the following expressions: \(6 x^2+3 x y, x y-y^2, 2 x^2+5 y^2-3 x y\).

Solution:

Given:

\(6 x^2+3 x y, x y-y^2, 2 x^2+5 y^2-3 x y\) \(\begin{gathered}
6 x^2+3 x y \\
\quad+x y-y^2 \\
2 x^2-3 x y+5 y^2 \\
\hline 8 x^2+x y+4 y^2 \\
\hline
\end{gathered}\)

 

∴ The required sum = \(8 x^2+x y+4 y^2\).

\(\)6X2+ 3xy, xy – y2, 2x2 + 5y2 – 3xy = 8x2 + xy + 4y2.

Example 8. Find the sum of the following expressions m2 – \(\frac{1}{2}\) mn, \(\frac{1}{3}\) mn + n2,-m2 – \(\frac{1}{4}\)n2 .

Solution:

\(\begin{aligned}
& m^2+\frac{1}{2} m n \\
& +\frac{1}{3} m n+n^2 \\
& -m^2 \quad-\frac{1}{4} n^2 \\
& \left\lvert\, \begin{array}{l}
\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}=\frac{5}{6} \\
1-\frac{1}{4}=\frac{3}{4}
\end{array}\right. \\
&
\end{aligned}\)

 

∴ The required sum = –\(\frac{5}{6}\) mn + \(\frac{3}{4}\)n2

Subtraction

The quantity from which the subtraction is to be made is called minuend.

The quantity which is to be subtracted is called subtrahend. The result obtained after subtraction is called the difference or remainder.

By subtraction of b from a we mean, addition of the negative of b with a.

It means that, a – b = a + (- b). Thus to subtract one expression from another (having any number of terms in either expression), we have to change the sign of each term of the subtrahend and they will be added to the like terms of the minuend.

West Bengal Board Class 7 Algebra Assistance

Algebra Chapter 4 Polynomials Exercise 4 Some Examples Of Subtraction

Example 1. Subtract 5xy from 12xy.

Solution: 12xy – 5xy = 7xy.

Example 2. Subtract 12x from 3x.

Solution: 3x – 12x = – 9x.

Example 3. Subtract – 9xy from 25xy.

Solution: 25xy – (- 9xy) = 25xy + 9xy = 34xy.

Example 4. Subtract \(5 a^2+\mathbf{4} b^2+2 c^2\)from \(7 a^2-3 b^2+8 c^2\).

Solution:

\(\begin{gathered}
7 a^2-3 b^2+8 c^2 \\
5 a^2+4 b^2+2 c^2 \\
-\quad-\quad- \\
\hline 2 a^2-7 b^2+6 c^2 \\
\hline
\end{gathered}\)

 

∴ \(2 a^2-7 b^2+6 c^2\).

\(5 a^2+4 b^2+2 c^2-7 a^2-3 b^2+8 c^2=2 a^2-7 b^2+6 c^2\).

WBBSE Class 7 Chapter 4 Polynomials Guide

Example 5. Subtract \(5 x^2-5\) from \(2 x^2+5 x y\).

Solution:

\(\begin{array}{r}
2 x^2+5 x y \\
-\quad 5 x^2+5 \\
-\quad+ \\
\hline-3 x^2+5 x y+5 \\
\hline
\end{array}\)

 

∴ \(-3 x^2+5 x y+5\).

\(5 x^2-5-2 x^2+5 x y=-3 x^2+5 x y+5\).

Example 6. What is to be added with \(a^2-\mathbf{a b}+2 \mathbf{b}^2\)a2 – ab + 2b2 to get \(a^2 + b^2\)?

Solution :

The required expression = \(\left(a^2+b^2\right)-\left(a^2+a b-2 b^2\right)\)

= \(\left(a^2+b^2\right)-\left(a^2+a b-2 b^2\right)\)

= \(\left(a^2-a^2\right)+\left(b^2-2 b^2\right)+a b\)

= \(-b^2 + a b\)

∴ \(-b^2 + a b\) is to be added.

Example 7. Subtract \(x^2+y^2+2 x\) from \(x^2+y^2+2 x y\).

Solution:

Given:

\(x^2+y^2-2 x y\) and \(x^2+y^2+2 x\)

\(\begin{array}{r}
x^2+y^2+2 x y \\
x^2+y^2-2 x y \\
-\quad-\quad+ \\
\hline 4 x y \\
\hline
\end{array}\)

 

∴ 4xy.

\(x^2+y^2-2 x y-x^2+y^2+2 x y=4 x y\).

Polynomials for Class 7 Students

Example 8. What is to be added to \(a^2+b^2+c^2\) to get \(b^2+c^2\)?

Solution:

Given

\(a^2+b^2+c^2\)

The required expression = \(\left(b^2+c^2\right)-\left(a^2+b^2+c^2\right)\)

= \(b^2+c^2-a^2-b^2-c^2=-a^2\).

∴ \(– a^2\)is to be added.

Example 9. The difference of two expressions is 3a + 6. If the greater one is \(8 a^2+5 a+9\), what will be the smaller one? (a > 0)

Solution:

\(\begin{array}{r}
8 a^2+5 a+9 \\
3 a+6 \\
-\quad- \\
\hline 8 a^2+2 a+3 \\
\hline
\end{array}\)

 

∴The smaller one is \(\)8a2 + 2a + 3.

Example 10. If x = 2a – b, y = b – 3a then prove that, x – y = – (y – x).

Solution:

L.H.S. = x-y = (2a-b)-(b-3a)

= 2a – b – b + 3a = 5a – 2b.

R.H.S. = – (y-x)

= -y+x = -(b- 3a) + 2a-b

= – b +3a + 2a- b = 5a-2b.

∴ L.H.S. = R.H.S. (Proved).

Concept of index

When the same number is multiplied several times then it may be expressed in brief.

Class 7 Maths Polynomial Exercises Explained

Example: 8 x 8 = 82

8 x 8 x 8 = 83

8 x 8 x 8 x 8 = 84

8 x 8 x 8 x 8 x 8 = 85.

Here 8 is called the base and the number written over 8 on the right side is the power or index.

Any number can be expanded in the index of 10.

Class Vii Math Solution Wbbse

Example: 742 = 7 x 100 + 4 x 10 + 2 = 7 x 102 + 4 x 10 + 2

4584 = 4 x 1000 + 5 x 100 + 8 x 10 + 4

= 4 x 103 + 5 x 102 + 8 x 10 + 4

Again, 81 = 3x3x3x3 = 34

Therefore, the index form of 81 is 34.

Here 3 is the base and 4 is the index.

243 = 3 x 3 x 3 x 3 x 3 x 35.

Therefore, index form of 243 is 35. Here 3 is the base and 5 is the power.

Similarly:

  1. 32= 2 x 2 x 2 x 2 x 2 = 25
  2. 27 = 3 x 3 x 3 = 33
  3. 625 = 5 x 5 x 5 x 5 = 54
  4. 729 = 3 x 3 x 3 x 3 x 3 x 3= 36
  5. 343= 7 x 7 x 7 = 73

Therefore, if x is any integer,

  1. x X x = x2
  2. x X x 2 x = x3
  3. x X x X x X x = x4
  4. x X x X x X x X x = x5

Again, 2 x2 x 3 x 3 = 22 x 32

2 x 2 x 2 x 5 x 5 = 23 x 52 7 x 7 x 7 x 5 x 5 x 5 x 5 = 73 x 54.

Any number may be written in index form after breaking it in prime factors,

Example: 36 = 2 x 2 x 3 x 3 = 22 x 32

75 = 3 x 5 x 5 = 3x 52

50 = 2 x 5 x 5 = 2 x 52

24 = 2 x 2 x 2 x 3 = 23 x 3

Determination of greater and smaller numbers:

Between 43 and 34 which is greater and which is smaller?

43 = 4 x 4 x 4 = 64

34 = 3x3x3x3 = 81

∴ 34 is greater than 43 i.e., 34 > 43.

Some formulae :

  1. xm  X xn = xm+n
  2. .xm ÷ xn = xmn
  3. (xm)n = xmn
  4. xn X yn = (xy)n  
  5. x0 = 1

Example:

1. \(2^3 \times 2^4=2^{3+4}\)

2. \(5^8 \div 5^3=5^{8-3}=5^5\)

3. \(\left(2^3\right)^2=2^{3 \times 2}=2^6\)

4. \(5^2 \div 5^2=1\)

5. \(8^{\circ}=1\)

6. \((-3)^5 \times(-3)^3=(-3)^{5+3}=(-3)^8\)

7. \(7^2 \times 3^2=(7 \times 3)^2=21^2\)

8. \(5^3 \times 9^3=(5 \times 9)^3=45^3\)

Example 1.

\(\frac{2^5 \times 2^7}{\left(2^5\right)^2}=\frac{2^{5+7}}{2^{10}}=2^{12-10}=2^2=4\)

 

Example 2.

\(\frac{2^3 \times 3^5 \times 16}{3 \times 32}=\frac{2^3 \times 3^5 \times 2^4}{3 \times 2^5}=2^{3+4-5} \times 3^{5-1}\)

 

= \(2^2 \times 3^4=4 \times 81=324\).

To write a number from the expansion:

4 x 103+ 2 x 102 + 3 x 10 + 5

= 4000 + 200 + 30 + 5 = 4235

5 x 104+ 2 x 103 + 1 x 102 + 2 x 10 + 3

= 50000 + 2000 + 100 + 20 + 3 = 52123.

 

Algebra Chapter 4 Polynomials Exercise 4 Simple Multiplication

In case of algebraic multiplication of two quantities, the sign of the product is + when the two quantities are of the same sign and the sign of the product is — when the quantities are of opposite signs.

This may be explained in brief as follows:

(+ x) x (+ y) = + xy

(+ x) x (- y) = – xy

(- x) x (+ y) = – xy

(- x) x (- y) = + xy

In case of finding the product of the same variable having different powers, we follow the rule xm x xn – xm+n.

When we multiply some algebraic expressions having numerical coefficients then the numerical co-efficient of the product is equal to the product of all the coefficients.

Example 1. Multiply 7x3 by 2x4.

Solution: \(7 x^3 \text { by } 2 x^4=7 \times 2 \times x^3 \times x^4\)

= \(14 X x^{3+4}=14 X x^7=14 x^7\)

∴ \(14 x^7\).

\(7 x^3 \times 2 x^4=14 x^7\).

Example 2. Multiply -7p2 by 8pq2.

Solution: \(\left(-7 p^2\right) \times\left(8 p q^2\right)=-56 p^{2+1} q^2\)

∴ \(-56 p^3 q^2\)

\(-7 p^2 \times 8 p q^2=-56 p^3 q^2\)

Example 3. Find the product \(\left(-5 p^2 q\right) \times\left(3 p q^2\right) \times\left(-2 p^2 q^2\right)\).

Solution: The required product = \(\left(-5 p^2 q\right) \times\left(3 p q^2\right) \times\left(-2 p^2 q^2\right)\).

= \((-5) \times 3 \times(-2) \times p^{2+1+2} \times q^{1+2+2}\)

∴ \(30 p^5 q^5\)

\(\left(-5 p^2 q\right) \times\left(3 p q^2\right) \times\left(-2 p^2 q^2\right)=30 p^5 q^5\)

Example 4. Multiply  \(3 p^2 q^5 r\) by \(-7 p^3 q^2 r^5\)

Solution: The required product = \(\left(-3 p^2 q^5 r\right) \times\left(-7 p^3 q^2 r^5\right)\)

= \((-3) \times(-7) x^{p 2+3} \times q^{5+2} \times r^{1+5}=21 p^5 q^7 r^6\)

∴ \(21 p^5 q^7 r^6\)

\(3 p^2 q^5 r x-7 p^3 q^2 r^5=21 p^5 q^7 r^6\)

Example 5.  Find the product : (2xyz) x (8pqr) x (- 4abc).

Solution: The required product = (2xyz) x (8pqr) x (- 4abc)

= 2 x 8 x (- 4) x xyz pqr abc = – 64 abc pqr xyz

∴ – 64 abc pqr xyz.

(2xyz) x (8pqr) x (- 4abc) = – 64 abc pqr xyz.

Example 6. Find the product \(\left(\frac{1}{2} x y^2\right) \times\left(-\frac{2}{3} x y^4\right) \times\left(3 x^3 y\right)\)

Solution: The required product

\(\left(\frac{1}{2} x y^2\right) \times\left(-\frac{2}{3} x y^4\right) \times\left(3 x^3 y\right)\)

 

= \(\frac{1}{2}\) X – (\(\frac{2}{3}\)) X 3 X xy2 X xy4 X x3y

= \((-1) x x^5 x y^7=-x^5 y^7\)

Multiplication of two or more terms with a single term

When we are to find the product of the form (a+b)x, then since a and b are unlike terms, it is not possible to find their sum in one term.

So we should proceed as follows: (a + b) x = a X x + b X x = ax + bx

Similarly, (a + b + c +…… ) = ax + bx + cx +

In fact, this is the distributive law of multiplication.

Wbbse Class 7 Maths Solutions

Example 1. Multiply 3a + 26 by – 2ab.

Solution: The required product

= (3a + 2b) x (- 2ab)

= 3a x (-2ab) + 2b x (- 2ab)

= \(-6 a^2 6-4 a b^2\).

∴ \(-6 a^2 6-4 a b^2\).

3a + 26 X – 2ab = \(-6 a^2 6-4 a b^2\).

Example 2. Multiply \(x^2 y z-2 x y^2 z+3 x y z^2 \text { by } 3 x y^2\).

Solution: The required product = \(\left(x^2 y z-2 x y^2 z+3 x y z^2\right) x 3 x y z\)

= \(x^2 y z \times 3 x y z-2 x y^2 z \times 3 x y z+3 x y z^2 \times 3 x y z\)

∴ \(3 x^3 y^2 z^2-6 x^2 y z^2+9 x^2 y^2 z^3\)

\(x^2 y z-2 x y^2 z+3 x y z^2 \text { by } 3 x y^2=3 x^3 y^2 z^2-6 x^2 y z^2+9 x^2 y^2 z^3\)

Example 3. Simplify: \(a^3(2 a-b)+2 b^2(a+2 b)+5\left(a^3-3 b^3\right)\).

Solution : The given expression = \(a^3(2 a-b)+2 b^2(a+2 b)+5\left(a^3-3 b^3\right)\).

= \(2 a^3-a^2 b+2 a b^2+4 b^3+5 a^3-15 b^3\)

= \(\left(2 a^3+5 a^3\right)+\left(4 b^3-15 b^3\right)-a^2 b+2 a b^2\)

∴ \(7 a^3-11 b^3-a^3 b+2 a b^2\)

\(a^3(2 a-b)+2 b^2(a+2 b)+5\left(a^3-3 b^3\right)=7 a^3-11 b^3-a^3 b+2 a b^2\)

Step-by-Step Solutions for Class 7 Algebra

Example 4. Simplify  2x (x – 3y) + 5y (2x + 3y) – 2x (y + 5x).

Solution: The given expression = 2x (x – 3y) + 5y (2x + 3y) -2x (y + 5x)

= \(2 x^2-6 x y+10 x y+15 y^2-2 x y-10 x^2\)

= \(\left(2 x^2-10 x^2\right)+(-6 x y+10 x y-2 x y)+15 y^2\)

∴ \(-8 x^2+2 x y+15 y^2\)

2x (x – 3y) + 5y (2x + 3y) – 2x (y + 5x) = \(-8 x^2+2 x y+15 y^2\)

Wbbse Class 7 Maths Solutions

Multiplication of a binomial expression by another binomial expression

When we are to find the product of two binomial expressions we proceed as follows (a + b) (c + d) = ac + ad + bc + bd

Thus, we see that when two binomial expressions are multiplied then each term of the multiplicand is multiplied by each term of the multiplier and all the products are finally added.

Example 1. Multiply : 7a + 36 by 2a – 6.

Solution: The required product = (7a + 36) (2a – 6)

= 7a.2a + 7a.(-b) + 3b.2a + 3b.(- b)

= \(14 a^2-7 a b+6 a b-3 b^2=14 a^2-a b-3 b^2\).

Alternative Method:

\(\begin{array}{r}
7 a+3 b \\
2 a-b \\
\hline 14 a^2+6 a b \\
-7 a b-3 b^2 \\
\hline 14 a^2-a b-3 b^2 \\
\hline
\end{array}\)

 

∴ The required product = \(14 a^2-a b-3 b^2\).

7a + 36 X 2a – 6 = \(14 a^2-a b-3 b^2\).

Example 2. Multiply \(a^2+a 6+6^2 \times a-6=a^3-6^3\) by a – 6 and find the product.

Solution:

\(\begin{aligned}
& a^2+a b+b^2 \\
& a-b \\
& \hline a^3+a^2 b+a b^2 \\
& \quad-a^2 b-a b^2-b^3 \\
& \hline a^3-b^3 \\
& \hline
\end{aligned}\)

 

∴ The required product = a3 – 63.

\(a^2+a 6+6^2 \times a-6=a^3-6^3\).

Example 3. Multiply a + x by a – x.

Solution:

\(\begin{aligned}
& a+x \\
& a-x \\
& \hline a^2+a x \\
& -a x-x^2 \\
& \hline a^2-x^2 \\
& \hline
\end{aligned}\)

 

a + x X a – x = \(a^2-x^2\)

Class Vii Math Solution Wbbse

Example 4. Multiply \(x^4+x^2 y^2+y^4 \text { by } x^2-y^2\).

Solution:

\(\begin{aligned}
& x^4+x^2 y^2+y^4 \\
& x^2-y^2 \\
& \hline x^6+x^4 y^2+x^2 y^4 \\
& -x^4 y^2-x^2 y^4-y^6 \\
& \hline x^6-y^6 \\
& \hline
\end{aligned}\)

 

Example 5. Simplify 3(x + y) – y(3 + x) + x(y – 3).

Solution: The given expression = 3(x +y)- y(3 + x) + x(y – 3)

= 3x + 3y – 3y – xy + xy – 3x = 0

∴ 0.

3(x + y) – y(3 + x) + x(y – 3)= 0

Example 6. Simplify : (2x + 9)(x – 7) – (x – 9)(2x + 7).

Solution: The given expression = (2x + 9)(x – 7) – (x – 9)(2x + 7)

= \(\left(2 x^2-14 x+9 x-63\right)-\left(2 x^2+7 x-18 x-63\right)\)

= \(\left(2 x^2-5 x-63\right)-\left(2 x^2-11 x-63\right)\)

= \(2 x^2-5 x-63-2 x^2+11 x+63=6 x\)

∴ 6x.

(2x + 9)(x – 7) – (x – 9)(2x + 7) = 6x.

Wbbse Class 7 Maths Solutions

Example 7. Simplify : a(b – c) + b(c – a) + c(a – b).

Solution : a(b – c) + b(c – a) + c(a – b)

= ab – ac + bc – ab + ca – bc = 0

∴ 0.

a(b – c) + b(c – a) + c(a – b) = 0

Class Vii Math Solution Wbbse

Example 8. Simplify : (x + 9)(x – 7) + (x + 9)(x + 10) – (2x +1)(x + 10).

Solution : (x + 9)(x -7) + (x + 9)(x + 10) -(2x + 1)(x + 10)

= \(\left(x^2-7 x+9 x-63\right)+\left(x^2+10 x+9 x+90\right)-\left(2 x^2+20 x+x+10\right)\)

= \(\left(x^2+2 x-63\right)+\left(x^2+19 x+90\right)-\left(2 x^2+21 x+10\right)\)

= \(x^2+2 x-63+x^2+19 x+90-2 x^2-21 x-10\)

∴ 17

(x + 9)(x – 7) + (x + 9)(x + 10) – (2x +1)(x + 10) = 17

Example 9. Simplify \(-5 a-[-3 b-\{-6 c-(-3 a-\overline{5 b-4 c})\}]\)

Solution:

\(-5 a-[-3 b-\{-6 c-(-3 a-\overline{5 b-4 c})\}]\)

 

= – 5a – [ – 36 – {- 6c – (- 3a – 56 + 4c)}] = – 5a – [ – 36 – { -6c + 3a + 56 – 4c}]

= – 5a – [ – 36 – { – 10c + 3a + 56}]

= – 5a – [ – 36 + 10c – 3a – 56]

= – 5a – [ – 86 + 10c – 3a]

= – 5a + 86 – 10c + 3a = – 2a + 86 – 10c

∴ – 2a + 86 — 10c.

Example 10. Simplify \(x-[y+\{x-(y-\overline{x-y})\}]\)

Solution:

\(x-[y+\{x-(y-\overline{x-y})\}]\)

 

= x- [y + {x- (y-x+y)}]

= x – [ y + {x – (2y – x)}]

= x – {y +{x – 2y + x}]

= x – [ y + 2x – 2y]

= x – [ – y + 2x] = x + y – 2x = y – x

∴ y -x.

Simple Division

Wbbse Class 7 Maths Solutions

In case of algebraic division of the form \(/frac{a}{b}\) = c, we call a the dividend, b the divisor and c the quotient.

If the dividend and the divisor are of the same sign then the sign of the quotient will be + and if the dividend and the divisor are of opposite signs then the sign of the quotient will be This may be explained in brief as follows :

\(\frac{(+x)}{(+y)}=+\frac{x}{y}\) \(\frac{(+x)}{(-y)}=-\frac{x}{y}\) \(\frac{(-x)}{(+y)}=-\frac{x}{y}\) \(\frac{(-x)}{(-y)}=+\frac{x}{y}\)

 

In case of finding the quotient of the same variable having different powers, we follow the rule xm ÷xn = xmn. We shall also take x° = 1.

Example 1. Divide \(35 a^4 b^8 \text { by } 5 a^2 b^2\).

Solution: The required quotient

\(\frac{35 a^4 b^8}{5 a^2 b^2}\)

= \(7 a^{4-2} b^{8-2}=7 a^2 b^6\)

Example 2. Divide \(36 x^{12} y^8 \text { by }\left(-9 x^5 y^4\right)\).

Solution: The required quotient

\(\frac{36 x^{12} y^8}{-9 x^5 y^4}\)

 

= \(-4 x^{12-5} y^{8-4}\)

= \(-4 x^7 y^4\).

Example 3. Divide \(\left(-20 p^8 q^6\right) \text { by } 5 p^4 q^2\).

Solution: The required quotient

\(\frac{-20 p^8 q^6}{5 p^4 q^2}\)

 

= \(-4 p^{8-4} q^{8-2}\) = \(-4 p^4 q^4\)

Example 4. Divide \(\left(-81 m^5 n^6\right) \text { by }\left(-27 m^2 n^2\right)\).

Wbbse Class 7 Maths Solutions

Solution: The required quotient

\(\frac{-81 m^5 n^6}{-27 m^2 n^2}\)

 

= \(3 m^{5-2} n^{6-2}=3 m^3 n^4\)

Class Vii Math Solution Wbbse

Division of a polynomial by a monomial

We know, from the distributive law of division that, (a+ b + c) ÷ x = a÷x + b ÷ x + c ÷ x

i.e.,  \(\frac{a+b+c}{x}\) = \(\frac{a}{x}\) + \(\frac{b}{x}\) + \(\frac{c}{x}\)

Thus, we see that when a polynomial is divided by a monomial then the quotient is obtained by dividing each term of the dividend by the divisor.

Example 1. Divide 15a – 25b + 35c by 5.

Solution: The required quotient

= \(\frac{15a-25b+35c}{5}\)

= \(\frac{15a}{5}\) – \(\frac{25ba}{5}\) + \(\frac{35a}{5}\)

∴ 3a – 5b + 7c

15a – 25b + 35c X 5 = 3a – 5b + 7c

Example 2. Divide \(4 x^5+3 x^4+8 x^3+7 x^2 \text { by } x^2\).

Solution: The required quotient

\(\frac{4 x^5+3 x^4+8 x^3+7 x^2}{x^2}\) \(\frac{4 x^5}{x^2}+\frac{3 x^4}{x^2}+\frac{8 x^3}{x^2}+\frac{7 x^2}{x^2}\)

 

= \(4 x^{5-2}+3 x^{4-2}+8 x^{3-2}+7 x^{2-2}=4 x^3+3 x^2+8 x^4+7 x^0\)

= \(4 x^3+3 x^2+8 x+7\)

= \(4 x^5+3 x^4+8 x^3+7 x^2 x x^2=4 x^3+3 x^2+8 x+7\)

= \(4 x^3+3 x^2+8 x+7\)4x3 + 3X2 + 8x + 7.

Wbbse Class 7 Maths Solutions

Example 3.  Divide \(12 m^2 n^4+16 m^3 n^3-20 m^4 n^2 \text { by } 4 m^2 n^2\).

Solution: The required quotient

\(\frac{12 m^2 n^4+16 m^3 n^3-20 m^4 n^2}{4 m^2 n^2}\) \(\frac{12 m^2 n^4}{4 m^2 n^2}+\frac{16 m^3 n^3}{4 m^2 n^2}-\frac{20 m^4 n^2}{4 m^2 n^2}\)

 

= \(3 m^{2-2} n^{4-2}+4 m^{3-2} n^{3-2}-5 m^{4-2} n^{2-2}=3 m^{\circ} n^2+4 m^1 n^1-5 m^2 n^{\circ}\)

∴ \(3 n^2+4 m n-5 m^2\)

\(12 m^2 n^4+16 m^3 n^3-20 m^4 n^2 \text { by } 4 m^2 n^2=3 n^2+4 m n-5 m^2\)

Example 4. Divide \(3 a b c-6 a^2 b c+5 a^2 b^2 c^2 \text { by } 3 a b c\).

Solution: The required quotient

\(\frac{3 a b c-6 a^2 b c+5 a^2 b^2 c^2}{3 a b c}\)

\(\frac{3 a b c}{3 a b c}-\frac{6 a^2 b c}{3 a b c}+\frac{5 a^2 b^2 c^2}{3 a b c}\)

∴ 1 – 2a + \(\frac{5}{3}\) abc

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