Maharashtra Board Class 9 Maths Chapter 3 Set 3.2 Algebra Standard Part 1 Polynomials Solutions

Question 1. Use the given letters to write the answer.
(i) There are ‘a’ trees in the village Lat. If the number of trees increases every year by ‘b’, then how many trees will there be after ‘x’ years?
(ii) For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all?
(iii) The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Answer:
(i) Number of trees after \( x \) years = \( a + bx \)
(ii) Total number of students in the parade = \( xy \)
(iii) The polynomial representing the two-digit number = \( 10m + n \)
In simple words: To find the total trees, we start with \( a \) and add \( b \) trees for each of the \( x \) years. For the parade, we multiply the number of rows by students per row. For the two-digit number, we multiply the tens digit by 10 and add the units digit.

🎯 Exam Tip: Always write the final algebraic expressions clearly with correct variables, and remember that the place value of the tens digit is always multiplied by 10.

Question 1. Write the polynomial for each of the given information:
(i) There are 'a' trees in the village Lat. If the number of trees increases every year by 'b', then how many trees will there be after 'x' years?
(ii) For the parade there are 'y' students in each row and 'x' such rows are formed. Then, how many students are there for the parade in all?
(iii) The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Answer:
(i) Number of trees in the village Lat = \( a \)
Number of trees increasing each year = \( b \)
\( \therefore \) Number of trees after \( x \) years = \( a + bx \)
\( \dots \) There will be \( a + bx \) trees in the village Lat after \( x \) years. This linear expression helps us calculate the future tree population easily.

(ii) Total rows = \( x \)
Number of students in each row = \( y \)
\( \therefore \) Total students = Total rows \( \times \) Number of students in each row
= \( x \times y \)
= \( xy \)
\( \therefore \) There are in all \( xy \) students for the parade.

(iii) Digit in units place = \( n \)
Digit in tens place = \( m \)
\( \therefore \) The two digit number = \( 10 \times \) digit in tens place + digit in units place
= \( 10m + n \)
\( \therefore \) The polynomial representing the two digit number is \( 10m + n \).
In simple words: We turn word problems into math expressions. For trees, we start with \( a \) and add \( b \) for each of the \( x \) years. For students, we multiply rows by students per row. For a two-digit number, we multiply the tens digit by 10 and add the units digit.

🎯 Exam Tip: Always define your variables clearly and write the final answer with appropriate units or concluding statements to secure full marks.

Question 2. Add the given polynomials.
(i) \( x^3 - 2x^2 - 9 \); \( 5x^3 + 2x + 9 \)
(ii) \( -7m^4 + 5m^3 + \sqrt{2} \); \( 5m^4 - 3m^3 + 2m^2 + 3m - 6 \)
(iii) \( 2y^2 + 7y + 5 \); \( 3y + 9 \); \( 3y^2 - 4y - 3 \)
Answer:
(i) \( (x^3 - 2x^2 - 9) + (5x^3 + 2x + 9) \)
= \( x^3 - 2x^2 - 9 + 5x^3 + 2x + 9 \)
= \( x^3 + 5x^3 - 2x^2 + 2x - 9 + 9 \)
= \( 6x^3 - 2x^2 + 2x \)

(ii) \( (-7m^4 + 5m^3 + \sqrt{2}) + (5m^4 - 3m^3 + 2m^2 + 3m - 6) \)
= \( -7m^4 + 5m^3 + \sqrt{2} + 5m^4 - 3m^3 + 2m^2 + 3m - 6 \)
= \( -7m^4 + 5m^4 + 5m^3 - 3m^3 + 2m^2 + 3m + \sqrt{2} - 6 \)
= \( -2m^4 + 2m^3 + 2m^2 + 3m + \sqrt{2} - 6 \)

(iii) \( (2y^2 + 7y + 5) + (3y + 9) + (3y^2 - 4y - 3) \)
= \( 2y^2 + 7y + 5 + 3y + 9 + 3y^2 - 4y - 3 \)
= \( 2y^2 + 3y^2 + 7y + 3y - 4y + 5 + 9 - 3 \)
= \( 5y^2 + 6y + 11 \)
This final simplified expression represents the sum of all three quadratic polynomials.
In simple words: To add polynomials, we group the like terms (terms with the same variable and exponent) together and then add their coefficients.

🎯 Exam Tip: When adding polynomials, group like terms together in descending order of their powers to avoid calculation errors.

Question 3. Subtract the second polynomial from the first.
(i) \( x^2 - 9x + \sqrt{3} ; -19x + \sqrt{3} + 7x^2 \)
(ii) \( 2ab^2 + 3a^2b - 4ab ; 3ab - 8ab^2 + 2a^2b \)
Answer:
(i) \( (x^2 - 9x + \sqrt{3}) - (-19x + \sqrt{3} + 7x^2) \)
\( = x^2 - 9x + \sqrt{3} + 19x - \sqrt{3} - 7x^2 \)
\( = x^2 - 7x^2 - 9x + 19x + \sqrt{3} - \sqrt{3} \)
\( = -6x^2 + 10x \)

(ii) \( (2ab^2 + 3a^2b - 4ab) - (3ab - 8ab^2 + 2a^2b) \)
\( = 2ab^2 + 3a^2b - 4ab - 3ab + 8ab^2 - 2a^2b \)
\( = 2ab^2 + 8ab^2 + 3a^2b - 2a^2b - 4ab - 3ab \)
\( = 10ab^2 + a^2b - 7ab \)
We group the like terms together and perform the subtraction carefully by changing the signs of the terms in the second polynomial.
In simple words: To subtract one polynomial from another, change the sign of every term in the second polynomial and then add the like terms together.

🎯 Exam Tip: Always remember to distribute the negative sign to every single term inside the bracket of the polynomial being subtracted to avoid sign errors.

Question 4. Multiply the given polynomials.
(i) \( 2x ; x^2 - 2x - 1 \)
(ii) \( x^5 - 1 ; x^3 + 2x^2 + 2 \)
(iii) \( 2y + 1 ; y^2 - 2y^3 + 3y \)
Answer:
(i) \( (2x) \times (x^2 - 2x - 1) = 2x^3 - 4x^2 - 2x \)

(ii) \( (x^5 - 1) \times (x^3 + 2x^2 + 2) \)
\( = x^5(x^3 + 2x^2 + 2) - 1(x^3 + 2x^2 + 2) \)
\( = x^8 + 2x^7 + 2x^5 - x^3 - 2x^2 - 2 \)

(iii) \( (2y + 1) \times (y^2 - 2y^3 + 3y) \)
\( = 2y(y^2 - 2y^3 + 3y) + 1(y^2 - 2y^3 + 3y) \)
\( = 2y^3 - 4y^4 + 6y^2 + y^2 - 2y^3 + 3y \)
\( = -4y^4 + 2y^3 - 2y^3 + 6y^2 + y^2 + 3y \)
\( = -4y^4 + 7y^2 + 3y \)
We multiply each term of the first polynomial by every term of the second polynomial and then simplify by combining the like terms.
In simple words: To multiply polynomials, distribute each term of the first expression to every term of the second expression, add the exponents of like bases, and combine any terms that are alike.

🎯 Exam Tip: When multiplying terms with exponents, remember the rule \( x^a \times x^b = x^{a+b} \) and write the final polynomial in descending order of powers.

Question 5. Divide first polynomial by second polynomial and write the answer in the form

Question 5. Divide the first polynomial by the second and write the answer in the form 'Dividend = Divisor × Quotient + Remainder'.
(i) \( x^3 - 64; x - 4 \)
(ii) \( 5x^5 + 4x^4 - 3x^3 + 2x^2 + 2; x^2 - x \)
Answer:

(i) \( x^3 - 64; x - 4 \)
Write the dividend in index form: \( x^3 + 0x^2 + 0x - 64 \)
Perform the polynomial division:

From the division:
Quotient = \( x^2 + 4x + 16 \)
Remainder = \( 0 \)
Using the formula: Dividend = Divisor × Quotient + Remainder
\( \therefore x^3 - 64 = (x - 4)(x^2 + 4x + 16) + 0 \)

(ii) \( 5x^5 + 4x^4 - 3x^3 + 2x^2 + 2; x^2 - x \)
Write the dividend in index form: \( 5x^5 + 4x^4 - 3x^3 + 2x^2 + 0x + 2 \)
Perform the polynomial division:

From the division:
Quotient = \( 5x^3 + 9x^2 + 6x + 8 \)
Remainder = \( 8x + 2 \)
Using the formula: Dividend = Divisor × Quotient + Remainder
\( \therefore 5x^5 + 4x^4 - 3x^3 + 2x^2 + 2 = (x^2 - x)(5x^3 + 9x^2 + 6x + 8) + (8x + 2) \)
In simple words: To divide polynomials, we perform long division step-by-step by dividing the highest power terms, and then express the final result as Dividend = (Divisor × Quotient) + Remainder.

🎯 Exam Tip: Always write the dividend in its complete index form by adding terms with zero coefficients for any missing powers to keep your columns aligned during division.

Question 6. Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length \( (2a^2 + 3b^2) \) m and breadth \( (a^2 + b^2) \) m. The farmer used a square shaped plot of the farm to build a house. The side of the plot was \( (a^2 - b^2) \) m. What is the area of the remaining part of the farm?
Answer:
Given:
Length of the rectangular farm = \( 2a^2 + 3b^2 \) m
Breadth of the rectangular farm = \( a^2 + b^2 \) m
Side of the square plot = \( a^2 - b^2 \) m

Step 1: Find the area of the rectangular farm.
Area of rectangular farm = Length × Breadth
\( = (2a^2 + 3b^2)(a^2 + b^2) \)
\( = 2a^2(a^2 + b^2) + 3b^2(a^2 + b^2) \)
\( = 2a^4 + 2a^2b^2 + 3a^2b^2 + 3b^4 \)
\( = 2a^4 + 5a^2b^2 + 3b^4 \) sq. m.

Step 2: Find the area of the square plot.
Area of square plot = \( (\text{side})^2 \)
\( = (a^2 - b^2)^2 \)
\( = a^4 - 2a^2b^2 + b^4 \) sq. m.

Step 3: Find the area of the remaining part of the farm.
Area of remaining part = Area of rectangular farm - Area of square plot
\( = (2a^4 + 5a^2b^2 + 3b^4) - (a^4 - 2a^2b^2 + b^4) \)
\( = 2a^4 + 5a^2b^2 + 3b^4 - a^4 + 2a^2b^2 - b^4 \)
\( = (2a^4 - a^4) + (5a^2b^2 + 2a^2b^2) + (3b^4 - b^4) \)
\( = a^4 + 7a^2b^2 + 2b^4 \) sq. m.

Thus, the area of the remaining part of the farm is \( a^4 + 7a^2b^2 + 2b^4 \) sq. m.
In simple words: To find the remaining area, we first calculate the total area of the rectangular farm and subtract the area of the square house plot from it.

🎯 Exam Tip: Be very careful with the negative sign when subtracting the area of the square plot; it changes the sign of every term inside the brackets.

Question. There is a rectangular farm with length \( (2a^2 + 3b^2) \) metre and breadth \( (a^2 + b^2) \) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was \( (a^2 - b^2) \) metre. What is the area of the remaining part of the farm?
Answer:
Length of the rectangular farm = \( (2a^2 + 3b^2) \) m
Breadth of the rectangular farm = \( (a^2 + b^2) \) m
Area of the farm = length \( \times \) breadth = \( (2a^2 + 3b^2) \times (a^2 + b^2) \)
\( = 2a^2(a^2 + b^2) + 3b^2(a^2 + b^2) \)
\( = 2a^4 + 2a^2b^2 + 3a^2b^2 + 3b^4 \)
\( = (2a^4 + 5a^2b^2 + 3b^4) \) sq. m ... (i)

The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = \( (a^2 - b^2) \) m
\( \therefore \) Area of the plot = \( (\text{side})^2 \)
\( = (a^2 - b^2)^2 \)
\( = (a^4 - 2a^2b^2 + b^4) \) sq. m ... (ii)

\( \dots \) Area of the remaining farm = Area of the farm \( - \) Area of the plot
\( = (2a^4 + 5a^2b^2 + 3b^4) - (a^4 - 2a^2b^2 + b^4) \) ... [From (i) and (ii)]
\( = 2a^4 + 5a^2b^2 + 3b^4 - a^4 + 2a^2b^2 - b^4 \)
\( = 2a^4 - a^4 + 5a^2b^2 + 2a^2b^2 + 3b^4 - b^4 \)
\( = a^4 + 7a^2b^2 + 2b^4 \)
\( \therefore \) The area of the remaining farm is \( (a^4 + 7a^2b^2 + 2b^4) \) sq. m. This algebraic calculation helps us determine the exact leftover space without needing numerical values for \( a \) and \( b \).
In simple words: To find the leftover area, we first calculate the total area of the rectangular farm and the area of the square house. Then, we subtract the house's area from the total farm area by combining like terms.

🎯 Exam Tip: Be very careful with the negative sign when subtracting the area of the square plot, as it changes the signs of all terms inside the parentheses.