krit.club logo

Introduction to Polynomials - Model real-life linear situations using linear polynomials and expressions

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A Linear Polynomial in one variable xx is an algebraic expression of the form P(x)=ax+bP(x) = ax + b, where aa and bb are real numbers and a≠0a \neq 0.

•

In real-life modeling, the constant term bb often represents a fixed value or initial starting point (e.g., a base fee or initial height).

•

The coefficient aa of the variable xx represents the rate of change (e.g., cost per km, speed, or hourly wage).

•

To model a situation, identify the independent variable (usually xx) and the dependent variable (the total result yy or P(x)P(x)).

•

Translating words to symbols: 'Sum' or 'More than' suggests addition (++), 'Difference' or 'Less than' suggests subtraction (−-), and 'Product' or 'Times' suggests multiplication (×\times).

📐Formulae

P(x)=ax+bP(x) = ax + b

Total Cost=(Rate×x)+Fixed Charge\text{Total Cost} = (\text{Rate} \times x) + \text{Fixed Charge}

Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

Perimeter of Rectangle=2(l+w)\text{Perimeter of Rectangle} = 2(l + w)

💡Examples

Problem 1:

A library charges a fixed membership fee of ₹200₹200 and an additional ₹5₹5 for every day a book is kept. Formulate a linear polynomial P(d)P(d) for the total cost after dd days.

Solution:

P(d)=5d+200P(d) = 5d + 200

Explanation:

The fixed fee is a constant 200200. The variable cost depends on the number of days dd, which is 5×d5 \times d. Adding them gives the total cost polynomial.

Problem 2:

The length of a rectangular park is 10 m10\text{ m} more than its width xx. Write a linear expression for the perimeter of the park.

Solution:

P(x)=4x+20P(x) = 4x + 20

Explanation:

Let width =x= x. Then length =x+10= x + 10. Perimeter is given by 2(length+width)2(\text{length} + \text{width}). So, P(x)=2(x+10+x)=2(2x+10)=4x+20P(x) = 2(x + 10 + x) = 2(2x + 10) = 4x + 20.

Problem 3:

A student has ₹500₹500 in a savings account and withdraws ₹40₹40 each week. Express the remaining balance BB as a polynomial of the number of weeks ww. Calculate the balance after 88 weeks.

Solution:

B(w)=500−40wB(w) = 500 - 40w. After 88 weeks: B(8)=500−40(8)=500−320=180B(8) = 500 - 40(8) = 500 - 320 = 180.

500−320180\begin{array}{r} 500 \\ - 320 \\ \hline 180 \end{array}

Explanation:

The starting amount 500500 is the constant. Since money is being removed, the rate 4040 is subtracted per week ww. Vertical subtraction confirms the final balance of ₹180₹180.