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Introduction to Linear Polynomials - Introduction

Grade 9CBSE

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

🔑Concepts

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A polynomial is an algebraic expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

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The highest power of the variable in a polynomial is called the degree of the polynomial.

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A Linear Polynomial is a polynomial of degree 1. It is of the form p(x)=ax+bp(x) = ax + b, where aa and bb are real numbers and a≠0a \neq 0.

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In the expression ax+bax + b, aa is the coefficient of xx and bb is the constant term.

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A linear polynomial can be a monomial (e.g., 3x3x) or a binomial (e.g., 3x+53x + 5).

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A value kk is called a zero of a polynomial p(x)p(x) if p(k)=0p(k) = 0. For a linear polynomial ax+bax + b, there is exactly one zero.

📐Formulae

p(x)=ax+b, where a≠0p(x) = ax + b, \text{ where } a \neq 0

Zero of p(x)=ax+b is x=−ba\text{Zero of } p(x) = ax + b \text{ is } x = -\frac{b}{a}

Value of p(x) at x=k is p(k)=ak+b\text{Value of } p(x) \text{ at } x = k \text{ is } p(k) = ak + b

💡Examples

Problem 1:

Identify which of the following are linear polynomials: (i) 4x+54x + 5, (ii) x2−3x^2 - 3, (iii) 3y3y, (iv) 2−z2 - z.

Solution:

The linear polynomials are (i) 4x+54x + 5, (iii) 3y3y, and (iv) 2−z2 - z.

Explanation:

A linear polynomial must have a degree of 1. In (i), (iii), and (iv), the highest power of the variable is 1. In (ii), the degree is 2, so it is a quadratic polynomial, not linear.

Problem 2:

Find the zero of the linear polynomial p(x)=5x−15p(x) = 5x - 15.

Solution:

To find the zero, set p(x)=0p(x) = 0: 5x−15=05x - 15 = 0 5x=155x = 15 x=155x = \frac{15}{5} x=3x = 3 The zero of the polynomial is 33.

Explanation:

The zero of a polynomial is the value of the variable that makes the entire expression equal to zero.

Problem 3:

Find the value of the polynomial q(y)=12y+4q(y) = \frac{1}{2}y + 4 at y=10y = 10.

Solution:

Substitute y=10y = 10 into the expression: q(10)=12(10)+4q(10) = \frac{1}{2}(10) + 4 q(10)=5+4q(10) = 5 + 4 q(10)=9q(10) = 9 The value is 99.

Explanation:

The value of a polynomial at a given point is found by replacing the variable with that specific number and simplifying.