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Polynomials - Introduction

Grade 10CBSE

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

🔑Concepts

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A polynomial p(x)p(x) in one variable xx is an algebraic expression of the form p(x)=anxn+an−1xn−1+⋯+a1x+a0p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0, where an,an−1,…,a0a_n, a_{n-1}, \dots, a_0 are real numbers and nn is a non-negative integer.

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The highest power of xx in p(x)p(x) is called the degree of the polynomial. Based on degree, polynomials are classified as: Linear (Degree 1, e.g., ax+bax + b), Quadratic (Degree 2, e.g., ax2+bx+cax^2 + bx + c), and Cubic (Degree 3, e.g., ax3+bx2+cx+dax^3 + bx^2 + cx + d).

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A real number kk is said to be a zero of a polynomial p(x)p(x) if p(k)=0p(k) = 0.

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Geometrically, the zeros of a polynomial p(x)p(x) are the xx-coordinates of the points where the graph of y=p(x)y = p(x) intersects the xx-axis. For a polynomial of degree nn, the graph can intersect the xx-axis at most at nn points.

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For a quadratic polynomial ax2+bx+cax^2 + bx + c, the sum of zeros is given by −ba-\frac{b}{a} and the product of zeros is given by ca\frac{c}{a}.

📐Formulae

p(x)=anxn+an−1xn−1+⋯+a1x+a0,an≠0p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0, a_n \neq 0

Sum of zeros (α+β)=−coefficient of xcoefficient of x2=−ba\text{Sum of zeros } (\alpha + \beta) = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{b}{a}

Product of zeros (αβ)=constant termcoefficient of x2=ca\text{Product of zeros } (\alpha \beta) = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a}

Quadratic polynomial with zeros α,β:p(x)=k[x2−(α+β)x+αβ]\text{Quadratic polynomial with zeros } \alpha, \beta: p(x) = k[x^2 - (\alpha + \beta)x + \alpha\beta]

For cubic ax3+bx2+cx+d:α+β+γ=−ba,αβ+βγ+γα=ca,αβγ=−da\text{For cubic } ax^3 + bx^2 + cx + d: \alpha + \beta + \gamma = -\frac{b}{a}, \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}, \alpha\beta\gamma = -\frac{d}{a}

💡Examples

Problem 1:

Find the zeros of the quadratic polynomial p(x)=x2−2x−8p(x) = x^2 - 2x - 8 and verify the relationship between the zeros and the coefficients.

Solution:

Step 1: Find zeros by factorizing: x2−4x+2x−8=0x^2 - 4x + 2x - 8 = 0 x(x−4)+2(x−4)=0x(x - 4) + 2(x - 4) = 0 (x−4)(x+2)=0(x - 4)(x + 2) = 0 So, zeros are α=4\alpha = 4 and β=−2\beta = -2.

Step 2: Verify relationships: Sum of zeros: α+β=4+(−2)=2\alpha + \beta = 4 + (-2) = 2. From formula: −ba=−−21=2-\frac{b}{a} = -\frac{-2}{1} = 2. Product of zeros: αβ=4×(−2)=−8\alpha \beta = 4 \times (-2) = -8. From formula: ca=−81=−8\frac{c}{a} = \frac{-8}{1} = -8.

Explanation:

We first use the splitting the middle term method to find the roots. Then we compare the sum and product of these roots with the values obtained from the coefficients a=1,b=−2,c=−8a=1, b=-2, c=-8.

Problem 2:

Find a quadratic polynomial, the sum and product of whose zeros are 14\frac{1}{4} and −1-1 respectively.

Solution:

Let the zeros be α\alpha and β\beta. Given: α+β=14\alpha + \beta = \frac{1}{4} and αβ=−1\alpha\beta = -1. The general form of a quadratic polynomial is: p(x)=k[x2−(α+β)x+αβ]p(x) = k[x^2 - (\alpha + \beta)x + \alpha\beta] Substituting the values: p(x)=k[x2−14x+(−1)]p(x) = k[x^2 - \frac{1}{4}x + (-1)] p(x)=k[x2−14x−1]p(x) = k[x^2 - \frac{1}{4}x - 1] If we take k=4k = 4 to remove the fraction: p(x)=4x2−x−4p(x) = 4x^2 - x - 4

Explanation:

A quadratic polynomial can be constructed if the sum and product of zeros are known using the standard identity. We can choose an appropriate value for kk to simplify the coefficients to integers.