 # AP SSC Class 10 Maths Chapter 3 Polynomials

The AP SSC Class 10 Maths Chapter 3 Polynomials discusses polynomials and its various properties. Here, in the article, we have provided a brief explanation of polynomials and methodical solutions to the chapter questions. Refer to the below given AP SSC 1oth Class Maths Chapter 3 Polynomials Notes and Solutions and learn how to give step-by-step correct solutions for the concepts taught under the subject in class.

## What is a Polynomial?

A polynomial is an algebraic expression constructed using constants and variables. Coefficients can be raised to various powers of non-negative integer exponents. A few examples of polynomials are

$$\begin{array}{l}3x^{2}+5x+6\end{array}$$
,
$$\begin{array}{l}x^{3}\end{array}$$
and
$$\begin{array}{l}2x+5\end{array}$$

### Degree of a Polynomial

If p(x) is a polynomial in x, the highest power of x in p(x) is known as the degree of the polynomial p(x).

• A polynomial of degree 1 is known as a linear polynomial. Example,
$$\begin{array}{l}4x+6\end{array}$$
• A polynomial of degree 2 is known as a quadratic polynomial. Example,
$$\begin{array}{l}x^{2}+3x+5\end{array}$$
• A polynomial of degree 3 is known as a cubic polynomial. Example,
$$\begin{array}{l}a^{3}+a^{2}+a+6\end{array}$$

If p(x) is a polynomial in x, and if a is a real number, then the value obtained by replacing x by a in p(x), is called the value of p(x) at x = a, and is denoted by p(a). The value of x that makes the polynomial equal to 0 is known as the zero of a polynomial.

## Class 10 Maths Chapter 3 Polynomials Questions

1. If
$$\begin{array}{l}p(x)=5x^{7}-4x^{6}+8x-6\end{array}$$
, find the
• Coefficient of
$$\begin{array}{l}x^7\end{array}$$
• Degree of p(x)
• Constant term

Solution:

1. The coefficient of
$$\begin{array}{l}x^7\end{array}$$
is 5
2. The degree of p(x) is 7
3. The constant term of p(x) is -6
1. If
$$\begin{array}{l}p(t)=t^{3}-1\end{array}$$
, find the values of p(1), p(–1), p(0), p(2), p(–2).

Solution:

1. p(1) =
$$\begin{array}{l}p(t)=1^{3}-1=0\end{array}$$
2. p(–1) =
$$\begin{array}{l}p(t)=(-1)^{3}-1=-2\end{array}$$
3. p(0) =
$$\begin{array}{l}p(t)=0^{3}-1=-1\end{array}$$
4. p(2)=
$$\begin{array}{l}p(t)=2^{3}-1=7\end{array}$$
5. p(–2) =
$$\begin{array}{l}p(t)=(-2)^{3}-1=-9\end{array}$$

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