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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