Zeros Of polynomial

General form of a polynomial in \(x\) is \(a_n x^n~ +~ a_{n-1} x^{n-1}~ + …..~+ ~a_1 x~+~a_0\), where \(a_n,~ a_{n-1},~ …..~,~a_1,~ a_0\) are constants, \(a_n~≠~0\) and \(n\) is a whole number.

Algebraic expressions such as \(√x + x + 5, ~x^2 + \frac{1}{x^2}\) are not polynomials because, all exponents of x in terms of the expressions are not whole numbers.

Degree of a polynomial is highest power of the variable \(x\).

  • Polynomial of degree 1 is known as linear polynomial.
    Standard form is \(ax + b\), where \(a\) and \(b\) are real numbers and \(a≠0\).
    \(2x + 3\) is a linear polynomial.
  • Polynomial of degree 2 is known as quadratic polynomial.
    Standard form is \(ax^2 + bx + c\), where \(a,~ b ~and ~c\) are real numbers and \(a ≠ 0\)
    \(x^2 + 3x + 4\) is an example for quadratic polynomial.
  • Polynomial of degree 3 is known as cubic polynomial.
    Standard form is \(ax^3 + bx^2 + cx + d\), where \(a,~ b,~ c ~and~ d\) are real numbers and \(a≠0\).
    \(x^3 + 4x + 2\) is an example for cubic polynomial.

Similarly,

\(y^6 + 3y^4 + y\) is a polynomial in y of degree 6.

Example: What is the value of a if degree of polynomial \(x^3 + x^{a-4} + x^2 + 1\) is 4?

Degree of a polynomial \(P(x)\) is the highest power of \(x\) in \(P(x)\).

Therefore, \(x^3 + x^{a-4} + x^2 + 1\), \(x^{a-4}\) = \(x^4\)

\(a-4\) = \(4\), \(a\) = \(4 + 4\) = \(8\)

Consider, \(P(x)\) = \(x^2 – 3x + 2\),

Put \(x\) = \(3\) in \(P(x)\) which gives,

\(P(3)\) = \(9 – 9 + 2\) = \(2\)

Replace \(x\) by 2 in the polynomial \(x^2 – 3x + 2\), which gives \(P(2)\) = \(4 – 6 + 2\) = \(0\).

Similarly, value of \(x^2 – 3x + 2\) at \(x\) = \(0\) is,

\(P(0)\) = \(0 – 0 + 2\) = \(2\)

In general; if P(x) is a polynomial in x and k is any real number, then value of \(P(k)\) at \(x\) = \(k\) is denoted by \(P(k)\) is found by replacing \(x\) by \(k\) in \(P(x)\).

In the polynomial \(x^2 – 3x + 2\),

Replacing \(x\) by 1 gives,

\(P(1)\) = \(1 – 3 + 2\) = \(0\)

Similarly, replacing \(x\) by 2 gives,

\(P(2)\) = \(4 – 6 + 2\) = \(0\)

For a polynomial \(P(x)\), real number k is said to be zero of polynomial \(P(x)\), if
\(P(k)\) = \(0\).

Therefore, 1 and 2 are the zeros of polynomial \(x^2 – 3x + 2\).

Consider, \(P(x)\) = \(4x + 5\) to be a linear polynomial in one variable;

Let \(a\) be zero of \(P(x)\), then,

\(P(a)\) = \(4k + 5\) = \(0\)

Therefore,\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\) \(k\) = \(-\frac{5}{4}\)

In general, If \(k\) is zero of the linear polynomial in one variable; \(P(x)\) = \(ax + b\), then

\(P(k)\) = \(ak + b\) = \(0\)

\(k\) = \(-\frac{b}{a}\)<

It can also be written as,

Zeros of polynomial

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Practise This Question

Find zeroes of the polynomial x22