Polynomials Notes

Polynomials

Introduction

• A polynomial in one variable and its degree.

• If $p(x)$ is a polynomial in $x$, the highest power of $x$ in $p(x)$ is called the degree of the polynomial $p(x)$.

  • Example: $4x + 2$ is a polynomial in the variable $x$ of degree 1.

  • $2y^2 – 3y + 4$ is a polynomial in the variable $y$ of degree 2.

  • $5x^3 – 4x^2 + x – 2$ is a polynomial in the variable $x$ of degree 3.

  • $7u^6 – \frac{3}{2}u^4 + 4u^2 – 8$ is a polynomial in the variable $u$ of degree 6.

• Expressions like $\frac{1}{x} – 1$, $x + \frac{2}{x}$, $\frac{2}{x^2} + 3$ etc., are not polynomials.

• A polynomial of degree 1 is called a linear polynomial.

  • Example: $2x – 3$, $x + y + 2$, $3z + 4$, etc., are all linear polynomials.

• Polynomials such as $2x + 5 – x^2$, $x^3 + 1$, etc., are not linear polynomials.

• A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’.

  • $2x^2 + 3$, $5x^2 + x – 1$,$y^2 – 2$, $2 – x – x^2$, $\frac{2}{5}v^2 – \frac{3}{7}u + 5$, $z^2 – \frac{1}{3}$ are some examples of quadratic polynomials (whose coefficients are real numbers).

• More generally, any quadratic polynomial in $x$ is of the form $ax^2 + bx + c$, where $a$, $b$, $c$ are real numbers and $a \neq 0$.

• A polynomial of degree 3 is called a cubic polynomial.

  • Some examples of a cubic polynomial are $2 – x^3$, $x^3$, $x^3 – x^2 + x$, $3x^3 – 2x^2 + x – 1$.

• In fact, the most general form of a cubic polynomial is $ax^3 + bx^2 + cx + d$, where $a$, $b$, $c$, $d$ are real numbers and $a \neq 0$.

Value of a Polynomial

• Consider the polynomial $p(x) = x^2 – 3x – 4$.

• Putting $x = 2$ in the polynomial, we get $p(2) = 2^2 – 3 \times 2 – 4 = –6$.

• The value ‘–6’, obtained by replacing $x$ by 2 in $x^2 – 3x – 4$, is the value of $x^2 – 3x – 4$ at $x = 2$.

• Similarly, $p(0)$ is the value of $p(x)$ at $x = 0$, which is – 4.

• If $p(x)$ is a polynomial in $x$, and if $k$ is any real number, then the value obtained by replacing $x$ by $k$ in $p(x)$, is called the value of $p(x)$ at $x = k$, and is denoted by $p(k)$.

Zeroes of a Polynomial

• What is the value of $p(x) = x^2 –3x – 4$ at $x = –1$?

  • $p(–1) = (–1)^2 –{3 \times (–1)} – 4 = 0$

• Also, note that $p(4) = 4^2 – (3 \times 4) – 4 = 0$.

• As $p(–1) = 0$ and $p(4) = 0$, –1 and 4 are called the zeroes of the quadratic polynomial $x^2 – 3x – 4$.

• More generally, a real number $k$ is said to be a zero of a polynomial $p(x)$, if $p(k) = 0$.

• If $k$ is a zero of $p(x) = 2x + 3$, then $p(k) = 0$ gives us $2k + 3 = 0$, i.e., $k = -\frac{3}{2}$.

• In general, if $k$ is a zero of $p(x) = ax + b$, then $p(k) = ak + b = 0$, i.e., $k = -\frac{b}{a}$.

• So, the zero of the linear polynomial $ax + b$ is $-\frac{\text{Constant term}}{\text{Coefficient of }x} = -\frac{b}{a}$.

• The zero of a linear polynomial is related to its coefficients.

Geometrical Meaning of the Zeroes of a Polynomial

• A real number $k$ is a zero of the polynomial $p(x)$ if $p(k) = 0$.

• The geometrical representations of linear and quadratic polynomials and the geometrical meaning of their zeroes.

• Consider first a linear polynomial $ax + b$, $a \neq 0$.

• The graph of $y = ax + b$ is a straight line.

• the graph of $y = 2x + 3$ is a straight line passing through the points (– 2, –1) and (2, 7).

• The graph of $y = 2x + 3$ intersects the x -axis mid-way between $x = –1$ and $x = – 2$, that is, at the point $\left(-\frac{3}{2}, 0\right)$.

• The zero of $2x + 3$ is $-\frac{3}{2}$.

• The zero of the polynomial $2x + 3$ is the x-coordinate of the point where the graph of $y = 2x + 3$ intersects the x-axis.

• In general, for a linear polynomial $ax + b$, $a \neq 0$, the graph of $y = ax + b$ is a straight line which intersects the x-axis at exactly one point, namely, $\left(-\frac{b}{a}, 0\right)$.

• The linear polynomial $ax + b$, $a \neq 0$, has exactly one zero, namely, the x-coordinate of the point where the graph of $y = ax + b$ intersects the x-axis.

• The geometrical meaning of a zero of a quadratic polynomial.

• Consider the quadratic polynomial $x^2 – 3x – 4$. The graph of $y = x^2 – 3x – 4$

• For any quadratic polynomial $ax^2 + bx + c$, $a \neq 0$, the graph of the corresponding equation $y = ax^2 + bx + c$ has one of the two shapes either open upwards like or open downwards like depending on whether a > 0 or a < 0. (These curves are called parabolas.)

• –1 and 4 are zeroes of the quadratic polynomial. Also note from Fig. 2.2 that –1 and 4 are the x-coordinates of the points where the graph of $y = x^2 – 3x – 4$ intersects the x-axis.

• The zeroes of the quadratic polynomial $x^2 – 3x – 4$ are x-coordinates of the points where the graph of $y = x^2 – 3x – 4$ intersects the x-axis.

• The zeroes of a quadratic polynomial $ax^2 + bx + c$, $a \neq 0$, are precisely the x-coordinates of the points where the parabola representing $y = ax^2 + bx + c$ intersects the x-axis.

Cases for Quadratic Polynomials

• Case (i): The graph cuts x-axis at two distinct points A and A. The x-coordinates of A and A are the two zeroes of the quadratic polynomial $ax^2 + bx + c$ in this case.

• Case (ii): The graph cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A of Case (i) coincide here to become one point A.

  • The x-coordinate of A is the only zero for the quadratic polynomial $ax^2 + bx + c$ in this case.

• Case (iii): The graph is either completely above the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point.

  • The quadratic polynomial $ax^2 + bx + c$ has no zero in this case.

• A quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero.

• A polynomial of degree 2 has atmost two zeroes.

• Consider the cubic polynomial $x^3 – 4x$. The graph of $y = x^3 – 4x$

• – 2, 0 and 2 are zeroes of the cubic polynomial $x^3 – 4x$.

• – 2, 0 and 2 are, in fact, the x-coordinates of the only points where the graph of $y = x^3 – 4x$ intersects the x-axis.

• The curve meets the x-axis in only these 3 points, their x-coordinates are the only zeroes of the polynomial.

• There are at most 3 zeroes for any cubic polynomial.

• Any polynomial of degree 3 can have at most three zeroes.

• In general, given a polynomial $p(x)$ of degree $n$, the graph of $y = p(x)$ intersects the x-axis at atmost $n$ points. Therefore, a polynomial $p(x)$ of degree $n$ has at most $n$ zeroes.

Examples

• Example 1: Look at the graphs in Fig. 2.9 given below. Each is the graph of $y = p(x)$, where $p(x)$ is a polynomial. For each of the graphs, find the number of zeroes of $p(x)$.

  • (i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.

  • (ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.

  • (iii) The number of zeroes is 3.

  • (iv) The number of zeroes is 1.

  • (v) The number of zeroes is 1.

  • (vi) The number of zeroes is 4.

Relationship between Zeroes and Coefficients of a Polynomial

• Zero of a linear polynomial $ax + b$ is $-\frac{b}{a}$.

• For the relationship between zeroes and coefficients of a quadratic polynomial.

• Let us take a quadratic polynomial, say $p(x) = 2x^2 – 8x + 6$.

  • $2x^2 – 8x + 6 = 2x^2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3) = (2x – 2)(x – 3) = 2(x – 1)(x – 3)$

• The value of $p(x) = 2x^2 – 8x + 6$ is zero when $x – 1 = 0$ or $x – 3 = 0$, i.e., when $x = 1$ or $x = 3$.

• The zeroes of $2x^2 – 8x + 6$ are 1 and 3.

  • Sum of its zeroes = $1 + 3 = 4 = -\frac{(-8)}{2} = -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}$

  • Product of its zeroes = $1 \times 3 = 3 = \frac{6}{2} = \frac{\text{Constant term}}{\text{Coefficient of }x^2}$

• Let us take one more quadratic polynomial, say, $p(x) = 3x^2 + 5x – 2$.

  • $3x^2 + 5x – 2 = 3x^2 + 6x – x – 2 = 3x(x + 2) –1(x + 2) = (3x – 1)(x + 2)$

• The value of $3x^2 + 5x – 2$ is zero when either $3x – 1 = 0$ or $x + 2 = 0$, i.e., when $x = \frac{1}{3}$ or $x = –2$.

• The zeroes of $3x^2 + 5x – 2$ are $\frac{1}{3}$ and – 2.

  • Sum of its zeroes = $\frac{1}{3} + (-2) = -\frac{5}{3} = -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}$

  • Product of its zeroes = $\frac{1}{3} \times (-2) = -\frac{2}{3} = \frac{\text{Constant term}}{\text{Coefficient of }x^2}$

• In general, if $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = ax^2 + bx + c$, $a \neq 0$, then you know that $x – \alpha$ and $x – \beta$ are the factors of $p(x)$.

  • $ax^2 + bx + c = k(x – \alpha) (x – \beta)$, where $k$ is a constant

  • $= k[x^2 – (\alpha + \beta)x + \alpha \beta] = kx^2 – k(\alpha + \beta)x + k \alpha \beta$

• Comparing the coefficients of $x^2$, $x$ and constant terms on both the sides, we get $a = k$, $b = – k(\alpha + \beta)$ and $c = k\alpha\beta$.

  • This gives $\alpha + \beta = -\frac{b}{a}$, $\alpha\beta = \frac{c}{a}$

• $\alpha, \beta$ are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively.

• sum of zeroes = $\alpha + \beta = -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2} = -\frac{b}{a}$ , product of zeroes = $\alpha\beta = \frac{\text{Constant term}}{\text{Coefficient of }x^2} = \frac{c}{a}$.

Examples

• Example 2: Find the zeroes of the quadratic polynomial $x^2 + 7x + 10$, and verify the relationship between the zeroes and the coefficients.

  • $x^2 + 7x + 10 = (x + 2)(x + 5)$

  • the value of $x^2 + 7x + 10$ is zero when $x + 2 = 0$ or $x + 5 = 0$, i.e., when $x = – 2$ or $x = –5$.

  • The zeroes of $x^2 + 7x + 10$ are – 2 and – 5.

    • Sum of zeroes = $(-2) + (-5) = -7 = -\frac{7}{1} = -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}$

    • product of zeroes = $(-2) \times (-5) = 10 = \frac{10}{1} = \frac{\text{Constant term}}{\text{Coefficient of }x^2}$

• Example 3: Find the zeroes of the polynomial $x^2 – 3$ and verify the relationship between the zeroes and the coefficients.

  • Recall the identity $a^2 – b^2 = (a – b)(a + b)$. Using it, we can write:

  • $x^2 – 3 = (x - \sqrt{3})(x + \sqrt{3})$

  • the value of $x^2 – 3$ is zero when $x = \sqrt{3}$ or $x = -\sqrt{3}$.

  • The zeroes of $x^2 – 3$ are $\sqrt{3}$ and $- \sqrt{3}$.

    • Sum of zeroes = $(\sqrt{3}) + (-\sqrt{3}) = 0 = -\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}$

    • product of zeroes = $(\sqrt{3}) \times (-\sqrt{3}) = -3 = \frac{-3}{1} = \frac{\text{Constant term}}{\text{Coefficient of }x^2}$

• Example 4: Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.

  • Let the quadratic polynomial be $ax^2 + bx + c$, and its zeroes be $\alpha$ and $\beta$.

  • $\alpha + \beta = – 3 = -\frac{b}{a}$, and $\alpha\beta = 2 = \frac{c}{a}$.

  • If $a = 1$, then $b = 3$ and $c = 2$.

  • one quadratic polynomial which fits the given conditions is $x^2 + 3x + 2$.

  • any other quadratic polynomial that fits these conditions will be of the form $k(x^2 + 3x + 2)$, where $k$ is real.

• Let us now look at cubic polynomials.

• Let us consider $p(x) = 2x^3 – 5x^2 – 14x + 8$.

  • $p(x) = 0$ for $x = 4, – 2, \frac{1}{2}$.

  • $p(x)$ can have atmost three zeroes, these are the zeores of $2x^3 – 5x^2 – 14x + 8$.

    • Sum of the zeroes = $4 + (-2) + \frac{1}{2} = \frac{5}{2} = -\frac{(-5)}{2} = -\frac{\text{Coefficient of }x^2}{\text{Coefficient of }x^3}$

    • product of the zeroes = $4 \times (-2) \times \frac{1}{2} = -4 = -\frac{8}{2} = -\frac{\text{Constant term}}{\text{Coefficient of }x^3}$

• Consider the sum of the products of the zeroes taken two at a time.

  • ${4 \times (-2) + (-2) \times \frac{1}{2} + \frac{1}{2} \times 4} = -8 - \frac{1}{2} + 2 = -\frac{14}{2} = -7 = \frac{\text{Coefficient of }x}{\text{Coefficient of }x^3}$

• In general, it can be proved that if $\alpha, \beta, \gamma$ are the zeroes of the cubic polynomial $ax^3 + bx^2 + cx + d$, then

  • $\alpha + \beta + \gamma = -\frac{b}{a}$,

  • $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$,

  • $\alpha \beta \gamma = -\frac{d}{a}$.

• Example 5: Verify that $3, –1, -\frac{1}{3}$ are the zeroes of the cubic polynomial $p(x) = 3x^3 – 5x^2 – 11x – 3$, and then verify the relationship between the zeroes and the coefficients.

  • Comparing the given polynomial with $ax^3 + bx^2 + cx + d$, we get $a = 3$, $b = – 5$, $c = –11$, $d = – 3$.

  • $p(3) = 3 \times 3^3 – (5 \times 3^2) – (11 \times 3) – 3 = 81 – 45 – 33 – 3 = 0$,

  • $p(–1) = 3 \times (–1)^3 – 5 \times (–1)^2 – 11 \times (–1) – 3 = –3 – 5 + 11 – 3 = 0$,

  • $p(-\frac{1}{3}) = 3 \times (-\frac{1}{3})^3 – 5 \times (-\frac{1}{3})^2 – 11 \times (-\frac{1}{3}) – 3 = -\frac{1}{9} – \frac{5}{9} + \frac{11}{3} – 3 = 0$

  • Therefore, $3, –1$ and $\frac{1}{3}$ are the zeroes of $3x^3 – 5x^2 – 11x – 3$.

  • $\alpha = 3, \beta = –1$ and $\gamma = -\frac{1}{3}$.

    • $\alpha + \beta + \gamma = 3 + (-1) + (-\frac{1}{3}) = \frac{5}{3} = -\frac{(-5)}{3} = -\frac{b}{a}$ ,

    • $\alpha\beta + \beta\gamma + \gamma\alpha = 3 \times (-1) + (-1) \times (-\frac{1}{3}) + (-\frac{1}{3}) \times 3 = -\frac{11}{3} = \frac{c}{a}$ ,

    • $\alpha\beta\gamma = 3 \times (-1) \times (-\frac{1}{3}) = 1 = -\frac{(-3)}{3} = -\frac{d}{a}$ .