Polynomials-Notes

If \(p(x)\) is a polynomial in \(x\), the highest power of \(x\) in \(p(x)\) is called the degree of the polynomial \(p(x)\).
A polynomial of degree 1 is called a Linear polynomial.
For example, \[2x – 3 \] A polynomial of degree 2 is called a Quadratic polynomial.
The name ‘quadratic’ has been derived from the word quadrate which means square $$\begin{aligned}2x^{2}+3x-\frac{2}{5},\\ y^{2}-2,\\ 2-x^{2}+\sqrt{3}x\end{aligned}$$ are some examples of quadratic polynomials (whose coefficients are real number)
More generally, any quadratic polynomial in \(x\) is of the form \(ax^2 + bx + c\), where \(a,\ b,\ c\) are real numbers and \(a\ne 0\).

A polynomial of degree 3 is called a Cubic polynomial. Some examples of a cubic polynomial are $$\begin{aligned}2-x^{3},\\ \sqrt{2}x^{3},\\ 3-x^{2}+x^{3}\end{aligned}$$ In fact, the most general form of a cubic polynomial is \(ax^3 + bx^2 + cx + d\)

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

\(k\) is a zero of \(p(x) = ax + b\), then \(p(k) = ak + b = 0\), i.e., \(k =-b/a\)

So, the zero of the linear polynomial \(ax + b\) is \(-b/a=\scriptsize\dfrac{-\text{Constant term}} {\text{Coefficient of x}}\)

The geometric meaning of a quadratic equation \(ax^2 + bx + c = 0 \) is best understood in terms of the graph of the quadratic function \( y = ax^2 + bx + c \), which is a parabola.

• The coefficients \(a, b,\text{ and }c\) determine the shape and position of this parabola.
• The roots (or zeros) of the quadratic equation correspond to the \(x\)-coordinates where the parabola intersects the \(x\)-axis. These are the solutions \(x_1\) and \(x_2\) of the equation.
• The value \(c\) represents the \(y\)-intercept of the parabola, i.e., the point where it crosses the \(y\)-axis (at \(x=0\)).
• The parabola's direction of opening (upward if \(a>0\), downward if \(a\lt 0\)) and its symmetry axis depend on \(a\) and \(b\).
• The discriminant \(b^2 - 4ac\) determines the nature of the roots and how the parabola intersects the \(x\)-axis (two points, one point, or none). Therefore, geometrically, the quadratic equation translates to finding the points where the parabola crosses the \(x\)-axis, and \(c\) specifically is the height (ordinate) of the parabola at \(x=0\).

The relationship between zeroes and coefficients of a polynomial is a fundamental concept in algebra that connects the roots (zeroes) of a polynomial to its coefficients.
This relationship varies depending on the degree of the polynomial:

Linear Polynomial

For a linear polynomial of the form \(ax + b\)


• The zero of the polynomial is \(-\frac{b}{a}\), i.e., the negative of the constant term divided by the coefficient of \(x\).

For a quadratic polynomial of the form \( ax^2 + bx + c \)


1. Let the zeroes be \(\alpha\) and \(\beta\).
2. Sum of zeroes:
   
   \(\alpha + \beta = -\frac{b}{a}\\\\\)(negative coefficient of \(x\) divided by coefficient of \(x^2\))
3. Product of zeroes:
   
   \(\alpha \beta = \frac{c}{a}\\\\\)(constant term divided by coefficient of \( x ^2\))

For a cubic polynomial of the form \(ax^3 + bx^2 + cx + d \)


• Let the zeroes be \(\alpha\), \(\beta\), and \(\gamma\).
• Sum of zeroes:
  
  \(\alpha + \beta + \gamma = -\frac {b}{a}\)
• Sum of the product of zeroes taken two at a time:
  
  \(\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}\)
• Product of zeroes:
  
  \(\alpha \beta \gamma = -\frac {d}{a}\)

General Understanding

• The number of zeroes of a polynomial equals its degree.
• These relationships allow determining the sum and product of the roots directly from the coefficients without solving the polynomial.
• This is particularly useful for verifying solutions and reconstructing polynomials from given roots.
• If further explanation or examples are needed, they can be provided.

Find the zeroes of the quadratic polynomial \(x^2 + 7x + 10\), and verify the relationship between the zeroes and the coefficients.

Solution:
\( x^2 + 7x + 10 \\\) Splitting Mid-term Method:
We will find coprime factors of 10 such that the sum is 7 and the product is 10
\(10=5\times 2\)
\[ \scriptsize\begin{aligned} x^2 + 7x + 10 &= x^2 +2x+5x+10\\ &=x(x+2)+5(x+2)\\ &=(x+2)(x+5)\\ \Rightarrow x&=-2\\\text{ and }x&=-5 \end{aligned} \\\text{are roots of equation}\] Sum of roots: \[ \scriptsize\begin{aligned} (-2) + (-5)&=-\frac{b}{a}\\\\&= -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}\\\\ &=-\frac{7}{1}\\\\&=-7 \end{aligned} \] Product of roots: \[ \scriptsize\begin{aligned} (-5)\times (-2)&=\frac{c}{a}\\\\&=\frac{\text{constant term}}{\text{coefficient of }x^2}\\\\ &=\frac{10}{1}\\\\&=10 \end{aligned} \]

Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.

Solution:

Let \( p(x) = x^{2} - 3 \)

Formula:
\[ x^{2} - a^{2} = (x + a)(x - a) \] \[ x^{2} - (\sqrt{3})^{2} = (x + \sqrt{3})(x - \sqrt{3}) \]

Hence roots of polynomial \(p(x)\) are:
\[ \begin{aligned} x &= \sqrt{3} \\ x &= -\sqrt{3} \end{aligned} \]

Sum of roots:
\[ \begin{aligned} \sqrt{3} + (-\sqrt{3}) &= \left( \frac{-b}{a}\right) \\ 0 &= \left( \frac{-b}{a}\right) \\ 0 &= \frac{0}{1} \\ 0 = 0 \end{aligned} \]

Product of roots:
\[ \begin{aligned} (\sqrt{3})(-\sqrt{3}) &= \left( \frac{c}{a}\right) \\ -3 &= \left( \frac{-3}{1}\right) \\ -3 = -3 \end{aligned} \]

Hence sum and product of roots verified

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

Solution:
Let \(p\left( x\right) =3x^{3}-5x^{2}-11x-3\)
Verify that \(3,\ -1,\ -\frac{1}{3}\) are zeros of \(p(x)\) $$\scriptsize\begin{aligned}p\left( x\right) &=3x^{3}-5x^{2}-11x-3\\ p\left( 3\right) &=3-3^{3}-5\cdot 3^{2}-11\left( 3\right) -3\\ &=3\cdot 27-5\cdot 9-33-3\\ &=81-45-33-3\\ &=81-81\\ &=0\end{aligned}$$ 3 is zero of polynomial \(p (x)\) $$\scriptsize\begin{aligned}p\left( x\right) &=3x^{3}-5x^{2}-11x-3\\ p\left( -1\right) &=3\left( -1\right) ^{3}-5\left( -1\right) ^{2}-11\left( -1\right) -3\\ &=3\left( -1\right) -5\left( 1\right) +11-3\\ &=-3-5+11-3\\ &=-11+11\\ &=0\end{aligned}$$ - 1 is root of polynomials \(p(x)\) $$\tiny \begin{aligned}p\left( x\right)&=3x^{3}-5x^{2}-11x-3\\\\ p\left( \frac{-1}{3}\right) & =3\left( \frac{-1}{3}\right) ^{3}-5\left( \frac{-1}{3}\right) ^{2}-11\times \left( \frac{-1}{3}\right) -3\\\\ &=3\left( -\dfrac{1}{27}\right) -5\left( \dfrac{1}{9}\right) +\dfrac{11}{3}-3\\\\ &=-3\times \dfrac{1}{27}-\dfrac{5}{9}-\dfrac{11}{3}-3\\\\ &=-\dfrac{1}{9}-\dfrac{5}{9}+\dfrac{11}{3}-3\\\\ &=\dfrac{-1-5+33-27}{9}\\\\ &=\dfrac{-33+33}{9}\\\\ &=\dfrac{0}{9}\\\\ &=0\end{aligned}$$ \(-\dfrac{1}{3}\) is root of the polynomial \(p(x)\)

Sum of roots = $$\scriptsize\begin{aligned}\alpha +\beta +r=3-1-\dfrac{1}{3}\\ \Rightarrow 2-\dfrac{1}{3}&=\dfrac{-b}{a}\\ \Rightarrow \dfrac{6-1}{3}&=-\left( \dfrac{-5}{3}\right) \\ & =\dfrac{5}{3}=\dfrac{5}{3}\end{aligned}$$ Product of roots $$\scriptsize\begin{aligned}\Rightarrow 3\cdot \left( -1\right) \left( \frac{-1}{3}\right) &=\dfrac{-d}{a}\\ \Rightarrow -3x-\dfrac{1}{3}&=-\dfrac{\left( -3\right) }{3}\\ \Rightarrow 1&=1\end{aligned}$$ sum of products of roots $$\tiny\begin{aligned}\alpha \beta +\beta r+r\alpha &=\dfrac{c}{a}\\\\ \Rightarrow 3\cdot \left( -1\right) +\left( -1\times \left( -\dfrac{1}{3}\right) \right) +\left( \frac{-1}{3}\times 3\right) &=\dfrac{-11}{3}\\\\ \Rightarrow -3+\dfrac{1}{3}-1&=\dfrac{-11}{3}\\\\ \Rightarrow -\dfrac{9+1-3}{3}&=\dfrac{-11}{3}\\\\ \Rightarrow \dfrac{-11}{3}&=\dfrac{-11}{3}\end{aligned}$$ Hence, the relationship between the zeroes and the coefficients are verified

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

Solution:
Let the quadratic polynomial be \(ax^2+bx+c\) and its zeroes be-3 and 2.\[\] Sum of roots = -3 $$\begin{aligned}\alpha +\beta &=-3\\ \Rightarrow -3&=\dfrac{-b}{a}\\ \Rightarrow 3a&=b\end{aligned}$$ Product of roots = 2 $$\begin{aligned}\alpha \beta &=2\\ \Rightarrow \dfrac{c}{a}&=2\\ \Rightarrow c&=2a\end{aligned}$$ if \(a=1\) then \(b=3,\ c=2\)\[\] Quadratic Polynomial will be $$\begin{aligned}ax^{2}+bx+c\\ \Rightarrow x^{2}+3x+2\end{aligned}$$

1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials respectively.
2. A quadratic polynomial in \(x\) with real coefficients is of the form \(ax^2 + bx + c\), where \(a,\ b,\ c\) are real numbers with \(a \ne 0\).
3. The zeroes of a polynomial \(p(x)\) are precisely the x-coordinates of the points, where the graph of \(y = p(x)\) intersects the x-axis.
4. A quadratic polynomial can have at most 2 zeroes, and a cubic polynomial can have at most 3 zeroes
5. If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(ax^2 + bx + c\), then \[\begin{aligned} \alpha + \beta &=-\dfrac{b}{a},\\\alpha \beta&=\dfrac{c}{a} \end{aligned}\]
6. If \(\alpha,\ \beta,\ \gamma\) are the zeroes of the cubic polynomial \(ax^3 + bx^2 + cx + d\), then \[\begin{aligned} \alpha + \beta + \gamma &= -\dfrac{b}{a}\\ \alpha \beta + \beta \gamma + \gamma + \alpha &= \dfrac{c}{a}\\ \alpha\beta\gamma&=-\frac{d}{a} \end{aligned}\]