POLYNOMIALS
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.
General form, $ax+b$
A polynomial of degree 2 is called a quadratic polynomial.
General form, $ax^2+bx+c$
A polynomial of degree 3 is called a cubic polynomial.
General form, $ax^3+bx^2+cx+d$
Value of a polynomial
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 $x=k$.Zero of a polynomial
A real number $k$ is said to be a zero of a polynomial $p(x)$, if $p(k)=0$.Zero of a linear polynomial $ax+b$ is, $\frac{-b}{a}=\frac{-(constant\: term)}{co-efficient \: of \: x}$
A quadratic polynomial has atmost two zeroes.
That is, it can have either two distinct zeroes, or two equal zeroes (i.e., one zero) or no zero.
A cubic polynomial has atmost 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 atmost $n$ zeroes.
In a quadratic polynomial $ax^2+bx+c$,
the parabola is open upwards, when $a>0$, i.e, when a is positive.
and the parabola is open downwards, when $a<0$, i.e., when a is negative.
Geometrically, the number points at which the graph of the given polynomial intersects the x-axis, is the number of zeroes of the given polynomial.
Number of zeroes = 1, as it intersects x-axis at one point.
Number of zeroes = 2, as it intersects x-axis at two points.
Number of zeroes = 2, as it intersects x-axis at two points.
Number of zeroes = 2, as it intersects x-axis at two points.
Number of zeroes = 1, as it intersects x-axis at one point.
Number of zeroes = 1, as it intersects x-axis at one point.
Number of zeroes = 0, as it intersects x-axis at no points.
Number of zeroes = 0, as it intersects x-axis at no points.
Number of zeroes = 3, as it intersects x-axis at three points.
Number of zeroes = 1, as it intersects x-axis at one point.
Number of zeroes = 2, as it intersects x-axis at two points.
(i) Number of zeroes = 1, as it intersects x-axis at one point.
(ii)Number of zeroes = 2, as it intersects x-axis at two points.
(iii)Number of zeroes = 3, as it intersects x-axis at three points.
(iv)Number of zeroes = 1, as it intersects x-axis at one point.
(v)Number of zeroes = 1, as it intersects x-axis at one point.
(vi)Number of zeroes = 4, as it intersects x-axis at four points.
Exercise - 2.1
(i) Number of zeroes = 0, as it intersects x-axis at no points.
(ii)Number of zeroes = 1, as it intersects x-axis at one point.
(iii)Number of zeroes = 3, as it intersects x-axis at three points.
(iv)Number of zeroes = 2, as it intersects x-axis at two points.
(v)Number of zeroes = 4, as it intersects x-axis at four points.
(vi)Number of zeroes = 3, as it intersects x-axis at three points.
Relationship between Zeroes and Co-efficients of a Polynomial
Linear polynomial $ax+b$
Zero=$\frac{-b}{a}=\frac{-(constant\:term)}{co-efficient\:of\:x}$Quadratic polynomial $ax^2+bx+c$
Sum of Zeroes, $\alpha + \beta =\frac{-b}{a}=\frac{-(co-efficient\:of\:x)}{co-efficient\:of\:x^2}$Product of Zeroes, $\alpha\beta = \frac{c}{a}=\frac{constant\:term}{co-efficient\:of\:x^2}$
Cubic polynomial $ax^3+bx^2+cx+d$
$\alpha + \beta +\gamma=\frac{-b}{a}=\frac{-(co-efficient\:of\:x^2)}{co-efficient\:of\:x^3}$$\alpha\beta+\beta\gamma +\gamma\alpha = \frac{c}{a}=\frac{co-efficient\:of\:x}{co-efficient\:of\:x^3}$
$\alpha \beta \gamma =\frac{-d}{a}=\frac{-(constant\:term)}{co-efficient\:of\:x^3}$
Example 2.
Find the zeroes of the quadratic polynomial $x^2+7x+10$, and verify the relationship between the zeroes and the coefficients.Solution:
$x^2+7x+10=(x+2)(x+5)$
When $x+2=0$, $x=-2$.
When $x+5=0$, $x=-5$.
The zeroes are -2 and -5.
Let $\alpha=-2$ and $\beta=-5$.
Verification:
Sum of Zeroes, $\alpha + \beta =-2+(-5)=-2-5=-7=\frac{-7}{1}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(-2)(-5)=10=\frac{10}{1}=\frac{c}{a}$
Hence verified.
Example 3.
Find the zeroes of the polynomial $x^2–3$ and verify the relationship between the zeroes and the coefficients.Solution :
$x^2–3=x^2–(\sqrt{3})^2=(x+\sqrt{3})(x-\sqrt{3})$
When $x+\sqrt{3}$, $x=-\sqrt{3}$.
When $x-\sqrt{3}$, $x=\sqrt{3}$.
The zeroes are $-\sqrt{3}$ and $\sqrt{3}$.
Let $\alpha=-\sqrt{3}$ and $\beta=\sqrt{3}$.
Verification:
Sum of Zeroes, $\alpha + \beta =-\sqrt{3}+\sqrt{3}=0=\frac{0}{1}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(-\sqrt{3} \times \sqrt{3})=-3=\frac{-3}{1}=\frac{c}{a}$
Hence verified.
Example 4.
Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.Solution :
Sum of Zeroes, $\alpha + \beta =-3$
Product of Zeroes, $\alpha\beta =2$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(-3)x+2=x^2+3x+2$.
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.Solution :
$p(x)=3x^3–5x^2–11x–3$
$p(3)=3(3)^3–5(3)^2–11(3)–3$
$p(3)=3(27)-5(9)-33-3$
$p(3)=81-45-33-3$
$p(3)=81-81$
$p(3)=0$
$p(-1)=3(-1)^3–5(-1)^2–11(-1)–3$
$p(-1)=3(-1)-5(1)+11-3$
$p(-1)=-3-5+11-3$
$p(-1)=-11+11$
$p(-1)=0$
$p(-\frac{1}{3})=3(-\frac{1}{3})^3–5(-\frac{1}{3})^2–11(-\frac{1}{3})–3$
$p(-\frac{1}{3})=3(-\frac{1}{27})-5(\frac{1}{9})-11(-\frac{1}{3})-3$
$p(-\frac{1}{3})=-\frac{1}{9}-\frac{5}{9}+\frac{11}{3}-3$
$p(-\frac{1}{3})=-\frac{1}{9}-\frac{5}{9}+\frac{33}{9}-\frac{27}{9}$
$p(-\frac{1}{3})=\frac{-1-5+33-27}{9}$
$p(-\frac{1}{3})=\frac{33-33}{9}$
$p(-\frac{1}{3})=\frac{0}{9}$
$p(-\frac{1}{3})=0$
Therefore, $3, –1, -\frac{1}{3}$ are the zeroes of the cubic polynomial $p(x)=3x^3–5x^2–11x–3$.
Let $\alpha=3$, $\beta=-1$ and $\gamma=-\frac{1}{3}$.
$\alpha + \beta +\gamma=3+(-1)+(-\frac{1}{3})=3-1-\frac{1}{3}=2-\frac{1}{3}$
$=\frac{6-1}{3}=\frac{5}{3}=\frac{-(-5)}{3}=\frac{-b}{a}$
$\alpha\beta+\beta\gamma +\gamma\alpha = (3)(-1)+(-1)(-\frac{1}{3})+(-\frac{1}{3})(3)$
$=-3+\frac{1}{3}-1=\frac{1}{3}-4=\frac{1-12}{3}=\frac{-11}{3}=\frac{c}{a}$
$\alpha \beta \gamma =(3)(-1)(-\frac{1}{3})=1=\frac{-(-3)}{3}=\frac{-d}{a}$
Hence verified.
Exercise 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.(i) $x^2–2x–8$
Solution :
$x^2–2x–8=(x-4)(x+2)$
When $x-4=0$, $x=4$.
When $x+2=0$, $x=-2$.
The zeroes are 4 and -2.
Let $\alpha=4$ and $\beta=-2$.
Verification:
Sum of Zeroes, $\alpha + \beta =4+(-2)=4-2=2=\frac{-(-2)}{1}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(4)(-2)=-8=\frac{-8}{1}=\frac{c}{a}$
Hence verified.
(ii) $4s^2–4s+1$
Solution :
$4s^2–4s+1=(2s-1)(2s-1)$
When $2s-1=0$, $s=\frac{1}{2}$.
When $2s-1=0$, $s=\frac{1}{2}$.
The zeroes are $\frac{1}{2}$ and $\frac{1}{2}$.
Let $\alpha=\frac{1}{2}$ and $\beta=\frac{1}{2}$.
Verification:
Sum of Zeroes, $\alpha + \beta =\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1=\frac{-(-4)}{4}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}=\frac{c}{a}$
Hence verified.
(iii) $6x^2–3–7x$
Solution :
$6x^2–3–7x=6x^2–7x-3$
$6x^2–7x-3=(3x+1)(2x-3)$
When $3x+1=0$, $x=-\frac{1}{3}$.
When $2x-3=0$, $x=\frac{3}{2}$.
The zeroes are $-\frac{1}{3}$ and $\frac{3}{2}$.
Let $\alpha=-\frac{1}{3}$ and $\beta=\frac{3}{2}$.
Verification:
Sum of Zeroes, $\alpha + \beta =-\frac{1}{3}+\frac{3}{2}=\frac{-2+9}{6}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =-\frac{1}{3} \times \frac{3}{2}=\frac{-1}{2}=\frac{-3}{6}=\frac{c}{a}$
Hence verified.
(iv) $4u^2+8u$
Solution :
$4u^2+8u=4u(u+2)$
$u=0$.
When $u+2=0$, $u=-2$.
The zeroes are 0 and -2.
Let $\alpha=0$ and $\beta=-2$.
Verification:
Sum of Zeroes, $\alpha + \beta =0+(-2)=0-2=-2=\frac{-(8)}{4}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(0)(-2)=0=\frac{0}{4}=\frac{c}{a}$
Hence verified.
(v) $t^2–15$
Solution :
$t^2–15=t^2–(\sqrt{15})^2=(t+\sqrt{15})(t-\sqrt{15})$
When $t+\sqrt{15}$, $t=-\sqrt{15}$.
When $t-\sqrt{15}$, $t=\sqrt{15}$.
The zeroes are $-\sqrt{15}$ and $\sqrt{15}$.
Let $\alpha=-\sqrt{15}$ and $\beta=\sqrt{15}$.
Verification:
Sum of Zeroes, $\alpha + \beta =-\sqrt{15}+\sqrt{15}=0=\frac{0}{1}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(-\sqrt{15} \times \sqrt{15})=-15=\frac{-15}{1}=\frac{c}{a}$
Hence verified.
(vi) $3x^2–x–4$
Solution :
$3x^2–x–4=(3x-4)(x+1)$
When $3x-4=0$, $x=\frac{4}{3}$.
When $x+1=0$, $x=-1$.
The zeroes are $\frac{4}{3}$ and -1.
Let $\alpha=\frac{4}{3}$ and $\beta=-1$.
Verification:
Sum of Zeroes, $\alpha + \beta =\frac{4}{3}+(-1)=\frac{4}{3}-1=\frac{4-3}{3}=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-b}{a}$
Product of Zeroes, $\alpha\beta =(\frac{4}{3})\times (-1)=\frac{-4}{3}=\frac{c}{a}$
Hence verified.
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\frac{1}{4},\:-1$
Solution :
Sum of Zeroes, $\alpha + \beta =\frac{1}{4}$
Product of Zeroes, $\alpha\beta =-1$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(\frac{1}{4})x+(-1)=\frac{1}{4}(4x^2-x-4)$.
The required polynomial is $4x^2-x-4$.
(ii) $\sqrt{2}, \: \frac{1}{3}$
Solution :
Sum of Zeroes, $\alpha + \beta =\sqrt{2}$
Product of Zeroes, $\alpha\beta =\frac{1}{3}$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(\sqrt{2})x+\frac{1}{3}=\frac{1}{3}(3x^2-3\sqrt{2}x+1)$.
The required polynomial is $3x^2-3\sqrt{2}x+1$.
(iii) 0, $\sqrt{5}$
Solution :
Sum of Zeroes, $\alpha + \beta =0$
Product of Zeroes, $\alpha\beta =\sqrt{5}$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(0)x+\sqrt{5}=x^2+\sqrt{5}$.
(iv) 1, 1
Solution :
Sum of Zeroes, $\alpha + \beta =1$
Product of Zeroes, $\alpha\beta =1$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(1)x+1=x^2-x+1$.
(v) $-\frac{1}{4}, \:\frac{1}{4}$
Solution :
Sum of Zeroes, $\alpha + \beta =-\frac{1}{4}$
Product of Zeroes, $\alpha\beta =\frac{1}{4}$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(-\frac{1}{4})x+\frac{1}{4}=\frac{1}{4}(4x^2+x+1)$.
The required polynomial is $4x^2+x+1$.
(vi) 4, 1
Solution :
Sum of Zeroes, $\alpha + \beta =4$
Product of Zeroes, $\alpha\beta =1$
Required polynomial is of the form, $x^2-(\alpha+\beta)x+\alpha\beta$
$\implies x^2-(4)x+1=x^2-4x+1$.