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Relationship Between Zeroes and Coefficients of a Polynomial

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p(x)= value of x at which p(x)=0 is called zero of p(x) ,which geometrically means x cordinate of point where graph of y= p(x) cuts x axis. Here in this concept we are going to study how by using coefficeint we can tell the zeroes of polynnomial p(x).
1) Linear Polynomial- p(x)= ax+b, a is not equal to 0 
                                            Let ax+b= 0
                                            ax=b
Hence x= `-b/a.` By substituing this in p(x) we get `p(-b/a)= a(-b/a)+b`
After observing carefully we can conclude that 
zero of linear polynomial= -constant/ coefficient      Example- Find the zero of `p(x)= 13x+9,` zero of `p(x)= -9/13`


Let's cross check is method, `p(-9/13)= 13(-9/13)+9= -9+9= 0` 


Thus in case of Linear Polynomial `p(x)=ax+b`, where a is not equal to 0 


Zero of p(x)= `-"constant"/ "coefficient"` also zero of `p(x)= "-b"/a`


2) Quadratic Polynomial- `p(x)= ax^2+bx+c`, here in quadratic polynomial the maximum numbers of zeros are two. So let's assume alpha and beta are two zeros of

`p(x)= ax^2+bx+c,`Then the `"Sum"  "of"  "zeros"= "-coefficient of x"/ "coefficient of x"^2`  i.e `alpha+ beta= "-b"/a`   

 

And `"Product"  "of"  "zeros"= "constant term"/ "coefficient of x"^2` i.e `alpha×beta= c/a`


Example 1- Find the zeros of the polynomial `p(x)= x^2+12x+35` and verfiy the relationship between the zeros and the coefficient.
Using Factorisation method to find factors of the given equation we get (x+7) and (x+5) as the factors of polynomial `x^2+12x+35`. If (x+7)= 0 then x= -7 also if (x+5)= 0 then x= -5, here if -7 and-5 are the zeros of the polynomial then let alpha be -7 and beta be -5 


As we know `alpha×beta= "-coefficient of x"/"coefficient of x"^2` = `x^2=-12/1= -12`


and `alpha×beta= "constant term"/" coefficient of x"^2= 35/1= 35`

 
As assumed above alpha is -7 and beta is -5, thus `alpha+beta= -7+(-5)= -7-5= -12`
and `alpha xx beta= -7 xx -5= 35`  Hence verified. 
Example 2- Find a quadratic polynomial, the sum and product of which zeros are -3 and 2 respectively.
General quadratic polynomial is `p(x)= ax^2+bx+c`, where a is not equal to 0 and alpha and beta are the zeros of this polynomial.
As per the given condition, `alpha+beta= -3 and alpha×beta= 2`
`alpha+beta= "-b"/a`


     `-3= "-b"/a`


         `b=3a`


 `alpha×beta= c/a`


         `2= c/a`


         `c= 2a`
by substitution we get, `p(x)= ax^2+bx+c= ax^2+ 3ax+ 2a`
taking a common `p(x)= a(x^2+3x+2)`                            Here, because a is not equal to zero, for different value of a we will have different value of polynomial. If take a=1 then `p(x)= x^2+3x+2`
Let's say for a=3, `p(x)= 3(x^2+3x+2)= 3x^2+9x+6`


`alpha+beta= "-b"/a= "-9"/3= -3`


`alpha×beta= c/a= 6/3= 2`     

Hence verified with the given condition above.
So if we take a as any value we get `alpha+beta` and `alpha×beta` verified.
This example could also be solved using a formula i.e 
p(x)= `x^2`- (sum of roots)x+ product of the roots
`p(x)= x^2+3x+2`
So we can say that if the sum and product of roots are given then by using p(x)= `x^2`- (sum of roots)x+ product of the roots, we could directly find the quadratic polynomial.
3) Cubic Polynomial- `p(x)= ax^3+bx^2+cx+d,` here in cubic  polynomial the maximum numbers of zeros are three. So let's assume `alpha, beta and gamma` are the three zeros of `p(x)= ax^3+bx^2+cx+d`, 
then `alpha+beta+gamma= "-b"/a,`            `alpha×beta+ beta×gamma+ gamma×alpha= c/a`,                                         and `alpha×beta×gamma= "-d"/a`
Let us try to understand how is it derived
In generic terms, `p(x)= ax^3+bx^2+cx+d`, where a is not equal to zero, and alpha, beta and gamma are the zero of the polynomial. If alpha, beta and gamma are the zeros of the polynomial then the roots of the polynomial will be `(x-alpha) (x-beta) and (x-gamma).`
Thus, `p(x)= k (x-alpha) (x-beta) (x-gamma)`, where k is any constant
`ax^3+bx^2+cx+d = k (x-alpha) (x-beta) (x-gamma)`
`ax^3+bx^2+cx+d =k [x^3- (alpha+beta+gamma)x^2+ (alpha×beta+ beta×gamma+gamma×alpha)x- alpha×beta×gamma]`
comparing the coefficient of LHS and RHS we get,
a=k
b= -k(`alpha+beta+gamma`) i.e sum of the roots 

`alpha+beta+gamma = -b/k`

`alpha+beta+gamma`= `"-b"/a`


c= k(`alpha×beta+ beta×gamma+ gamma×alpha`) i.e product of zeros taken two at a time= `alpha×beta+ beta×gamma+ gamma×alpha`

= `c/k`

`alpha×beta+ beta×gamma+ gamma×alpha= c/a`


d= -k(`alpha×beta×gamma`) i.e product of zeros= `alpha×beta×gamma`= `-d/k`

`alpha×beta×gamma`= `"-d"/a`


This is the relationship betweem coefficient and zeros of cubic polynomials.
Example- `2x^3-5x^2-14x+8=0`


                    `alpha+beta+gamma= "-b"/a= "-(-5)"/2= 5/2`


                    `alpha×beta+ beta×gamma+ gamma×alpha= c/a= "-14"/2= -7`


                   `alpha×beta×gamma= "-d"/a= "-8"/2= -4`

                    

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