Before understanding operations on polynomial we have to learn the concept of zeros of a polynomial. A polynomial f(x) = anxn + an-1xn-1 + … + a1x +a0 has as many zeros as it intersects x-axis.
Note: A polynomial has number of zeros equal to its degree.
Zero of a polynomial: A real number ? is a zero of a polynomial f(x), if (?) = 0.
Finding zeros of a polynomial f(x) means solving the polynomial equation f(x) = 0.
We will understand it by some examples:
1. f(x) = 3x+5 , It is a linear polynomial because it has its degree 1. So, it will intersects x-axis at one place.
2. f(x) = 4x2 - 3x+5 , It is a quadratic polynomial because it has its degree 2. So, it will intersects x-axis at two places.
3. f(x) = 7x3 + 4x2 - 3x+5 , It is a cubic polynomial because it has its degree 3 polynomial . So, it will intersects x-axis at three places.
4. f(x) = 9x4+ 7x3 + 4x2 - 3x+5 , It is a bi-quadratic polynomial because it has its degree 4. So, it will intersects x-axis at four places.
Value of a Polynomial
If f(x) is a polynomial and ? is any real number, then the real number obtained by replacing x by ? in f(x), is called the value of f(x) at x = ? and is denoted by f(?).
The values of the quadratic polynomial f(x) = 2x2 – 3x – 2 at x = 1 and x = - 2 are given by
1. f(1) = 2 (1)2 – 3 ( 1) – 2 = 2 – 3 – 2 = -3
2. f(-2) = 2 (-2)2 - 3 (-2) – 2 = 8 + 6 – 2 = 12
Consider the cubic polynomial f(x) = x3 – 6x2 + 11x – 6. The value of this polynomial at x = 2 is given by
f(2) = (2)3 - 6(2)2 + 11(2) – 6 = 8 – 24 + 22 – 6 = 0
Also, f(1) = (1)3 - 6(1)2 + 11(1) –6 = 1 – 6 + 11 – 6 = 0
And, f(3) = (3)3- 6(3)2 + 11(3) – 6 = 27 – 54 + 33 – 6 = 0
1, 2 and 3 are called the zeros of the cubic polynomial f(x) = x3 – 6x2 + 11x – 6.
Relationship between the Zeros and Coefficients of a Polynomial
In the previous section, we have studied that a polynomial of degree n has exactly n zeros (real or imaginary). The zeros of a polynomial are closely connected to its coefficients, Factoring Polynomials. In this section, we will find out relationship between the zeros and coefficients of a polynomial.
Consider the quadratic polynomial f(x) = 6x2 – x – 2. By the method of splitting the middle term, we have
f(x) = 6x2 – x – 2
= 6x2 – 4x + 3x – 2
= 2x (3x – 2) + 1(3x – 2)
= (3x – 2) (2x + 1)
Or,f(x) = (3x – 2) (2x + 1)
Putting f(x) = 0 ,we get
(3x – 2) (2 x + 1) = 0
3x – 2 = 0 or, 2 x + 1 = 0
X = $\frac{2}{3}$ or, x = $\frac{-1}{2}$
Hence, the zeros of 6x2 – x – 2 are ? = $\frac{2}{3}$ and ? = $\frac{-1}{2}$ .
We observe that
Sum of its zeros ? + ? = $\frac{2}{3}$ - $\frac{-1}{2}$ = $\frac{-1}{6}$ = $\frac{-(1)}{6}$ = $\frac{coefficient of x}{coefficient of x^2}$
Product of its zeros ? ? = $\frac{2}{3}$ ×$\frac{-1}{2}$ = $\frac{-1}{3}$ = $\frac{-(2)}{6}$ = $\frac{constant term}{coefficient of x^2}$
Let us now consider a cubic polynomial p(x) given by
P(x) = 6x3 + 5x2 – 12x + 4
P(x) = (x +2) (2x – 1) (3x – 2) [ By using factorization ]
P(x) = 0
(x +2) (2x – 1) (3x – 2) = 0
X = - 2, $\frac{1}{2}$, $\frac{2}{3}$
Hence, the zeros of p(x) = 6x3 + 5x2 – 12x + 4 are ?= - 2, ?= $\frac{1}{2}$ and ?= $\frac{2}{3}$
Also, p(x) is a cubic polynomial. So, p(x) can have at most three real zeros.
Now,
Sum of the zeros ? + ? +?= -2 + $\frac{1}{2}$ + $\frac{2}{3}$ = $\frac{-(5)}{6}$ = $\frac{coefficient of x^2}{coefficient of x^3 }$
Sum of the products of zeros taken two at a time
= ?? + ?? + ??
= (-2) × $\frac{1}{2}$ + $\frac{1}{2}$ × $\frac{2}{3}$ + $\frac{2}{3}$ ×(-2)
= - 1 + $\frac{1}{2}$ -- $\frac{-4}{3}$ = - 2 = $\frac{-12}{6}$ = $\frac{coefficient of x}{coefficient of x^3 }$
Product of all zeros ??y = (-2) × $\frac{1}{2}$ × $\frac{2}{3}$= $\frac{-2}{3}$ = $\frac{-4}{6}$= $\frac{constant term}{coefficient of x^3 }$
Let us now find a formal relation between zeros and coefficients of a polynomial.
Relationship between Zeros and Coefficients of a Quadratic Polynomial
Let ? and ? be the zeros of a quadratic polynomial f(x) = ax2 + bx + c. By factor theorem (x -?) and (x -?) are the factors of f(x).
F(x) = k(x -?) (x -?), where k is a constant
Ax2 + bx + c = k {x2 – ( ?+?) x +??}
Ax2 + bx + c = k x2 – k( ?+?) x + k??
Comparing the coefficients of x2, x and constant terms on both sides, we get
A = k, b = – k( ?+?) and c = k??
( ?+?) = $\frac{-b}{a}$ and ?? = $\frac{c}{a}$
( ?+?) = $\frac{coefficient of x}{coefficient of x^2}$ and ??= $\frac{constant term}{coefficient of x^2}$
Hence,
Sum of the zeros = $\frac{-b}{a}$ = $\frac{coefficient of x}{coefficient of x^2}$
Product of the zeros = $\frac{c}{a}$ = $\frac{constant term}{coefficient of x^2}$
Relationship between Zeors and Coefficient of a Cubic Polynomial
Let ?, ?, y be the zeros of a cubic polynomial f(x) = ax3 + bx2 + cx + d , a? 0.
By factor theorem, (x -?), (x -?) and (x – y) are factors of f(x). Also f(x), being a cubic polynomial, cannot have more than three linear factors.
F(x) = k (x -?)(x -?)(x – y)
Ax3 + bx2 + cx + d =k (x -?)(x -?)(x – y)
Ax3 + bx2 + cx + d = k {x3 – (?+?+y) x2 + (??+ ?y +y?) x - ??y}
Ax3 + bx2 + cx + d = kx3 – k (?+?+y)) x2 + k (??+ ?y +y?) x - k??y
Comparing the coefficients of x3, x2, x and constant terms on both sides, we get
A = k, b = – k (?+?+y)),c = k(??+ ?y +y?) and d = - k??y
?+?+y = $\frac{-b}{a}$
??+ ?y +y? = $\frac{c}{a}$
And,??y= $\frac{-d}{a}$
Sum of the zeros = $\frac{-b}{a}$ = $\frac{coefficient of x^2}{coefficient of x^3 }$
Sum the of the products of the zeros taken two at a time = $\frac{c}{a}$ = $\frac{coefficient of x}{coefficient of x^3 }$
Product of the zeros = $\frac{-d}{a}$ = $\frac{constant term}{coefficient of x^3 }$
Remark: It follows from the above discussion that a cubic polynomial having ?, ? and y as its zeros is given by
F(x) = k (x -?)(x -?)(x – y)
F(x) = k {x3 – (?+?+y) x2 + (??+ ?y +y?) x - ??y}, where k is any non-zero real number.
Operations on Polynomials - Example
Example1. Find the zeros of the quadratic polynomial x2 + 7x + 12, and verify the relation between the zeros and its coefficients.
Solution: We have,
F(x) = x2 + 7x + 12 = x2 + 4x + 3x + 12
F(x) = x(x + 4) +3(x + 4)
F(x) = (x + 4) (x+ 3)
The zeros of f(x) are given by ,F(x) = 0
X2 + 7x + 12 = 0
(x+4) (x+3) = 0
X+4 = 0 or, x+3 = 0
X = -4 or, x = -3
Thus, the zeros of f(x) = x2 + 7x + 12 are ? = -4 and ?= -3
Now,
Sum of the zeros = ? + ? = (-4) + (-3) = -7
Sum of the zeros = - $\frac{-7}{1}$ = $\frac{coefficient of x}{coefficient of x^2 }$
Product of the zeros = = (-4) (-3) = 12
Product of the zeros = $\frac{12}{1}$ = $\frac{constent term}{coefficient of x^2 }$
Example 2. Verify that 3, -1 and - $\frac{-1}{3}$ are the zeros of the cubic polynomial p(x) = 3x2 - 5x2 – 11x – 3 and then verify the relationship between the zeros and its coefficients.
Solution: We have, P(x) = 3x3 – 5x2 – 11x – 3
P (3) = 3 (3)3 -5 (3)2 -11(3) -3 = 81 – 45 – 33 – 3 = 0
P (-1) = 3(-1)3 – 5 (-1)2 - 11 (-1) -3 = -3 -5 + 11 -3 = 0
P ($\frac{-1}{3}$) = 3($\frac{-1}{3}$)3 -5 ($\frac{-1}{3}$)2 -11 ($\frac{-1}{3}$) -3 = $\frac{-1}{9}$ - $\frac{5}{9}$ + $\frac{11}{3}$ - 3 = 0
So, 3, -1 and $\frac{-1}{3}$ are the zeros of polynomial p(x).
Sum of the zeros ? + ? +?= 3 -1 + $\frac{-1}{3}$ = $\frac{-(-5)}{3}$ = -$\frac{coefficient of x^2}{coefficient of x^3 }$
Sum of the products of zeros taken two at a time
= ?? + ?? + ??
= (3) × (-1) + (-1) × $\frac{-1}{3}$ + $\frac{-1}{3}$ ×(3)
= - 3 + $\frac{1}{3}$ -1 = $\frac{-11}{3}$ = $\frac{coefficient of x}{coefficient of x^3 }$
Product of all zeros ??y = (3) × (-1) × $\frac{-1}{3}$= 1 = $\frac{-(-3)}{3}$= $\frac{constant term}{coefficient of x^3 }$
ILLUSTRATION 1 Divide the polynomial f(x) = 6x3 + 11x2 – 39x - by the polynomial g(x) = x2 – 1 +x. Also, find the quotient and remember.
Solution: We have,
Clearly, quotient q(x) = 6x + 5 and remember r(x) = - 38x – 60.
Also, 6x3 + 11x2 – 39x – 65 = (x2 + x – 1) (6x + 5) + (- 38x – 60)
F(x) = g(x) q(x) + r(x)
Dividend = Quotient x Divisor + Remainder