Relationship of Zeros and Coefficients of Polynomials

1) Verify its relationship between coefficient of polynomial and its zeros for a polynomial x² – 9.

Solution
Here p(x) = x² – 9
compare it with ax² + bx + c
p(x) = x² + 0x – 9
Here a = 1, b = 0, c = -9
$\mathrm{Sum\; of\; zeros}=\frac{–b}{a}$
-3 + 3 = 0
$=–\frac{0}{1}$
$=–\frac{b}{a}$
$\mathrm{Product\; of\; zeros}=\frac{c}{a}$
= (-3)(3) = -9
$=\frac{-9}{1}$
$=\frac{c}{a}$

2) Form a quadratic polynomial whose sum of zeros is 5 and product of zeros is 6.

Solution
Sum of zeros = 5
Product of zeros = 6
As we know, quadratic polynomial is the form of x² – (sum of zeros)x + product of zeros
By putting the above values, it becomes x² – 5x + 6
Hence, x² – 5x + 6 is a quadratic polynomial.

3) Verify relationship between coefficient of polynomial and its zeros if x³ – 9x² – 12x + 20 has zeros -2, 1 and 10.

Solution
Here, compare x³ – 9x² – 12x + 20 with ax³ + bx² + cx + d
a = 1, b = -9, c = -12, d = 20
Zeros are -2, 1 and 10 (given)
α = -2
β = 1
γ = 10
$\text{Sum of zeros}=–\frac{{\text{coefficient of x}}^2}{{\text{coefficient of x}}^3}$
∴ α + β + γ = -2 + 1 + 10
= 9
$=–\frac{\mathrm{(-9)}}{1}$
$=–\frac{b}{a}$
Sum of product of zeros taken two at a time $=\frac{\text{coefficient of x}}{{\text{coefficient of x}}^3}$
αβ + βγ + γα = (-2)(1) + (1)(10) + (10)(-2)
= -2 + 10 -20
= -12
$=\frac{\mathrm{(-12)}}{1}$
$=\frac{c}{a}$
$\text{Product of zeros}=–\frac{\text{constant of term}}{{\text{coefficient of x}}^3}$
αβγ = (-2)(1)(10)
= -20
$=–\frac{20}{1}$
$=–\frac{d}{a}$