Find the zeroes of the quadratic polynomial. Find the sum and product

1.	Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 + 2x + 1
Answer» Given polynomial p(x) = x² + 2x + 1 We have x² + 2x + 1 = x² + x + x + 1 = x(x + 1) + l(x + 1) = (x + 1) (x + 1) So, the value of x² + 2x + 1 is zero when x + 1 = 0 (or) x + 1 = 0, i.e., when x = – 1 or – 1 So, the zeroes of x² + 2x + 1 are – 1 and – 1. ∴ Sum of the zeroes = (-1) + (-1) = -2 \(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-2}{1}={-2}\) And product of the zeroes = (-1) × (-1) = 1 \(=\frac{Constant}{Coefficient\,of\,x^2}=\frac{1}{1}=1\)

Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 + 2x + 1

Answer»

Given polynomial p(x) = x² + 2x + 1

We have x² + 2x + 1 = x² + x + x + 1

= x(x + 1) + l(x + 1)

= (x + 1) (x + 1)

So, the value of x² + 2x + 1 is zero

when x + 1 = 0 (or) x + 1 = 0, i.e.,

when x = – 1 or – 1

So, the zeroes of x² + 2x + 1 are – 1 and – 1.

∴ Sum of the zeroes = (-1) + (-1) = -2

\(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-2}{1}={-2}\)

And product of the zeroes = (-1) × (-1) = 1

\(=\frac{Constant}{Coefficient\,of\,x^2}=\frac{1}{1}=1\)