Question

Find the zeroes of the polynomial p(x) = 9x² – 6x – 35 and verify the relationship between zeroes and its coefficients.

Answer 1

p(x) = 9x² – 6x – 35

Zero of the polynomial is the value of x where p(x) = 0.

Putting p(x) = 0

9x² – 6x – 35 = 0

Splitting the middle term method,

⇒ 9x² – 21x + 15x – 35 = 0

⇒ 3x(3x – 7) + 5(3x – 7) = 0

⇒ (3x – 7) (3x + 5) = 0

⇒ (3x – 7) = 0 or (3x + 5) = 0

⇒ 3x = 7

x = `7/3`

⇒ 3x = –5

x = `(-5)/3`

Verification:

In p(x) = 9x² – 6x – 35

a = 9, b = (– 6), c = –35

Zeroes of the polynomial are `7/3` and `(-5)/3`.

Sum of zeroes = `(-("Coefficient of"  x))/("Coefficient of"  x^2) = (-b)/a`

⇒ `7/3 + ((-5)/3) = (-(-6))/9`

⇒ `(7 - 5)/3 = 6/9`

⇒ `2/3 = 2/3`

Product of zeroes = `"Constant term"/("Coefficient of"  x^2) = c/a`

⇒ `7/3 xx ((-5)/3) = (-35)/9`

⇒ `(-35)/9 = (-35)/9`