Find the zeroes of the polynomial t3 - 2t2 - 15t, and verify the relat

1.	Find the zeroes of the polynomial t3 - 2t2 - 15t, and verify the relation between the coefficients and the zeroes of the polynomial.
Answer» t³ – 2t² – 15t Taking t common, we get, t ( t² -2t -15) Splitting the middle term of the equation t² -2t -15, we get, t( t² -5t + 3t -15) Taking the common factors out, we get, t (t (t-5) +3(t-5) On grouping, we get, t (t+3)(t-5) So, the zeroes are, t=0 t+3=0 ⇒ t= -3 t -5=0 ⇒ t=5 Therefore, zeroes are 0, 5 and -3 Verification: Sum of the zeroes = – (coefficient of x²) ÷ coefficient of x³ α + β + γ = – b/a (0) + (- 3) + (5) = – (- 2)/1 = 2 = 2 Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³ αβ + βγ + αγ = c/a (0)(- 3) + (- 3) (5) + (0) (5) = – 15/1 = – 15 = – 15 Product of all the zeroes = – (constant term) ÷ coefficient of x³ αβγ = – d/a (0)(- 3)(5) = 0 0 = 0

Find the zeroes of the polynomial t3 - 2t2 - 15t, and verify the relation between the coefficients and the zeroes of the polynomial.

Answer»

t³ – 2t² – 15t

Taking t common, we get,

t ( t² -2t -15)

Splitting the middle term of the equation t² -2t -15, we get,

t( t² -5t + 3t -15)

Taking the common factors out, we get,

t (t (t-5) +3(t-5)

On grouping, we get,

t (t+3)(t-5)

So, the zeroes are,

t=0

t+3=0 ⇒ t= -3

t -5=0 ⇒ t=5

Therefore, zeroes are 0, 5 and -3

Verification:

Sum of the zeroes = – (coefficient of x²) ÷ coefficient of x³

α + β + γ = – b/a

(0) + (- 3) + (5) = – (- 2)/1

= 2 = 2

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a

(0)(- 3) + (- 3) (5) + (0) (5) = – 15/1

= – 15 = – 15

Product of all the zeroes = – (constant term) ÷ coefficient of x³

αβγ = – d/a

(0)(- 3)(5) = 0

0 = 0