Question

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

t³ – 2t² – 15t

Answer 1

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

`\implies` t = – 3

t – 5 = 0

`\implies` 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 

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

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

αβγ = `- d/a`

(0)(– 3)(5) = 0

= 0