Question

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

5t² + 12t + 7

Answer 1

5t² + 12t + 7

Splitting the middle term, we get,

5t² + 5t + 7t + 7

Taking the common factors out, we get,

5t(t + 1) + 7(t + 1)

On grouping, we get,

(t + 1)(5t + 7)

So, the zeroes are,

t + 1 = 0

`\implies` y = –1

5t + 7 = 0

`\implies` 5t = –7

`\implies` t = `-7/5`

Therefore, zeroes are `(-7/5)` and –1

Verification:

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

α + β = `- b/a`

`(-1) + (-7/5) = - (12)/5`

= `-12/5 = -12/5`

Product of the zeroes = constant term ÷ coefficient of x²

αβ = `c/a`

`(-1)(-7/5) = 7/5`

`7/5 = 7/5`