Question

Find the zeroes of the polynomial p(x) = 3x² − 4x − 4. Hence, write a polynomial whose each of the zeroes is 2 more than the zeroes of p(x).

Answer 1

We are given the polynomial:

p(x) = 3x² − 4x − 4

Using the quadratic formula:

x = `(-b ± sqrt(b^2 - 4ac))/(2a)`

where a = 3, b = −4, c = −4.

b² − 4ac = (−4)² − 4(3)(−4)

=16 + 48

= 64

x = `(4 ± sqrt64)/(2(3))`

x = `(4 ± 8)/6`

Solving for x:

x₁ = `(4 + 8)/6`

= `12/6`

= 2

x₂ = `(4 - 8)/6`

= `-4/6`

= `-2/3`

So, the zeroes are 2 and `-2/3`.

We need to form a new polynomial whose zeroes are 2 more than the original zeroes.

α' = 2 + 2 = 4,

β' = `-2/3 + 2 = 4/3`

The new polynomial is formed as:

(x - α')(x - β') = `(x - 4)(x - 4/3)`

Multiplying:

`(x - 4)(x - 4/3) = x^2 - 4/3x - 4x + 16/3`

= `x^2 - 12/3x - 4/3x - 16/3`

= `x^2 - 16/3x + 16/3`

Multiplying by 3 to remove fractions: 

3x² − 16x + 16