Question

find the zeroes of the polynomial
$3 {x}^{2} - 4x - 4$
​

Answer 1

Answer:

$\large{\underline{\boxed{\sf x = 2 \: \: or \: \: x = -\dfrac{2}{3}}}}$

Step-by-step explanation:

Given that -

• 3x² - 4x - 4

We need to split its middle term such that product of the two numbers be {3 * (-4} = (-12) and sum of numbers be (-4).

Such two numbers can be (-6) and (2).

3x² - 4x - 4

On splitting it's middle term,

⇒ 3x² - 6x + 2x - 4

⇒ 3x (x - 2) + 2 (x - 2)

⇒ (x - 2)(3x + 2)

By zero product rule -

∴ x = 2 or x = -2/3

Hence, the two zeroes of the polynomial are 2 and (-2/3).

Answer 2

Answer:

${\huge \boxed {x =2 \: or \: \frac{ - 2}{3} }}$

Step-by-step explanation:

$\mathbf{3x {}^{2} - 4x - 4 = 0} \\ \\ \mathbf{\implies \: 3x {}^{2} - 4x - 4 = 0} \\ \\ \mathbf{ \implies \: 3x {}^{2} - 6x + 2x - 4 = 0} \\ \\ \mathbf{ \implies \: 3x(x - 2) + 2(x - 2) = 0} \\ \\ \mathbf{ \implies \: (3x + 2)(x - 2) = 0} \\ \\ \mathbf{\implies \:3x + 2 = 0 \: \: or \: \: x - 2 = 0} \\ \\ \mathbf{ \implies \: x = 2 \: \: or \: \: x = \frac{-2}{3} }$

Zeros of given equation are 2 and -2/3