Question

why 2/3 and -2 are zeros of polynomials of p(x)=3xsq+4x-4​

Answer 1

Step-by-step explanation:

3x^2+4x-4=0

3x^2+6x-2x-4=0

3x(x+2)-2(x+2)=0

(3x-2)(x+2)=0

x=2/3 or x=-2

I hope this answer will help you

Answer 2

Answer:

Sum of zeroes =

$\frac{2}{3} - 2 \\ = \frac{2 - 6}{3} \\ = - \frac{4}{3}$

Product of zeroes =

$\frac{2}{3} \times ( - 2) \\ = - \frac{4}{3}$

For given equation

$p(x) = 3 {x}^{2} + 4x - 4 \\ sum \: of \: \: zeroes \: = - \frac{b}{a} = - \frac{4}{3} \\ product \: of \: zeroes \: = \frac{c}{a} = - \frac{4}{3}$

Hence proved