Question

P (x) = 3y²-x-4 , alfa and bita are the Zeros of p(x) , alfa2bita + alfabita2​

Answer 1

Answer:

$\boxed{\bf \frac{-4}{9} \ is \ the \ answer}$

Step-by-step explanation:

$\text{Given polynomial,} \\ p(x)=3x^2-x-4 \\\\ \alpha \ and \ \beta \ are \ zeroes \ of \ the \ polynomial. \\\\ let's \ find \ the \ zeroes \ first, \\\\ 3x^2-x-4=0 \\3x^2+3x-4x-4=0\\3x(x+1)-4(x+1)=0 \\ (x+1)(3x-4)=0\\=>(x+1)=0 \ ; \ x=-1\\=>(3x-4)=0 \ ; \ x=\frac{4}{3} \\\\ the \ zeroes \ are \ -1 \ and \ \frac{4}{3}$

$let \ \alpha=-1 \ and \ \beta=\frac{4}{3} \\\\ => \alpha^2\beta+\alpha\beta^2 \\\\ =>(-1)^2(\frac{4}{3}) \ + \ (-1)(\frac{4}{3})^2 \\ \\ =>(1)(\frac{4}{3}) \ + \ (-1)(\frac{16}{9}) \\\\ =>\frac{4}{3}-\frac{16}{9} \\\\=>\frac{12-16}{9} \\\\ =>\frac{-4}{9}$