Question

find a quadratic polynomial whose zeros are -1 and 2/3​

Answer 1

ɢɪᴠᴇɴ :

Zeroes of the quadratic polynomial are -1 & 2/3.

$i.e. \: \: \: \alpha = - 1 \: \\ \: \: \: \: \: \: \beta = \frac{2}{3}$

ᴛᴏ ꜰɪɴᴅ:

Quadratic polynomial

ꜱᴏʟᴜᴛɪᴏɴ:

Quadratic polynomial is

$➡ {x}^{2} - (\alpha + \beta) x + \alpha \beta = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ➡ {x}^{2} - ( - 1 + \frac{2}{3} )x + ( - 1)( \frac{2}{3} ) = 0 \\ \\ ➡ {x }^{2} - ( \frac{ - 3 + 2}{3} )x - \frac{2}{3} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ➡ {x}^{2} -( \frac{ - 1}{3} )x - \frac{2}{3} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ➡ {x}^{2} + \frac{1}{3} x - \frac{2}{3} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ➡3({x}^{2} + \frac{1}{3} x - \frac{2}{3} ) = 3(0) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ ➡3 {x}^{2} + x - 2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 2

Answer:

3x²+x-2

Step-by-step explanation:

General form is

p(x)=k{x²-(sum of zeroes)x+(product of zeroes)

k{x²-(-1+2/3)x+2/3×(-1)}

k{x²-(-3+2/3)x-2/3

k{x²-(-1/3)x-2/3} [denominator value is 3 so k=3]

3{x²-(-1/3)x-2/3}

3x²+x-2 is the answer you want...

Hope it helps...please mark my answer as brainliest