Question

Obtain zeros of p(x) = 6x2

-x -2. Also verify the relationship between the zeros and the coefficients

of p(x).​

Answer 1

$\bold\colorbox{green}{Step-by-step explanation:}$

p(x) = 6x² - x - 2

0 = 6x² + 3x - 4x - 2

0 = 3x ( 2x + 1 ) - 2 ( 2x + 1 )

0 = ( 3x - 2 ) ( 2x + 1 )

Hence,

3x - 2 = 0 or 2x + 1 = 0

So,

x = 2/3 or x = -1/2

Therefore, there are 2 zeros of the given polynomial i.e., (2/3) & (-1/2).

Hᴏᴘᴇ Tʜɪs Hᴇʟᴘs Yᴏᴜ ❤️

Sᴜʀᴇʟʏ, Tʜᴀɴᴋ Mʏ Aɴsᴡᴇʀ✌️❣️

Answer 2

$\huge\tt\colorbox{orange}{Answer}$

6x² - x - 2 = 0

6x² + 3x - 4x - 2 = 0

3x (2x + 1) - 2(2x + 1) = 0

(3x - 2 ) (2x + 1)

3x - 2 = 0

3x = 2

x = 2/3

and

2x + 1 = 0

2x = -1

x = -1/2

$\alpha + \beta = \frac{ - b}{a} \\ \frac{2}{3} + ( \frac{ - 1}{2} ) = \frac{ - ( - 1)}{6} \\ \frac{2 - 1}{6} = \frac{1}{6} \\ \frac{1}{6} = \frac{1}{6}$

$\alpha \times \beta = \frac{c}{a} \\ \frac{2}{3} \times \frac{ - 1}{2} = \frac{ - 2}{6} \\ \frac{ - 2}{6} = \frac{ - 2}{6}$