Question

find the zeros of the polynomial 6x³-7x²-11x+12

pls someone help me... but pls ans it correctly if any irrelevant answer will be posted then I'll willingly report that person ​

Answer 1

$\sf\huge\blue{\underline{\underline{ Question : }}}$

Find the zeros of the polynomial 6x³-7x²-11x+12

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• P(x) = 6x³ - 7x² - 11x + 12

★ To find,

• Zeroes of p(x).

★ Let,

According to the form of cubic polynomial ax³ + bx² + cx + d = 0, consider this polynomial.

• a = 6
• b = - 7
• c = - 11
• d = 12

$\bf\red{ :\implies \alpha + \beta + \gamma = \frac{-b}{a}}$

$\bf\:\implies \frac{-(-7)}{6}$

$\bf\:\implies \frac{7}{6}$

$\bf\red{:\implies (\alpha\beta) + (\beta\gamma) + (\alpha\gamma) = \frac{c}{a} }$

$\bf\:\implies \frac{-11}{6}$

$\bf\red{:\implies \alpha\beta\gamma = \frac{d}{a}}$

$\bf\:\implies \frac{12}{6} = 2$

Hence,it was solved...