Question

4. Find all the zeroes of 4x3 -12x2 -9x+27

Answer 1

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

Find all the zeroes of 4x³ - 12x² - 9x + 27

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

★ Given that,

• P(x) = 4x³ - 12x² - 9x + 27

★To find,

• Zeroes of P(x).

★ Let,

According to the form of Cubic Polynomial :

ax³ + bx² + cx + d

• a = 4
• b = - 12
• c = - 9
• d = 27

$\sf\green{ \implies \alpha + \beta + \gamma = \frac{-b}{a} }$

$\sf\:\implies \frac{-(-12)}{4}$

$\sf\:\implies \frac{12}{4}$

$\sf\:\implies 3$

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

$\sf\:\implies \frac{-9}{4}$

$\sf\green{:\implies \alpha\beta\gamma = \frac{-d}{a} }$

$\sf\:\implies \frac{-27}{4}$

Hence,it was solved...