Question

If x=-1/3 is a 0 of a polynomial p(x)=27x^3-ax^2-x+3 find a

Answer 1

HEYA !

Given Polynomial :

p(x) = 27x³ - ax² - x + 3

And given that,

- 1/3 is zero of polynomial p(x)

→ p ( -1/3 ) = 0

→ 27 (-1/3)³ - a (-1/3)² - (-1/3) + 3 = 0

→ 27 × -1/27 - a × 1/9 + 1/3 + 3 = 0

→ - 1 - a/9 + (1 + 9) / 3 = 0

→ - a/9 + 10/3 - 1 = 0

→ - a/9 + (10 - 3)/3 = 0

→ - a/9 + 7/3 = 0

→ - a/9 = - 7/3

→ a = 9 × 7/3

→ a = 3 × 7

$\boxed{ \Rightarrow \: \text{a = 21}}\: \: \: \: \: \: \: \mathbf{Ans.}$

Answer 2

Heya !!!



X = -1/3



P(X) => 0


P(-1/3) => 0



27X³-AX²-X+3 = 0




27 × (-1/3)³ - A × (-1/3)² - (-1/3) + 3 = 0


27 × -1/27 - A × 1/9 + 1/3 + 3 = 0



-1 - A/9 + 1/3 + 3 = 0


-9 - A + 3 + 3/9 = 0



-9 - A +6 = 0 × 9


-A - 3 = 0



A = -3




HOPE IT WILL HELP YOU....... :-)