Question

if x-1 is zero of p(x) =ax^3-x^2+x++4,then find the value of 'a'

Answer 1

Hello friends!!

Here is your answer :

P(x) =
$a {x}^{3} - {x}^{2} + x + 4$

x - 1 is a zero of p(x).

P(1) = 0

$a {(1)}^{3} - {(1)}^{2} + 1 + 4 = 0$

$a - 1 + 1 + 4 = 0$

$a - 1 + 5 = 0$

$a + 4 = 0$

$a = - 4$

Therefore, value of a is - 4.

Hope it helps you.. ☺️☺️☺️☺️

Answer 2

Hii...☺

Here is your answer.....☺

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Value of "a" = -4.

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Thanks....☺