Question

if
$x = -1/3 is a zero of polynomial p(x) [tex]= 27 {x}^{3} - a {x}^{2} - x+ 3$
find the value of 'a'
​

Answer 1

Answer:

a=-18

Step-by-step explanation:

Given ,

Given , p(x) = 27x^3-ax^2+3

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)So,

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)So,p(-1/3)=0

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)So,p(-1/3)=027(-1/3)^3-a(-1/3)^2+3=0

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)So,p(-1/3)=027(-1/3)^3-a(-1/3)^2+3=0-1+a/9+3=0

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)So,p(-1/3)=027(-1/3)^3-a(-1/3)^2+3=0-1+a/9+3=0a/9= = -2

Given , p(x) = 27x^3-ax^2+3x=-1/3 is zero of p(x)So,p(-1/3)=027(-1/3)^3-a(-1/3)^2+3=0-1+a/9+3=0a/9= = -2a = -18 ( Answer)