Answer:
If x-2 is a factor of p(x) = x^{4} - 3 x^{2} + 2 x - 8 ,then p(x) = 0
Solution:
According to factor theorem of polynomials,
x-a is a factor of polynomial p(x), if and only p(a)=0
Step 1:
Since x-2 is factor of p(x),
x - 2 = 0
x = 2
Step 2:
[tex]p(x) = x^{4} - 3 x^{2} + 2 x - 8 ------ (equation 1)[/tex]
By substituting x = 2 in equation 1,
[tex]p(x) = 2^{4} - 3(2)^{2} + 2(2) - 8[/tex]
p(x) = 16 - 3(4 ) + 4 -8
p(x) = 16 - 12 - 4
p(x) = 16 - 16
p(x) = 0
Since x - 2 is a factor of p(x) , we get p(x)=0
