Question

the value of the polynomial x8-x5+x2-x+1 is​

Answer 1

Answer:

The given polynomial can take any value from 0 to infinity as X which is a variable can take up any value of real number, so thus the polynomial given to us do, it can take up values depended on the value of x provided, For a sample I am considering the value of x to be 1, so the polynomial becomes

1-1+1-1+1=1, which is a simple value, by substituting different values for the variable X we can get different values.

Answer 2

Step-by-step explanation:

ANSWER

By factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x 

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x 

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)Case II : [x∈(0,1)]

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)Case II : [x∈(0,1)]Maximum value of [−x(x−1)] is 1/4 at x=1/2.

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)Case II : [x∈(0,1)]Maximum value of [−x(x−1)] is 1/4 at x=1/2.And maximum value of [x4(x2+x+1)+1] is 4 at x=1 as its derivative is greater than zero in the range.

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)Case II : [x∈(0,1)]Maximum value of [−x(x−1)] is 1/4 at x=1/2.And maximum value of [x4(x2+x+1)+1] is 4 at x=1 as its derivative is greater than zero in the range.∴x(x−1)(x4(x2+x+1)+1)>−(1/4)×4=(−1) ...... [as maxima of the terms are not coincident]

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)Case II : [x∈(0,1)]Maximum value of [−x(x−1)] is 1/4 at x=1/2.And maximum value of [x4(x2+x+1)+1] is 4 at x=1 as its derivative is greater than zero in the range.∴x(x−1)(x4(x2+x+1)+1)>−(1/4)×4=(−1) ...... [as maxima of the terms are not coincident]⇒p(x)>0, x∈(0,1)

ANSWERBy factorisation of polynomial p(x)=x8−x 5+x−x+1, we get:p(x)=x 5(x3−1)+x(x−1)+1=x(x−1)(x 4(x 2+x+1)+1We know that, x 4≥0 and x2 +x+1>0 ∀ x∈R.Case I : x  ∈(0,1)As x(x−1)≥0,⇒p(x)≥1>0 ...... (positive in range)Case II : [x∈(0,1)]Maximum value of [−x(x−1)] is 1/4 at x=1/2.And maximum value of [x4(x2+x+1)+1] is 4 at x=1 as its derivative is greater than zero in the range.∴x(x−1)(x4(x2+x+1)+1)>−(1/4)×4=(−1) ...... [as maxima of the terms are not coincident]⇒p(x)>0, x∈(0,1)∴p(x)>0 ∀ x∈R

hope it will help u....