Math, asked by hktsks, 11 months ago

the value of the polynomial x power 8 minus x power 5 + x square minus x + 1 is
1) positive for all the real numbers
2)negative for all the real numbers
3)0
4) depends on value of x

Answers

Answered by Anonymous

$\huge{\red{\mathbb{Answer }}}$

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$\huge{\green{\mathfrak{Equation }}}$

${x}^{8} - {x}^{5} + {x}^{2} - x + 1$

the value of the polynomial x power 8 minus x power 5 + x square minus x + 1 is

1) positive for all the real numbers

2)negative for all the real numbers

3)0

$\huge{\purple{\mathfrak{Solution }}}$

By factorisation of polynomial p(x)=

${x}^{8} - {x}^{5} + {x}^{2} - x + 1 \:$

we get:

p(x)= {x}^{5} ( {x}^{3} −1)+x(x−1)+1=x(x−1)( {x}^{4} ( {x}^{4} +x+1)+1)+1

$p(x) = {x}^{5} ( {x}^{3} - 1) + x(x - 1) + 1 = x(x - 1)( {x}^{4} ( {x}^{4} + x + 1) + 1) + 1$

We know that, x 4 ≥0 and x 2 +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 [x

4 (x^2 +x+1)+1] is 4 at x=1 as its derivative is greater than zero in the range.

∴x(x−1)(x

4 (x ^2 +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

Answered by dhayaengineering

Step-by-step explanation:

d is the correct opition

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