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Byju's Answer

Standard VIII

Mathematics

Polynomial

Find the valu...

Question

Find the value of polynomial $s (z) = z^{3} - 1$ at $z = 1$ .

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Solution

The correct option is A $0$
Given, $s (z) = z^{3} - 1$
We need to find value of given polynomial at $z = 1$ .
Put the value $z = 1$ , we get
$s (z) = (1)^{3} - 1$
$\Rightarrow s (z) = 1 - 1$
$\Rightarrow s (z) = 0$
So, value is $s (z) = 0$ .

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Similar questions

If $(z^{3}$ + $\frac{1}{z^{3}})$ - $(z^{2}$ + $\frac{1}{z^{2}})$ + $(z + \frac{1}{z})$ - 1 = $(z + \frac{1}{z} - 2 α$ ) $(z + \frac{1}{z} - 2 β$ ) $(z + \frac{1}{z} - 2 γ$ ) Find the value of $α$ . $β$ . $γ$ .

Q. If

z^{2} + z + 1 = 0,

where

z

is complex number, then the value of

(z + \frac{1}{z}) + (z^{2} + \frac{1}{z^{2}}) + (z^{3} + \frac{1}{z^{3}}) + \dots + (z^{6} + \frac{1}{z^{6}})

Q. Let

z = \frac{1 - i \sqrt{3}}{2}, i = \sqrt{- 1} .

Then the value of

21 + {(z + \frac{1}{z})}^{3} + {(z^{2} + \frac{1}{z^{2}})}^{3} + {(z^{3} + \frac{1}{z^{3}})}^{3} + \dots + {(z^{21} + \frac{1}{z^{21}})}^{3}

a ε R

such that

z = - i

is a root for the polynomial

P (z) = z^{3} - z^{2} + z + 1 + a

, then the value of

a

- 2

Q. If

z_{1}, z_{2}, z_{3}

and

z_{4}

are the roots of the equation

z^{4} + z^{3} + z^{2} + z + 1 = 0

, then

\begin{matrix} Column I & Column II (A) & ∣ ∣ ∣ ∣ 4 \sum i = 1 (z_{i})^{4} ∣ ∣ ∣ ∣ is equal to & (p) & 0 (B) & 4 \sum i = 1 (z_{i})^{5} is equal to & (q) & 4 (C) & 4 \prod i = 1 (z_{i} + 2) is equal to & (r) & 1 (D) & Least value of [| z_{1} + z_{2} |] is & (s) & 11 \end{matrix}

where []

represents greatest integer function

Which of the following is the correct combination

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Find the value of polynomial s(z)=z3−1 at z=1.

The correct option is A 0Given, s(z)=z3−1We need to find value of given polynomial at z=1.Put the value z=1, we gets(z)=(1)3−1⇒s(z)=1−1⇒s(z)=0So, value is s(z)=0.

Find the value of polynomial $s (z) = z^{3} - 1$ at $z = 1$ .

The correct option is A $0$
Given, $s (z) = z^{3} - 1$
We need to find value of given polynomial at $z = 1$ .
Put the value $z = 1$ , we get
$s (z) = (1)^{3} - 1$
$\Rightarrow s (z) = 1 - 1$
$\Rightarrow s (z) = 0$
So, value is $s (z) = 0$ .