Question

find the zero of the polynomial x^3-8​

Answer 1

Answer:

answer is 2

Step-by-step explanation:

x3-8 =. 0

x3=8

x3=2^3

then power will eliminated

x=2

Answer 2

The zeros of $\bf {x}^{3} - 8$ are $\bf 2, \: - 1 \pm \: i \sqrt{3}$.

Given:

• A polynomial.
• ${x}^{3} - 8$

To find:

• Find the zeros of polynomial.

Solution:

Concept to be used:

$\bf {a}^{3} - {b}^{3} = (a - b)( {a}^{2} +a b + {b}^{2} )$

Step 1:

Simplify the polynomial.

${x}^{3} - 8 = {x}^{3} - {2}^{3}$

According to the identity.

$\bf {x}^{3} - {2}^{3} = (x - 2)( {x}^{2} + 2x + 4)$

Step 2:

Find the zeros of the polynomial.

Put the polynomial equal to zero.

$(x - 2)( {x}^{2} + 2x + 4) = 0$

So,

$x - 2 = 0$

$\bf x = 2$

or

${x}^{2} + 2x + 4 = 0$

Apply quadratic formula

$\bf x_{1,2} = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$x_{1,2} = \frac{ - 2\pm \sqrt{ {( -2 )}^{2} - 4 \times 4 } }{2}$

$x_{1,2} = \frac{ - 2\pm \sqrt{ 4 - 16 } }{2}$

$x_{1,2} = \frac{ - 2\pm \sqrt{ - 12} }{2}$

$x_{1,2} = \frac{ - 2\pm 2i \sqrt{3} }{2}$

$\bf x_{1,2} = - 1\pm i\sqrt{ 3}$

Thus,

The zeros of $\bf {x}^{3} - 8$ are $\bf 2, \: - 1 \pm \: i \sqrt{3}$

Learn more:

1) Find all the zeroes of the polynomial.

P(x) = x4 – 4x3 – 4x2 + 8x + 7 , if two of its zeroes are 

(3+root2)(3-root2)

https://brainly.in/question/18403070

2)The zeroes of the polynomial are p(x) = x² – 10x – 75

https://brainly.in/question/47993088

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