Question

How many real roots does the polynomial 2x^3+8x-7 have

Answer 1

Answer:

Step-by-step explanation:

No real roots

Answer 2

It has 1 real root

Step-by-step explanation:

Given polynomial,

$f(x) = 2x^3+8x-7$

$f(x) = 2x^3 + 0x^2 + 8x -7$

Since, the polynomial is arranged in descending powers of the variable and coefficients of our variable in f(x),

2, 0, 8, -7

Variables goes from positive(2) to 0 to positive(8) to negative(-7),

i.e. number of change in sign = 1,

Thus, there is only one positive real root,

Now,

$f(-x) = -2x^3 + 0x^2 - 8x -7$

Variables goes from negative(2) to 0 to negative(-8) to negative(-7),

i.e number of change in sign = 0,

Thus there is no negative real root.

∵ Degree of the polynomial = 3,

Hence, By Descartes's rule of sign,

The polynomial has 1 real roots and 2 imaginary root.

#Learn more:

What are polynomials? What is the degree of a polynomial?

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