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9. Show that the polynomial 3x3 + 8x2 – 1 = 0 has no integral zeroes.​

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Can you show that the polynomial 3x^3+8x^2-1 has no integral zeros?

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Martyn Hathaway, BSc Mathematics, University of Southampton (1986)

Answered August 12, 2018

Can you show that the polynomial 3x^3+8x^2-1 has no integral zeros?

So, you want me to show that there are no integer roots of the cubic equation:

3x3+8x2−1=03x3+8x2−1=0

From the integral zero theorem, any integer roots must be factors of -1. This means we have two possibilities: i) x=1x=1; and ii) x=−1x=−1

Let’s evaluate your cubic expression at these values.

i): 3×13+8×12−1=3+8−1=10≠03×13+8×12−1=3+8−1=10≠0

ii): 3×−13+8×−12−1=−3+8−1=4≠03×−13+8×−12−1=−3+8−1=4≠0

Thus, there are no integer roots.

A little bit of trial and error, and you should be able to find oou that one root is:

x=13x=13

We can thus factorise the cubic as (3x−1)(x2+3x+1)(3x−1)(x2+3x+1)

Using the formula for finding the roots of the quadratic term, we have:

x=−32±5√2x=−32±5/2

So, we have three distinct roots, none of which are integers.