Question

Zeros of linear polynomial is 8x+3

Answer 1

8x+3 = 0
x = -3/8

Thanks

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Answer 2

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

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2 ANSWERS



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

Answered Aug 12, 2018 · Author has 1.1kanswers and 311.1k answer views

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:

[math]3x^3 + 8x^2 - 1 = 0[/math]

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

Let’s evaluate your cubic expression at these values.

i): [math]3 \times 1^3 + 8 \times 1^2 - 1 = 3 + 8 - 1 = 10 \neq 0[/math]

ii): [math]3 \times -1^3 + 8 \times -1^2 - 1 = -3 + 8 - 1 = 4 \neq 0[/math]

Thus, there are no integer roots.

[math]\\[/math]

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

[math]x = \frac {1}{3}[/math]

We can thus factorise the cubic as [math](3x - 1)(x^2 + 3x + 1)[/math]

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

[math]x = -\frac {3}{2} \pm \frac {\sqrt{5}}{