Finding Real Zeros of g(x) = 2x^3 - x^2 - 10x + 5

[Sweetnlow]

Hi I have a question about finding all real zeros of the problem:

g(x) = 2x^3 - x^2 - 10x + 5

I believe the answer is 1/2, - squareroot 5, + squareroot 5
I know that there are 3 Real Zeros because of the highest degree (3).

I need help in finding out how to find the real zeros, or the steps to find them.

Thanks

 

[galactus]

Re: Finding Real Zeros Help

This is easy to factor. Let's do that.

\(\displaystyle 2x^{3}-x^{2}-10x+5\)

Rearrange and group:

\(\displaystyle (2x^{3}-10x)-(x^{2}-5)\)

\(\displaystyle 2x(x^{2}-5)-(x^{2}-5)\)

\(\displaystyle (2x-1)(x^{2}-5)\)

Now, set each to 0 and find your 0's.

 

[Sweetnlow]

Re: Finding Real Zeros Help

For the (x^2-5) part, how would you get a plus or negative equal to x?

 

[tkhunny]

Re: Finding Real Zeros Help

Factor it as a difference of squares.

\(\displaystyle a^{2}-b^{2} = (a+b)(a-b)\)