Let q(x)=2x^3-3x^2-10x+25. Show q(-5/2)=0 and find the other roots of q(x)=0

Answer 1

Answer 2

Hello, Alisa,

Input the value (-5/2) into each x in the equation:

q(-5/2)=2(-5/2)^3-3(-5/2)^2-10(-5/2)x+25

Now calculate each of the terms;

q(-5/2)= -(31.25)-(18.75)+(25)+25

q(-5/2)= 0

The second part of the question is difficult. Find other roots.

We can factor the equation. I'm not sure this is complete, but one I found is:

q(x) = (2x+5)((x-4)x + 5)

The (2x+5) = 0 provides the solution already given: - 5/2

(x-4)=0 provides a second root: 4

((x-4)x + 5) can also be written as (x^2 -4x + 5)

This would have roots of (2+i) and (2-i), I believe.

So -(5/2), (2+i) and (2-i) are all roots.

Please check my work carefully. The imaginary roots may be just that.

Bob

Answer 3

Let q(x)=2x^3-3x^2-10x+25. Show q(-5/2)=0 and find the other roots of q(x)=0

You can show that -5/2 is a root by substituting this into the original equation which gives 0 as a result; this it is a root. Divide the original equation by (X + 5/2) and you get X^2 -4X +5. = 0. This is a quadratic equation that can be solved by the quadratic formula. This gives results of 2+i and 2-i. Prove that these are roots by substituting into the original equation which should give zero.

Answer 4

q(-5/2) = 0 means that (x + 5/2) is a factor of q(x). If you wanted whole numbers, you can say that (2x+5) is a factor of q(x). Divide that by q(x) and then find the other roots. You already have that -5/2 is a root.