## Precalculus: Mathematics for Calculus, 7th Edition

1. We must factor this function if we want to find all real solutions of the function. 2. Since this is a cubic polynomial, we will need to find a way to make it into a second degree polynomial to make it easier for us to factor. 3. Since their is a x in common between both terms, we can factor out an x, giving us x$(4x^{2} - 25)$. 4. Now you can factor this form, factoring gives us x(2x + 5)(2x-5) = 0. 5. Now set all 3 terms equal to 0. 6. x = 0 is out first term (this is the x we factored out from beginning or problem). 7. 2x - 5 = 0 : 2x = 5 : x = 5/2 8. 2x + 5 = 0 : 2x = -5 : x = -5/2 9. This is how you get your solutions of 0, 5/2 and -5/2.