show that for the equation y^3 - 3y + 4 = 0 , one of the root lies between -3 and -2 i don't know how to show that one of the root lies between -3 and -2, but i can show that one of the root ( or more ) is smaller than - 3^(1/2), pay in mind that -3 and -2 are also smaller than -3^(1/2). Here is my method, but it doesn't solve the question, i wrote it just for your reference so you got more idea to solve it. U help is meaningful to me, thanks you ! y^3 - 3y + 4 = (y)(y^2 - 3) + 4 = 0 so this mean in order to make the equation becomes zero, the term (y)( y^2 - 3) must equal -4, in order word, it must less than zero... hence it is right to write (y)(y^2 - 3) < 0 by number line method or graph method, we know that the range of y for this inequality is y < -3^(1/2) or 0 < y < 3^(1/2) so this imply that one of the root ( or more ) is smaller than -3^(1/2). But, this is not the correct answer, we need to show it lies between -3 and -2 , not the negative squate root of 3 !!!! Please, any expert, if u know the method, please show me as soon as possible, u help is meaningful to me!