Answer to Question #245319 in Calculus for minu

Question

Let h(x) =f(x)/g(x) where h(x) is polynomial of x. f f(x) = x³-2x²-3x and g(x)=x-3,then find the number of real solutions of the equation h(x)=0 in the interval [-2.0, 2.0]

Answer 1

$h(x)=\dfrac{x^3-2x^2-3x}{x-3}=\dfrac{x(x^2-2x-3)}{x-3}$

$=\dfrac{x(x+1)(x-3)}{x-3}=x(x+1), x\not=3$

$h(x)=0=>x(x+1)=0, x\not=3$

$x_1=-1, x_2=0$

There are two real solutions of the equation h(x)=0 in the interval [-2.0, 2.0].