Question

one zero is given find all others P of x equals x to the third power minus 4x squared minus 25 x + 100 ; -5

Answer 1

The expression is:

$x^3-4x^2-25x+100$

We know that one of the zeros of this expression is x=-5. We have to find the others.

Then, we can write this expression as:

$\begin{gathered} x^3-4x^2-25x+100=(x+5)(ax^2+bx+c) \\ ax^2+bx+c=\frac{x^3-4x^2-25x+100}{x+5} \end{gathered}$

We have to perform the polynomial division:

$\begin{gathered} \frac{x^3-4x^2-25x+100}{x+5}=\frac{x^3-4x^2-25x+100-5x^2+5x^2}{x+5} \\ \frac{(x^3+5x^2)-4x^2-25x+100+5x^2}{x+5} \\ \frac{x^2(x+5)}{x+5}+\frac{-4x^2-25x+100-5x^2}{x+5} \\ x^2+\frac{-9x^2-25x+100}{x+5} \\ x^2+\frac{-9x^2-25x+100+45x-45x}{x+5} \\ x^2+(\frac{-9x^2-45x}{x+5}+\frac{-25x+45x+100}{x+5}) \\ x^2-9x+\frac{20x+100}{x+5} \\ x^2-9x+20 \end{gathered}$$x^3-4x^2-25x+100=(x+5)(x^2-9x+20)$

Now, we can calculate the zeros of the quadratic polynomial as:

$\begin{gathered} x=\frac{-(-9)\pm\sqrt[]{(-9)^2-4\cdot1\cdot(20)}}{2\cdot1} \\ x=\frac{9\pm\sqrt[]{81-80}}{2} \\ x=\frac{9\pm\sqrt[]{1}}{2} \\ x_1=\frac{9-1}{2}=\frac{8}{2}=4 \\ x_2=\frac{9+1}{2}=\frac{10}{2}=5 \end{gathered}$

Answer: the other two zeros are x=4 and x=5