Question

Use the given factor to solve the equationx^3+11x^2+38x+40=0 ; (x+2)

Answer 1

The solutions to this equation are: x = -2, x = -4 and x = -5

Step-by-step explanation

If (x + 2) is a factor, then x = -2 is one of the solutions to the equation. To find the other solutions we can divide the polynomial by this factor to find a quadratic equation that can be easily solved:

This means that this polynomial can be written as:

`x^3+11x^2+38x+40=(x+2)(x^2+9x+20)`

Now we can solve this equation to find the other two solutions:

`x^2+9x+20=0`

Using the following formula:

`\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}`

For this problem:

`\begin{gathered} x=\frac{-9\pm\sqrt[]{9^2-4\cdot1\cdot20}}{2\cdot1} \\ x=\frac{-9\pm\sqrt[]{81-80}}{2} \\ x=(-9\pm1)/(2) \\ x_1=(-9+1)/(2)=(-8)/(2)=-4 \\ x_2=(-9-1)/(2)=(-10)/(2)=-5 \end{gathered}`

So finally, the polynomial has 3 factors and can be written as:

`x^3+11x^2+38x+40=(x+2)(x+4)(x+5)`