Question

If f(x) = x3 + 8x2 + 4x – 48 and x + 4 is a factor zeros of f(x) algebraically. how would I solve this problem

Answer 1

We want to find the solutions for the following equation:

`x^3+8x^2+4x-48=0`

Since (x + 4) is a factor, we can rewrite this equation as a product of this factor by a second degree equation.

`\begin{gathered} x^3+8x^2+4x-48=(x+4)(ax^2+bx+c) \\ x^3+8x^2+4x-48=ax^3+bx^2+cx+4ax^2+4bx+4c \\ x^3+8x^2+4x-48=ax^3+(4a+b)x^2+(c+4b)x+4c \end{gathered}`

Comparing the coefficients from the left side with the coefficients from the right side, we get the following equations:

`\begin{gathered} a=1 \\ 4a+b=8 \\ c+4b=4 \\ 4c=-48 \end{gathered}`

Solving this system, we get the following results:

`\begin{gathered} a=1 \\ b=4 \\ c=-12 \end{gathered}`

Rewriting our polynomial, we have

`x^3+8x^2+4x-48=(x+4)(x^2+4x-12)`

We can rewrite our original equation as

`(x+4)(x^2+4x-12)=0`

We have a product of two terms, then, if one is zero the product is zero.

To find the roots of this polynomial, we have to solve them individually

`\begin{cases}x+4=0 \\ x^2+4x-12=0\end{cases}`

The solution for the first equation and the first root, is x = -4. Solving the second one we get the two remaining roots. We could solve it using the Bhaskara Formula, but I'm going to factorize it again using the same process.

`\begin{gathered} x^2+4x-12=(x+k)(x+l) \\ x^2+4x-12=x^2+(k+l)x+kl \end{gathered}`
`\begin{cases}k+l=4 \\ kl=-12\end{cases}\Rightarrow\begin{cases}k=-2 \\ l=6\end{cases}`

Our second degree polynomial can be rewritten as

`x^2+4x-12=(x-2)(x+6)`

This means our other roots are x = 2 and x = -6.

`\begin{gathered} x_1=-4 \\ x_2=2 \\ x_3=-6 \end{gathered}`