How to expand equations of bigger degree

Question

I am 8th grade, and I have to solve this equation (I got it from a problem when I simplified): $$ x^4 + 6x^3 + 11x^2 + 6x - 24 = 0 $$

I looked at some solutions in online solvers, but I don't understand how I should guess that $6x^3 = -x^3 + 7x^3$ or $11x^2 = -7x^2 + 18x^2$. I would be very grateful if you can explain to me how to expand it and solve it.

Answer 1

One can easily see that $x=1$ is a solution to the given equation. Thus, the given expression will be given as a product of $(x-1)$ with a polynomial of 3rd degree. More specifically :

$$x^4 + 6x^3 + 11x^2 + 6x - 24 = (x-1)(ax^3 + bx^2 + cx + d)$$

Now, you can expand the right hand side and calculate $a,b,c$ and $ d$.

$$x^4 + 6x^3 + 11x^2 + 6x - 24 = ax^4 + bx^3 + cx^2 + dx - ax^3 - bx^2 - cx - d$$ $$\Leftrightarrow$$ $$x^4 + 6x^3 + 11x^2 + 6x - 24 = ax^4+ (b-a)x^3 + (c-b)x^2 + (d-c)x-d$$

For the equation to hold, each polynomial term must be equal, thus the coefficient of each of the $x^4,x^3,x^2,x$ terms must be equal and of course the constants. Thus, $a,b,c$ and $d$ are given as : $$\begin{cases} x^4 = ax^4 \\ 6x^3=(b-a)x^3 \\ 11x^2 = (c-b)x^2 \\ 6x = (d-c)x \\ -24=-d\end{cases} \implies \begin{cases} a=1 \\ b=7 \\ c= 18\\ d = 24\end{cases}$$

Doind that, gives $a=1, b=7, c=18$ and $d=24$. Thus :

$$ax^3 + bx^2 + cx + d = x^3+7x^2+18x+24$$

You can apply any of the ways you've been taught to solve polynomials of 3rd degree and you shall get :

$$x^3+7x^2+18x+24 = 0 \Leftrightarrow (x+4)(x^2+3x+6) = 0$$

Thus the initial expression can be written as :

$$x^4 + 6x^3 + 11x^2 + 6x - 24 = 0 \Leftrightarrow (x-1)(x+4)(x^2+3x+6) =0$$

This means that the solutions to the equation are $x=1$ and $x=-4$, since the quadratic involved has a negative discriminant, thus no real solutions. Of course, if you're interested to solve over complex numbers, then you'd consider the complex numbers that satisfy the quadratic as well.

Answer 2

Hint

$$24=x(x+1)(x+2)(x+3)$$

Set $x(x+3)=y$ to form a quadratic equation in $y$

Answer 3

Tiny improvement to Rebellos' solution:

Once you see that $$x^4 + 6x^3 + 11x^2 + 6x - 24 = (x-1)(ax^3 + bx^2 + cx + d)$$

it's more efficient to do polynomial division rather than multiplying the RHS and forming linear equations.

        x^3 + 7x^2 + 18x + 24
       -------------------
x - 1 | x^4 + 6x^3 + 11x^2 + 6x - 24
        x^4 - x^3
        ----------
              7x^3
              7x^3 - 7x^2
              -----------
                     18x^2
                     18x^2 - 18x
                     ------------
                            24x
                            24x - 24
                            --------
                                   0