Question

How do you solve `x^3+4x^2+x-6=0` ??

Answer 1

`x=1, -2, and -3`

Explanation:

.

`y=x^3+4x^2+x-6=0`

For higher order functions like this, you can start trying `x=+-1, +-2, +-3, ....` to see if any of them is a root.

Let's try `x=1` and plug it in:

`y=(1)^3+4(1)^2+(1)-6=1+4+1-6=6-6=0`

This shows `x=1` is a solution for the function. Now, we can divide the function by`x-1`. You can use long division or synthetic division to do this.

The result is:

`(x^3+4x^2+x-6)/(x-1)=x^2+5x+6`

Therefore,

`y=(x-1)(x^2+5x+6)`

We can factor the second part if it is factorable. Otherwise, we can use the quadratic formula to solve it. The general form of a quadratic function is:

`y=ax^2+bx+c` and the quadratic formula is:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

But this function is factorable and is equal to :

`(x+2)(x+3)`

Therefore, our original function factors to:

`y=(x-1)(x+2)(x+3)`

Setting it equal to `0`, we can solve for `x` and get three answers:

`x=1, -2, and -3`