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

1 Answer
Jun 3, 2018

x=1, -2, and -3

Explanation:

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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 byx-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