How Do You Solve X^3+4x^2+x-6=0 ?? | Socratic

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

1 Answer

Sean Jun 3, 2018

#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#

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