1. ## Solving Polynomials

The problems is:
One factor of x^3 + 2x^2 - 23x - 60 is (x=4). Find the remianing factors.

I tried to group them and factor but the second binomial wouldn't factor.

2. Hello, duckduckgoose!

One factor of x³ + 2x² - 23x - 60 is: (x + 4).
Find the remianing factors.

Use long division: .(x³ + 2x² - 23x - 60) ÷ (x + 4) .= .x² - 2x - 15

Therefore: .x³ + 2x² - 23x - 60 .= .(x + 4)(x² - 2x - 15) .= .(x + 4)(x + 3)(x - 5)

3. Originally Posted by duckduckgoose33
The problems is:
One factor of x^3 + 2x^2 - 23x - 60 is (x=4). Find the remianing factors.

I tried to group them and factor but the second binomial wouldn't factor.
They tell you x=4 is a solution,
But that is not true!
Perhaps you mean x=-4
That means
(x-(-4))=(x+4)
Is a factor.

Thus, use long division,
Code:
x^3 + 2x^2 - 23x - 60 : x + 4=x^2 - 2x-15
x^3 +  4x^2
-------------
-2x^2 - 23x - 60
-2x^2 - 8x
------------
-15x - 60
-15x - 60
----------
0
Now we need to factor,
x^2-2x-15
Which factors as,
(x-5)(x+3)
Thus, the full factorization is,
(x-5)(x+3)(x+4)