To solve this, we can use the remainder theorem.
The theorem says:
Given a polynomial P(x), the remainder of
[tex]\frac{P(x)}{x-a}[/tex]
Is equal to P(a)
This means, that we are looking for a value of x such as P(a) = 0
We need to find the roots of the polynomial. We can do this, by trying values of x.
Let's use:
x = 0, 1, 2, 3
[tex]x^3+3x^2-16x-48[/tex]
Then:
[tex]\begin{gathered} x=0\Rightarrow0^3+3\cdot0^2-16\cdot0-48=-48 \\ x=1\Rightarrow1^3+3\cdot1^2-16\cdot1-48=1+3-16-48=-60 \\ x=2\Rightarrow2^3+3\cdot2^2-16\cdot2-48=8+12-32-48=-60 \\ x=3\Rightarrow3^3+3\cdot3^2-16\cdot3-48=27+27-48-48=-42 \end{gathered}[/tex]
Let's try negative values,
x = -1, -2, -3
[tex]\begin{gathered} x=-1\Rightarrow(-1)^3+3(-1)^2-16(-1)-48=-1+3+16-48=-30 \\ x=-2\Rightarrow(-2)^3+3(-2)^2-16(-2)-48=-8+12+32-48=-12 \\ x=-3\Rightarrow(-3)^3+3(-3)^2-16(-3)-48=-27+27+48-48=0 \end{gathered}[/tex]
We have found that the polynomial evaluated in x = -3 is equal to zero, which means:
[tex]\frac{x^3+3x^2-16x-48}{x+3}[/tex]
has remainder zero.
The answer is (x + 3)
