Question

Very confused on question 5 need help as soon as possible

Answer 1

To solve this, we can use the remainder theorem.

The theorem says:

Given a polynomial P(x), the remainder of

`(P(x))/(x-a)`

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

`x^3+3x^2-16x-48`

Then:

`\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}`

Let's try negative values,

x = -1, -2, -3

`\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}`

We have found that the polynomial evaluated in x = -3 is equal to zero, which means:

`(x^3+3x^2-16x-48)/(x+3)`

has remainder zero.

The answer is (x + 3)