Answer :
We are given the polynomial
[tex]$$
f(x)=x^3-3x^2+16x-48,
$$[/tex]
and the information that [tex]$3$[/tex] is a zero (that is, [tex]$f(3)=0$[/tex]). We will use this fact to factor the polynomial into factors of the form [tex]$(x-c)$[/tex].
Step 1. Since [tex]$3$[/tex] is a zero, we know that [tex]$(x-3)$[/tex] is a factor of [tex]$f(x)$[/tex]. To find the other factor, we divide [tex]$f(x)$[/tex] by [tex]$(x-3)$[/tex].
Step 2. Use synthetic division with the coefficients of [tex]$f(x)$[/tex], which are [tex]$1$[/tex], [tex]$-3$[/tex], [tex]$16$[/tex], and [tex]$-48$[/tex].
[tex]\[
\begin{array}{r|rrrr}
3 & 1 & -3 & 16 & -48 \\[6mm]
& & 3 & 0 & 48 \\
\hline
& 1 & 0 & 16 & 0
\end{array}
\][/tex]
Here is what happens:
- Bring down the first coefficient [tex]$1$[/tex].
- Multiply [tex]$1$[/tex] by [tex]$3$[/tex] to get [tex]$3$[/tex], and add this to [tex]$-3$[/tex], resulting in [tex]$0$[/tex].
- Multiply [tex]$0$[/tex] by [tex]$3$[/tex] to get [tex]$0$[/tex], and add this to [tex]$16$[/tex], resulting in [tex]$16$[/tex].
- Multiply [tex]$16$[/tex] by [tex]$3$[/tex] to get [tex]$48$[/tex], and add this to [tex]$-48$[/tex], resulting in [tex]$0$[/tex].
Since the remainder is [tex]$0$[/tex], the division is exact, and the quotient is
[tex]$$
x^2 + 16.
$$[/tex]
Thus, we have
[tex]$$
f(x) = (x-3)(x^2+16).
$$[/tex]
Step 3. The quadratic factor [tex]$x^2+16$[/tex] may now be factored further into linear factors over the complex numbers. Set
[tex]$$
x^2+16=0.
$$[/tex]
This gives
[tex]$$
x^2 = -16,
$$[/tex]
so
[tex]$$
x = \pm \sqrt{-16} = \pm 4i,
$$[/tex]
where [tex]$i = \sqrt{-1}$[/tex].
Thus, [tex]$x^2+16$[/tex] factors as
[tex]$$
x^2+16 = (x-4i)(x+4i).
$$[/tex]
Step 4. Therefore, the complete factorization of [tex]$f(x)$[/tex] into factors of the form [tex]$(x-c)$[/tex] is
[tex]$$
f(x) = (x-3)(x-4i)(x+4i).
$$[/tex]
To summarize:
(a) The factorization of [tex]$f(x)$[/tex] is
[tex]$$
x^3-3x^2+16x-48 = (x-3)(x^2+16),
$$[/tex]
or, equivalently, into linear factors:
[tex]$$
(x-3)(x-4i)(x+4i).
$$[/tex]
(b) The solutions to the equation
[tex]$$
x^3-3x^2+16x-48=0
$$[/tex]
are
[tex]$$
x = 3,\quad x = 4i,\quad x = -4i.
$$[/tex]
[tex]$$
f(x)=x^3-3x^2+16x-48,
$$[/tex]
and the information that [tex]$3$[/tex] is a zero (that is, [tex]$f(3)=0$[/tex]). We will use this fact to factor the polynomial into factors of the form [tex]$(x-c)$[/tex].
Step 1. Since [tex]$3$[/tex] is a zero, we know that [tex]$(x-3)$[/tex] is a factor of [tex]$f(x)$[/tex]. To find the other factor, we divide [tex]$f(x)$[/tex] by [tex]$(x-3)$[/tex].
Step 2. Use synthetic division with the coefficients of [tex]$f(x)$[/tex], which are [tex]$1$[/tex], [tex]$-3$[/tex], [tex]$16$[/tex], and [tex]$-48$[/tex].
[tex]\[
\begin{array}{r|rrrr}
3 & 1 & -3 & 16 & -48 \\[6mm]
& & 3 & 0 & 48 \\
\hline
& 1 & 0 & 16 & 0
\end{array}
\][/tex]
Here is what happens:
- Bring down the first coefficient [tex]$1$[/tex].
- Multiply [tex]$1$[/tex] by [tex]$3$[/tex] to get [tex]$3$[/tex], and add this to [tex]$-3$[/tex], resulting in [tex]$0$[/tex].
- Multiply [tex]$0$[/tex] by [tex]$3$[/tex] to get [tex]$0$[/tex], and add this to [tex]$16$[/tex], resulting in [tex]$16$[/tex].
- Multiply [tex]$16$[/tex] by [tex]$3$[/tex] to get [tex]$48$[/tex], and add this to [tex]$-48$[/tex], resulting in [tex]$0$[/tex].
Since the remainder is [tex]$0$[/tex], the division is exact, and the quotient is
[tex]$$
x^2 + 16.
$$[/tex]
Thus, we have
[tex]$$
f(x) = (x-3)(x^2+16).
$$[/tex]
Step 3. The quadratic factor [tex]$x^2+16$[/tex] may now be factored further into linear factors over the complex numbers. Set
[tex]$$
x^2+16=0.
$$[/tex]
This gives
[tex]$$
x^2 = -16,
$$[/tex]
so
[tex]$$
x = \pm \sqrt{-16} = \pm 4i,
$$[/tex]
where [tex]$i = \sqrt{-1}$[/tex].
Thus, [tex]$x^2+16$[/tex] factors as
[tex]$$
x^2+16 = (x-4i)(x+4i).
$$[/tex]
Step 4. Therefore, the complete factorization of [tex]$f(x)$[/tex] into factors of the form [tex]$(x-c)$[/tex] is
[tex]$$
f(x) = (x-3)(x-4i)(x+4i).
$$[/tex]
To summarize:
(a) The factorization of [tex]$f(x)$[/tex] is
[tex]$$
x^3-3x^2+16x-48 = (x-3)(x^2+16),
$$[/tex]
or, equivalently, into linear factors:
[tex]$$
(x-3)(x-4i)(x+4i).
$$[/tex]
(b) The solutions to the equation
[tex]$$
x^3-3x^2+16x-48=0
$$[/tex]
are
[tex]$$
x = 3,\quad x = 4i,\quad x = -4i.
$$[/tex]
