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Answer :
To factor the polynomial [tex]\(x^3 + 3x^2 + 16x + 48\)[/tex] completely, we can use the method of synthetic division and the Rational Root Theorem.
1. Use the Rational Root Theorem: This theorem suggests that any rational solution, or root, of the polynomial equation is a factor of the constant term. In this polynomial, the constant term is 48. Therefore, we consider possible rational roots from the factors of 48, which are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.
2. Test possible roots using synthetic division: We’ll start by testing some of the simple factors to see if any of them are roots.
Let's test [tex]\(x = -3\)[/tex].
- Set up synthetic division using [tex]\(-3\)[/tex]:
[tex]\[
\begin{array}{r|rrrr}
-3 & 1 & 3 & 16 & 48 \\
& & -3 & 0 & -48 \\
\hline
& 1 & 0 & 16 & 0 \\
\end{array}
\][/tex]
The last number is 0, which confirms that [tex]\(-3\)[/tex] is a root of the polynomial.
3. Factor the polynomial using the root: Since [tex]\(-3\)[/tex] is a root, [tex]\(x + 3\)[/tex] is a factor of the polynomial. The result of the division is the polynomial [tex]\(x^2 + 16\)[/tex].
4. Factor the quadratic [tex]\(x^2 + 16\)[/tex]: Notice that [tex]\(x^2 + 16\)[/tex] can be expressed as a sum of squares:
[tex]\[
x^2 + 16 = x^2 + (4i)^2
\][/tex]
This factors into:
[tex]\[
(x + 4i)(x - 4i)
\][/tex]
So, the complete factorization of the polynomial [tex]\(x^3 + 3x^2 + 16x + 48\)[/tex] is:
[tex]\[
(x + 3)(x + 4i)(x - 4i)
\][/tex]
Note: The factors [tex]\(x + 4i\)[/tex] and [tex]\(x - 4i\)[/tex] are complex conjugate pairs, as they arise from the sum of squares factorization [tex]\(x^2 + 16\)[/tex].
1. Use the Rational Root Theorem: This theorem suggests that any rational solution, or root, of the polynomial equation is a factor of the constant term. In this polynomial, the constant term is 48. Therefore, we consider possible rational roots from the factors of 48, which are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.
2. Test possible roots using synthetic division: We’ll start by testing some of the simple factors to see if any of them are roots.
Let's test [tex]\(x = -3\)[/tex].
- Set up synthetic division using [tex]\(-3\)[/tex]:
[tex]\[
\begin{array}{r|rrrr}
-3 & 1 & 3 & 16 & 48 \\
& & -3 & 0 & -48 \\
\hline
& 1 & 0 & 16 & 0 \\
\end{array}
\][/tex]
The last number is 0, which confirms that [tex]\(-3\)[/tex] is a root of the polynomial.
3. Factor the polynomial using the root: Since [tex]\(-3\)[/tex] is a root, [tex]\(x + 3\)[/tex] is a factor of the polynomial. The result of the division is the polynomial [tex]\(x^2 + 16\)[/tex].
4. Factor the quadratic [tex]\(x^2 + 16\)[/tex]: Notice that [tex]\(x^2 + 16\)[/tex] can be expressed as a sum of squares:
[tex]\[
x^2 + 16 = x^2 + (4i)^2
\][/tex]
This factors into:
[tex]\[
(x + 4i)(x - 4i)
\][/tex]
So, the complete factorization of the polynomial [tex]\(x^3 + 3x^2 + 16x + 48\)[/tex] is:
[tex]\[
(x + 3)(x + 4i)(x - 4i)
\][/tex]
Note: The factors [tex]\(x + 4i\)[/tex] and [tex]\(x - 4i\)[/tex] are complex conjugate pairs, as they arise from the sum of squares factorization [tex]\(x^2 + 16\)[/tex].
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