Question
\( 1 \leftarrow \) Fing Polynomilals \[ x^{3}-2 x^{2}-4 x+8 \] Select the correct choice below and, if necessary, fill in the answer box within your choice. A. \( x^{3}-2 x^{2}-4 x+8=\square \) B. The polynomial is prime.
Ask by Garrett Rose. in the United States
Mar 19,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( x^{3} - 2x^{2} - 4x + 8 = (x-2)^{2}(x+2) \), choice A.
Solution
We start with the polynomial
\[
x^{3} - 2x^{2} - 4x + 8.
\]
**Step 1: Grouping Terms**
Group the polynomial into two pairs:
\[
\begin{aligned}
x^{3} - 2x^{2} - 4x + 8 &= (x^{3} - 2x^{2}) + (-4x + 8).
\end{aligned}
\]
**Step 2: Factor Each Group**
Factor out the greatest common factor (GCF) from each group:
\[
\begin{aligned}
x^{3} - 2x^{2} &= x^{2}(x - 2), \\
-4x + 8 &= -4(x - 2).
\end{aligned}
\]
Now the expression becomes:
\[
x^{2}(x-2) - 4(x-2).
\]
**Step 3: Factor Out the Common Binomial**
Notice that \((x-2)\) is a common factor:
\[
x^{2}(x-2) - 4(x-2) = (x-2)(x^{2} - 4).
\]
**Step 4: Factor the Quadratic Expression**
Recognize that \(x^{2} - 4\) is a difference of squares:
\[
x^{2} - 4 = (x-2)(x+2).
\]
**Step 5: Write the Final Factored Form**
Substitute back into the expression:
\[
(x-2)(x^{2} - 4) = (x-2)(x-2)(x+2) = (x-2)^{2}(x+2).
\]
Thus, the factorization is
\[
x^{3} - 2x^{2} - 4x + 8 = (x-2)^{2}(x+2).
\]
Since the polynomial factors completely, the correct choice is:
A. \( x^{3} - 2x^{2} - 4x + 8 = (x-2)^{2}(x+2) \).
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution Bonus Knowledge
The polynomial \( x^{3}-2 x^{2}-4 x+8 \) can actually be factored! By using methods such as synthetic division or the Rational Root Theorem, one can discover that \( x = 2 \) is a root. Thus, we can rewrite the polynomial as \( (x - 2)(x^2 + 0x - 4) \), leading to \( (x - 2)(x^2 - 4) \). In case you find it challenging to factor polynomials, a common mistake is overlooking rational roots or failing to check for simple factors. Always remember to use synthetic division to simplify the polynomial, and don’t forget the possibility of factoring by grouping when facing more complex expressions!
