Question

What are the linear and quadratic factors of 8x4−32x3−5x2+20x
8
𝑥
4
−
32
𝑥
3
−
5
𝑥
2
+
20
𝑥
?(1 point)
Responses

x(x−4)(8x2+5)
𝑥
(
𝑥
−
4
)
(
8
𝑥
2
+
5
)
x open paren x minus 4 close paren open paren 8 x squared plus 5 close paren

x(x−4)(8x2−5)
𝑥
(
𝑥
−
4
)
(
8
𝑥
2
−
5
)
x open paren x minus 4 close paren open paren 8 x squared minus 5 close paren

x(x+4)(8x2−5)

Answer 1

To factor the polynomial \( 8x^4 - 32x^3 - 5x^2 + 20x \), we start by looking for common factors.

1. **Factor out the greatest common factor (GCF):**
The GCF of the terms is \( x \). So, we can factor \( x \) out:
\[
8x^4 - 32x^3 - 5x^2 + 20x = x(8x^3 - 32x^2 - 5x + 20)
\]

2. **Factor the cubic polynomial \( 8x^3 - 32x^2 - 5x + 20 \):**
We can use the rational root theorem or synthetic division to find potential rational roots. Testing \( x = 4 \):
\[
8(4)^3 - 32(4)^2 - 5(4) + 20 = 512 - 512 - 20 + 20 = 0
\]
Since \( x = 4 \) is a root, we can factor \( 8x^3 - 32x^2 - 5x + 20 \) using synthetic division by \( x - 4 \).

Performing synthetic division, we divide \( 8x^3 - 32x^2 - 5x + 20 \) by \( x - 4 \):
```
4 | 8 -32 -5 20
| 32 0 -20
-----------------
8 0 -5 0
```
The result is:
\[
8x^2 - 5
\]
So,
\[
8x^3 - 32x^2 - 5x + 20 = (x - 4)(8x^2 - 5)
\]

3. **Combining everything together:**
\[
x(8x^3 - 32x^2 - 5x + 20) = x(x - 4)(8x^2 - 5)
\]

Thus, the complete factorization of the original polynomial \( 8x^4 - 32x^3 - 5x^2 + 20x \) is:
\[
x(x - 4)(8x^2 - 5)
\]

**Final answer:**
\( x(x−4)(8x^2−5) \) is the correct choice.