To factor the polynomial \(8x^4 - 32x^3 - 5x^2 + 20x\), we can start by factoring out the greatest common factor (GCF).
-
Factor out the GCF: The GCF of the coefficients \(8, -32, -5, 20\) is \(1\), and all terms contain at least one \(x\). So we can factor out \(x\):
\[ 8x^4 - 32x^3 - 5x^2 + 20x = x(8x^3 - 32x^2 - 5x + 20) \]
-
Factor the cubic polynomial \(8x^3 - 32x^2 - 5x + 20\): We will use synthetic division or polynomial long division to help find the factors. We can also use the Rational Root Theorem to test for possible rational roots. The possible rational roots for the cubic polynomial could be \( \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20\) (factors of \(20\) over \(8\), which is \(1, 2, 4, 8\)).
After testing, it turns out \(x = 2\) is a root:
Using synthetic division to divide \(8x^3 - 32x^2 - 5x + 20\) by \(x - 2\):
\[ \begin{array}{r|rrrr} 2 & 8 & -32 & -5 & 20 \ & & 16 & -32 & -10 \ \hline & 8 & -16 & -37 & 10 \ \end{array} \]
Therefore, we get:
\[ 8x^3 - 32x^2 - 5x + 20 = (x - 2)(8x^2 - 16x - 10) \]
-
Factor the quadratic polynomial \(8x^2 - 16x - 10\): Now we will factor \(8x^2 - 16x - 10\). We can use the quadratic formula since it might not factor nicely:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 8\), \(b = -16\), and \(c = -10\):
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 8 \cdot (-10)}}{2 \cdot 8} \]
\[ = \frac{16 \pm \sqrt{256 + 320}}{16} \]
\[ = \frac{16 \pm \sqrt{576}}{16} \]
\[ = \frac{16 \pm 24}{16} \]
This gives us two roots:
\[ x = \frac{40}{16} = \frac{5}{2} \quad \text{and} \quad x = \frac{-8}{16} = -\frac{1}{2} \]
So we can express \(8x^2 - 16x - 10\) as:
\[ 8(x - \frac{5}{2})(x + \frac{1}{2}) \]
-
Combine all factors: Putting it all together, we have:
\[ 8x^4 - 32x^3 - 5x^2 + 20x = x(x - 2)\left(8\left(x - \frac{5}{2}\right)\left(x + \frac{1}{2}\right)\right) \]
This simplifies to:
\[ = x(x - 2)\left(8x - 20\right)\left(2x + 1\right) \]
Finally, the polynomial can be factored fully into linear and quadratic factors as:
\[ = x(x - 2)(2x - 5)(2x + 1) \]
Answer: The linear factors are \(x\), \(x - 2\), \(2x - 5\), and \(2x + 1\). There are no quadratic factors remaining since the quadratic polynomial was factored completely.
1 answer