Question

what is the quadratic factor of the expression 4x^4-x^3+20x^2-5x

Answer 1

To factor the expression \( 4x^4 - x^3 + 20x^2 - 5x \), we can first take out the common factor, which is \( x \):

\[
x(4x^3 - x^2 + 20x - 5)
\]

Next, we need to factor the cubic polynomial \( 4x^3 - x^2 + 20x - 5 \). To factor a cubic polynomial, we can try synthetic division or factor by grouping.

Let's check for rational roots using the Rational Root Theorem. The potential rational roots can be factors of the constant term (-5) divided by the leading coefficient (4), giving us the possible roots:

\[
\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}
\]

Testing \( x = 1 \):

\[
4(1)^3 - (1)^2 + 20(1) - 5 = 4 - 1 + 20 - 5 = 18 \quad (\text{not a root})
\]

Testing \( x = -1 \):

\[
4(-1)^3 - (-1)^2 + 20(-1) - 5 = -4 - 1 - 20 - 5 = -30 \quad (\text{not a root})
\]

Testing \( x = \frac{1}{2} \):

\[
4\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 20\left(\frac{1}{2}\right) - 5 = 4\left(\frac{1}{8}\right) - \frac{1}{4} + 10 - 5 = \frac{1}{2} - \frac{1}{4} + 10 - 5 = \frac{1}{4} + 5 = \frac{21}{4} \quad (\text{not a root})
\]

Testing \( x = \frac{5}{4} \):

\[
4\left(\frac{5}{4}\right)^3 - \left(\frac{5}{4}\right)^2 + 20\left(\frac{5}{4}\right) - 5 = 4\left(\frac{125}{64}\right) - \frac{25}{16} + 25 - 5 = \frac{500}{64} - \frac{100}{64} + \frac{1600}{64} - \frac{320}{64} = \frac{500 - 100 + 1600 - 320}{64} = \frac{1680}{64} = 26.25 \quad (\text{not a root})
\]

Since none of these are roots, let's try to factor \( 4x^3 - x^2 + 20x - 5 \) using grouping.

Grouping the terms we get:

\[
(4x^3 - x^2) + (20x - 5)
\]

Factoring out common factors:

\[
x^2(4x - 1) + 5(4x - 1) = (4x - 1)(x^2 + 5)
\]

Now substituting that back in:

\[
x (4x - 1)(x^2 + 5)
\]

Thus, the quadratic factor we are interested in is:

\[
x^2 + 5
\]

So the complete factorization is:

\[
4x^4 - x^3 + 20x^2 - 5x = x(4x - 1)(x^2 + 5)
\]

The quadratic factor of the expression is **\(x^2 + 5\)**.