Question

the function $f(x) = 4x^4 - 4x^3 - 13x^2 + 5x + 10$ has at least two rational roots. use the rational root theorem to find those roots, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.)

answer attempt 1 out of 2

there is one root :

Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that any possible rational root, \( \frac{p}{q} \), of a polynomial equation \( a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0 = 0 \) is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \). For \( f(x)=4x^4 - 4x^3 - 13x^2 + 5x + 10 \), the constant term \( a_0 = 10 \) and the leading coefficient \( a_n = 4 \). The factors of \( 10 \) are \( \pm1,\pm2,\pm5,\pm10 \) and the factors of \( 4 \) are \( \pm1,\pm2,\pm4 \). So the possible rational roots are \( \pm1,\pm2,\pm5,\pm10,\pm\frac{1}{2},\pm\frac{5}{2},\pm\frac{1}{4},\pm\frac{5}{4} \).

Step2: Test Possible Rational Roots

• Test \( x = 1 \): \( f(1)=4 - 4 - 13 + 5 + 10 = 2

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• Test \( x = - 1 \): \( f(-1)=4 + 4 - 13 - 5 + 10 = 0 \). So \( x=-1 \) is a root.
• Test \( x = 2 \): \( f(2)=4\times16 - 4\times8 - 13\times4 + 5\times2 + 10 = 64 - 32 - 52 + 10 + 10 = 0 \). So \( x = 2 \) is a root.

Step3: Factor the Polynomial

Since \( x=-1 \) and \( x = 2 \) are roots, \( (x + 1)(x - 2) \) are factors. Multiply \( (x + 1)(x - 2)=x^2 - x - 2 \). Now divide \( f(x) \) by \( x^2 - x - 2 \):

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So \( f(x)=(x + 1)(x - 2)(4x^2 - 5) \)

Step4: Find Remaining Roots

Solve \( 4x^2 - 5 = 0 \). Then \( 4x^2=5 \), \( x^2=\frac{5}{4} \), \( x=\pm\frac{\sqrt{5}}{2} \)

Answer:

The rational roots are \( x=-1 \) and \( x = 2 \), and the complex (including real and irrational) roots are \( x=-1 \), \( x = 2 \), \( x=\frac{\sqrt{5}}{2} \), \( x=-\frac{\sqrt{5}}{2} \)