Final answer to the problem
Step-by-step Solution
How should I solve this problem?
- Choose an option
- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
- Integrate by substitution
- Integrate by parts
- Integrate using tabular integration
- Integrate by trigonometric substitution
- Weierstrass Substitution
- Prove from LHS (left-hand side)
- Load more...
We can factor the polynomial $2x^4+x^3-8x^2-x+6$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $6$
Learn how to solve factorization problems step by step online.
$1, 2, 3, 6$
Learn how to solve factorization problems step by step online. Factor the expression 2x^4+x^3-8x^2-x+6. We can factor the polynomial 2x^4+x^3-8x^2-x+6 using the rational root theorem, which guarantees that for a polynomial of the form a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 there is a rational root of the form \pm\frac{p}{q}, where p belongs to the divisors of the constant term a_0, and q belongs to the divisors of the leading coefficient a_n. List all divisors p of the constant term a_0, which equals 6. Next, list all divisors of the leading coefficient a_n, which equals 2. The possible roots \pm\frac{p}{q} of the polynomial 2x^4+x^3-8x^2-x+6 will then be. Trying all possible roots, we found that -2 is a root of the polynomial. When we evaluate it in the polynomial, it gives us 0 as a result.
