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We can factor the polynomial $y^3+15y^2+76y+140$ 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 $140$
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$1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140$
Learn how to solve problems step by step online. Find the integral piint((y(y^2+12y+61))/(y^3+15y^276y+140))dy&-6&k. We can factor the polynomial y^3+15y^2+76y+140 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 140. Next, list all divisors of the leading coefficient a_n, which equals 1. The possible roots \pm\frac{p}{q} of the polynomial y^3+15y^2+76y+140 will then be. Trying all possible roots, we found that -7 is a root of the polynomial. When we evaluate it in the polynomial, it gives us 0 as a result.
