 # A quartic polynomial has 2 distinct real roots at x=1 and x=−3/5. If the function has a y-intercept at−1and has f(2)=2 and f(3)=3. How to determine the remaining roots when two distinct real roots and y-intercept are given. Khalfanebw 2022-09-19 Answered
A quartic polynomial has $2$ distinct real roots at $x=1$ and $x=-3/5$. If the function has a $y$-intercept at $-1$ and has $f\left(2\right)=2$ and $f\left(3\right)=3$. How to determine the remaining roots when two distinct real roots and y-intercept are given.
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You can sum up everything you know in a system like this:
$\left\{\begin{array}{l}f\left(1\right)=0\\ f\left(-\frac{3}{5}\right)=0\\ f\left(0\right)=-1\\ f\left(2\right)=2\\ f\left(3\right)=3\end{array}$
Then it's very simple to find all the coefficients of $f\left(x\right)=a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e$, because you have a $5$-variable system and $5$ equations, so the solution is unique! For example, from the $3$rd equation we get $f\left(0\right)=0+0+0+0+e=-1⇒e=-1$, and so on. After some calculus we obtain:
$\left\{\begin{array}{l}a=\frac{5}{156}\\ b=-\frac{41}{78}\\ c=\frac{289}{156}\\ d=-\frac{14}{39}\\ e=-1\end{array}$
Finally we manage to find the polynomail, and the associated function:
$f\left(x\right)=\frac{5}{156}{x}^{4}-\frac{41}{78}{x}^{3}+\frac{289}{156}{x}^{2}-\frac{14}{39}x-1$
So all the real roots are ${x}_{1}=-\frac{3}{5},{x}_{2}=1,{x}_{3}=8-2\sqrt{3},{x}_{4}=8+2\sqrt{3}$.