So, I have a question in a textbook that asks the following: $$3x^3 + 4x^2 - x + m$$ It says this equation has two identical linear factors. It asks for the possible real values of m, and the zeroes for each case.
So what I figured out is that two identical linear factors would be $$(ax+b)^2$$ and FOILing that gets $$a^2x^2 + 2abx + b^2$$ From there I would have to figure out a binomial that when multiplied by the above gets the original equation. So I wrote it out like this: $$(a^2x^2 + 2abx + b^2)(cx+d) = 3x^3 + 4x^2 - x + m$$ After that, I mutliplied the two equations to get this massive, confusing thing: $$ca^2x^3 + da^2x^2 + 2a2bcx^2 + 2a2bdx + b^2cx + db^2 = 3x^3 + 4x^2 - x + m$$
So finally, I made a system of equations for this which turned out to be: $$2a2bd + b^2c = -1$$ $$ca^2 = 3$$ $$da^2 + 2a2bc = 4$$ $$db^2 = m$$
And this is where I'm stuck. I can't find a single variable, let alone m. Did I do something wrong? Is the system of equations even possible? I can elaborate on my process getting the equations if that helps.
$3x^3+4x^2−x+m$ is a polynomial of degree $3$, and thus should have three roots, counting duplicates. Since it has two identical linear factors, $$3x^3+4x^2−x+m=3(x-a)^2(x-b)$$ From then, expand and simplify to get $$x^3+\frac{4x^2}{3}−\frac{x}{3}+\frac{m}{3}=(x^3+(-2a-b)x^2+(a^2-2ab)x+a^2b)$$ Thus by comparing coefficients of $x$ and $x^2$, $(a,b)=(-1/3,-2/3),(-1/5,-14/15)$. These are the zeroes of the equation for $m=-2/9$ and $-14/125$ respectively.