Multiplicity roots q

[Xibu4]

If [math]ax^4+3x^3-48x^2+60x=0[/math] has a double root at x=2 find:
a) The value of a
b) the sum of the other two roots

so far I've only found that [math]\beta+\gamma =1/a[/math]I've tried plugging [math]\beta =\frac{1}{a} - \gamma[/math] into the sum of two roots at a time but to no avail

 

[BigBeachBanana]

Let [imath]\alpha,\beta,\gamma,\delta[/imath] be the roots of the 4th-degree polynomial. You were told that two of the roots are 2. Thus, [imath]\alpha=\beta=2[/imath].
Next, notice you can factor out an [imath]x[/imath]:
[math]ax^4+3x^3-48x^2+60x=x(ax^3+3x^2-48x+60)=0[/math]Clearly, [imath]x=0[/imath] is another root, therefore [imath]\gamma=0[/imath].
Now, apply the sum and product of roots to the cubic, which has the roots of 2,2, and [imath]\delta[/imath]. The problem reduces to a system of 2 equations and 2 unknowns [imath]a, \delta[/imath].
This solves both questions simultaneously.

 

[JeffM]

I would proceed a bit differently from BBB (not that he is in any way wrong). His way is more direct because it relies on the constant term being zero; mine is more general because it allows for the constant term to have any value.

[math]\text {Given } f(x) = ax^4 + 3x^3 - 48x^2 + 60x \text { and } (x - 2)^2 \ | \ f(x),\\ \text {Find } a \text { and the sum of the other two zeroes of } f(x).[/math]
First, I note that the problem as stated by the OP does not say that f(x) is a fourth degree polynomial. However, that makes “the sum of the other two roots” meaningless so I agree with BBB that we are to assume that f(x) is a quartic. Then I would apply the fundamental theorem of algebra to get

[math]ax^4 + 3x^3 - 48x^2 + 60x + 0 = (x^2 - 4x + 4)(\alpha x^2 + \beta x + \gamma).[/math]
Now I would equate coefficients. That technically gives a system of five equations, but two are soluble by inspection.