# Solve a cubic polynomial?

I've been having trouble with this question:

Solve the equation, $$5x^3 - 24x^2 + 9x + 54 = 0$$ given that two of its roots are equal.

I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: $a$, $a$, $b$, but I am still unsuccessful.

Any help would be greatly appreciated.

• Hint: if two roots are equal to $a$, then the derivative of your polynomial has $a$ as a root. To see why, just write $P=(x-a)^2 Q$ and differentiate. Hence, you have only an equation of degree 2 to solve. – Jean-Claude Arbaut May 10 '13 at 11:41
• Sorry, but I have to use a method not involving calculus. – missiledragon May 10 '13 at 11:44
• @missiledragon In that case you can use Horners Synthetic Division method. – lsp May 10 '13 at 11:45
• Then the simplest who be to develop $5(x-a)^2(x-b)$ and identify coefficients. – Jean-Claude Arbaut May 10 '13 at 11:45
• I got up to that, but I'm not sure how to identify the coefficients. – missiledragon May 10 '13 at 11:48

$$(x-a)^2(x-b) = (x^2-2ax+a^2)(x-b)=x^3-(2a+b)x^2+a(a+2b)x-a^2b$$ $$=x^3-\frac{24}{5}x^2+\frac{9}{5}x+\frac{54}{5}$$
Then $$2a+b=\frac{24}{5}$$ $$a+2b=\frac{9}{5a}$$
Multiply the first by 2 and subtract: $$3a=\frac{48}{5}-\frac{9}{5a}$$
Now it should not be difficult to find $a$. Of course this equation gives two possible values of $a$, and only one is right (you have to check with your degree three equation).