Great. But here's another thought. Did we need to find the roots of the first equation? What if the roots were something like $4+3i$ and $4-3i$? That wouldn't be a neat numbers to use in a synthetic division. Is there anyway we can do this problem WITHOUT using synthetic division?

We could simply do a long division $\frac{2x^3+9x^2-6x-5}{x^2+4x-5}=2x+1$

Yes. Although it doesn't hurt you to find the roots, frt's method is quicker and what the CMO solution said (and how I did.. I actually did frt's method and used synthetic division to check my answer).

i think that's similar to what 236factorial did.

That's not a strange way. It is pretty clever actually. Noticing the first equation is MONIC (leading coefficent is 1) and the second equation has coefficent of 2 helps you to know that it's in form of $(2x+d)$. From here, like you did, it's not that hard to figure it out. Good job!

Yeah, I think that's actually faster than doing a long division or synthetic division.

for synthetic division I thought that you can only divide by (x-a), how do you divide by $x^2+4x+5$?

ajai wrote: for synthetic division I thought that you can only divide by (x-a), how do you divide by $x^2+4x+5$? Do you mean $x^2+4x-5$? Factoring, that makes (x+5) and (x-1). Now you can use synthetic division.

I used Long Division.

Silverfalcon wrote: The roots of the equation $x^2+4x-5 = 0$ are also the roots of the equation $2x^3+9x^2-6x-5 = 0$. What is the third root of the second equation? Please post full solution to the problem. If you consider this problem easy, leave to other people who can benefit from trying this. $x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$.

Let me add that this question can be solved by another method too. Solution by Viete formulas 1) Sum of roots of quadratic is -4. Sum of roots of cubic is $-4+r=-\frac92$. Third root is $r=-\frac12$ 2) Product of roots of quadratic is -5. Product of roots of cubic is $-5r=\frac52$. Third root is $r=-\frac12$

Iversonfan2005 wrote: $x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$. I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything.

AntonioMainenti wrote: Iversonfan2005 wrote: $x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$. I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything. Does it really matter?

Iversonfan2005 wrote: AntonioMainenti wrote: Iversonfan2005 wrote: $x^2+4x-5 \Longrightarrow (x-1)(x+5) \Longrightarrow x=1, -5$, now use one of the roots for the second equation with synthetic division, I chose $1$ and then I factored the second equation to: $(x-1)(2x^2+11x+5) \Longrightarrow 2x^2+11x+5 \Longrightarrow (2x+1)(x+5) \Longrightarrow x=-5,\boxed{-\frac{1}{2}}$. I hate to say this over and over, but this is a misuse of the $\Longrightarrow$ symbol. Statements imply other statements. An expression on its own cannot imply anything. Does it really matter? It doesn't matter a lot if used in forums, but I think it matters elsewhere OK here is one suggestion to use it better: $x^2+4x-5=(x-1)(x+5)=0\Longrightarrow x=1, -5$ $(x-1)(2x^2+11x+5)=0\Longrightarrow x-1=0\text{ or }2x^2+11x+5=0\Longrightarrow x=1\text{ or }(2x+1)(x+5)=0\Longrightarrow x=1,-5,\boxed{-\frac{1}{2}}$

Iversonfan2005 wrote: Does it really matter? Well, you're using an implication symbol where you really mean $=$. What if someone used $+$ where they meant $-$, or $\in$ where they meant $\le$. It's not any different to me at least. Anyway, I'm sure your math professors won't like it either, as 10000th User hinted at.

what is a third root.

anirudh wrote: what is a third root. Cubic equations have $3$ roots. In the question it gives you the first two roots (same as the quadratic equation's) and asks for the remaining (3rd) root.

What is synthetic divisioin?

anirudh wrote: What is synthetic divisioin? It's a way of factoring by using trial and error using the last term over the first term to find the roots.

but i don't like trial and error.

anirudh wrote: but i don't like trial and error. You must like spamming though.. Go read about synthetic division, it's very useful.

Iversonfan2005 wrote: anirudh wrote: but i don't like trial and error. You must like spamming though.. Go read about synthetic division, it's very useful. Synthetic division isn't all about guess and checking though (especially when you have a range of numbers using another theroem I can't remember, rational root theorem?). In this case, you knew the roots, so synthetic division is perfect. Just remember that the divisor has to be in the form x-a.

the equation is x^2 +4x-5 roots r 1 and -5 concerning 2nd equation 2x^3+9x^2-6x-5=0 the sum of the roots is 1-5+p=4.5 p-4=4.5 therefore p=9.5 ( i may be mistaken though if p-4 =-4.5 then p=-0.5)

*Casually revives 14-year-old thread*

if you have a new solution to a problem, it is encouraged to share it.

...But it's not new...

The simplest solution: Product of roots in cubic =$\frac{-5}{2}$ Product of roots in quadratic = $5$ Remaining root = $\frac{\frac{-5}{2}}{5}= \frac{-1}{2}$