$\displaystyle \displaystyle f(x)=-4.9x^2+19.6x+58.8$
I'm a little confused as using the quadratic formula would leave me with $\displaystyle x=6$ and $\displaystyle x=-2$, but $\displaystyle (x-6)(x+2)$ does not equal the original function.
$\displaystyle \displaystyle f(x)=-4.9x^2+19.6x+58.8$
I'm a little confused as using the quadratic formula would leave me with $\displaystyle x=6$ and $\displaystyle x=-2$, but $\displaystyle (x-6)(x+2)$ does not equal the original function.
Multiplying out will give you $\displaystyle x^2-4x-12$ and if you multiply this by a factor of -4.9 you get the original f(x) you had.
You could also do this the other way, you are looking for roots to the equation f(x)=0 so you get $\displaystyle -4.9 x^2 +19.6x+58.8=0$. Divide through by 4.9 and you get the equation I gave above.
If we assume that f(x)=0
Then we assume that:$\displaystyle -4.9x^2+19.6x+58.8=0$
My $\displaystyle -4.9$ times table is a little rusty, so allow me to tidy this up a little:
$\displaystyle -49x^2+196x+588=0$
Times both sides by $\displaystyle -1$:
$\displaystyle 49x^2-196x-588=0$
Divide both sides by $\displaystyle 49$:
$\displaystyle x^2-4x-12=0$
$\displaystyle (x-6)(x+2)=0$
$\displaystyle x=6$ or $\displaystyle x=-2$
If you substitute those back into the original function, you will find that they give a solution of $\displaystyle f(x)=0$.
If $\displaystyle f(x)=0$, then $\displaystyle 4.9f(x)=0$ too. Did you follow why?
Edit: Beaten again!