Math question

[pchan]

Factorized completely 2x^3 + 3x^2 - 23x - 12
Hence solve the equation x^3 + 3x^2 -46x - 48

The solution for the first equation is (2x+1)(x+4)(x-3).
The solution for the second equation is (x+1)(x+8)(x-6).

The problem is that I am having trouble using the first equation to selve the second.

Thanks in advance for your help.

[DB83]

I do not see even ONE equation {confused}

[deadrats]

as DB83 alluded to, you do not have 2 equations, you have 2 expressions, equations as the name implies require that an expression be equal to some value, for instance if both expressions had a "=1" at the end of them then they would be equations, since neither is equal to some value there is no way to isolate the variable thus you can't use one to solve the other.

[pchan]

Sorry for the confusion. Please allow me to clarify.

f(x)=2x^3 + 3x^2 - 23x - 12. When f(x)=0, the solutions are (2x+1)(x+4)(x-3)=0, i.e. x=-1/2, x=-4, and x=3.

Given that f(x)=x^3 + 3x^2 -46x - 48, solves f(x)=0 using the above solution. That is, I have to modify the above solution so that I can find the roots of this cubic equation. Of course, I could have find the roots directly. But the question specifically stated that I have to utilize the first cubic equation. This really stumps me.

Thanks.

[lordsmurf]

The answer is always 42.

[DB83]

The answer is always 42.

But only if the square root of the hippopotamus is stuck in a mud-hole much to small for him.

Sorry.

My knowledge of algebra left me many years ago. I did google quadratic equations but soon got a head-ache.

[gadgetguy]

The answer is always 42.

Unless you forget your towel. (Then it doesn't matter 'cause you're doomed.)

[pchan]

Hi folks,
Made some progress by using graph.
f1(x)=2x^3 + 3x^2 - 23x - 12
f2(x)=x^3 + 3x^2 -46x - 48
When I take the first cubic equation and subtract by the second equation, I have this.
f3(x)=x^3+23x+36.

If I let f1(x)=f2(x), then I will get f2(x). The x-axis locations where f1(x) and f3(x) intersect, are the x-axis location of f2(x) crossing the x-axis.


Just having trouble solving it algebraically.

[lordsmurf]

The answer is always 42.

Unless you forget your towel. (Then it doesn't matter 'cause you're doomed.)

I always forget that one.

However, you may be able to swing a squirrel by its tail, and calculate the square root of the distance of the throw by the length of tail, and then divide by 2. You'll end up with a number divisible into 42, which halts the doom clock.

Or cut the red wire.