Having trouble solving this equation: (x + 1)(x^2 + 2) + (x+2)(x^2 + 1) = 2

[Alqus]

[Solved]Having trouble solving this equation: (x + 1)(x^2 + 2) + (x+2)(x^2 + 1) = 2

Hi, I have an equation and I can't find a way to solve it :

(x + 1)(x^2 + 2) + (x+2)(x^2 + 1) = 2

I tried expanding it, but no luck. I suspect that it should be grouped somehow, but numbers aren't exactly the same. Any help would be appreciated.

 

[stapel]

Hi, I have an equation and I can't find a way to solve it :

(x + 1)(x^2 + 2) + (x+2)(x^2 + 1) = 2

I tried expanding it, but no luck....

Please reply showing your work. You expanded the left-hand side, you subtracted 2 from each side, you applied the Rational Roots Test (here) to obtain a (short) list of possible roots, you plugged them into synthetic division (here), found one root, applied the Quadratic Formula to what remained, and... then what?

Please be complete. Thank you!

 

[Alqus]

After expanding, I've got 2x^3 + 3x^2 + 3x + 2 = 0.

But I don't know about Rational Roots Test, we haven't touched it in my class.

 

[Deleted member 4993]

After expanding, I've got 2x^3 + 3x^2 + 3x + 2 = 0.

But I don't know about Rational Roots Test, we haven't touched it in my class.

By inspection (or by rational root theorem), one of the roots of (2x^3 + 3x^2 + 3x + 2) is x = -1

Now continue....

 

[stapel]

After expanding, I've got 2x^3 + 3x^2 + 3x + 2 = 0.

But I don't know about Rational Roots Test, we haven't touched it in my class.

Then do a quick graph in your calculator to see where the line crosses the x-axis. When x = a is an intercept, then x - a is a factor, which you can divide out. (here)

 

[Alqus]

Do you maybe see another way to do this ? I don't feel very comfortable doing homework with a method we haven't studied yet.

 

[Deleted member 4993]

Do you maybe see another way to do this ? I don't feel very comfortable doing homework with a method we haven't studied yet.

Okay... let's step back then...

What are the methods that you have been taught in class to solve a cubic-equation?

 

[Alqus]

We haven't touched cubic equations at all. What we currently do sometimes is, for example :

x^3 + 2x^2 + x + 2 = 0
-->
x^2(x + 2) + (x + 2) = 0
-->
(x+2)(x^2 + 1) = 0
-->
x = -2

 

[stapel]

Do you maybe see another way to do this ? I don't feel very comfortable doing homework with a method we haven't studied yet.

Your class hasn't covered the relationship (usually taught back in "solving linear equations", maybe "by graphing") between x-intercepts, solutions, and factors? :shock:

 

[Alqus]

Your class hasn't covered the relationship (usually taught back in "solving linear equations", maybe "by graphing") between x-intercepts, solutions, and factors? :shock:

I haven't seen my teacher doing something like Rational Roots Test, and we haven't graphed a cubic equation yet. But I guess I can give it a try and learn it. My teacher is currently ill so I won't be able to ask soon how would she have done it.

 

[Alqus]

I tried another part of my homework, and got even more complicated equation, I am not sure how is teacher expecting us to do this :

(x(x + 2))^2 - (x+1)^2 = 55
expand -->
x^4 + 4x^3 + 3x^2 -2x - 1 = 55

I quickly checked over other exercises, and they all had a similar pattern : a number is lower or higher by 1.
Looking at my first equation I had problems with :

(x + 1)(x^2 + 2) + (x+2)(x^2 + 1) = 2

We can see the same pattern : (x + 1) and (x + 2), (x^2 + 2) and (x^2 + 1).
Is there something we can do with this ?

 

[Steven G]

After expanding, I've got 2x^3 + 3x^2 + 3x + 2 = 0.

But I don't know about Rational Roots Test, we haven't touched it in my class.

As already mentioned, x=-1 is a solution and this was obtained by inspection. Now if x=-1 is a root then x+1 is a factor. Now do long division or synthetic division to find the other factor which will be a quadratic equation. Then try to factor the quadratic.

 

[Alqus]

Ok, I was able to ask another teacher about this and he gave me this solution :

After expanding : 2x^3 + 3x^2 + 3x + 2 = 0 -->
2x^3 + 2x^2 + x^2 + 2x + x + 2 = 0 -->
2x^2(x + 1) + x(x + 1) + 2(x + 1) = 0 -->
(x + 1)(2x^2 + x + 2) = 0
x = -1

Thanks for everyone who tried to help me

 

[stapel]

Ok, I was able to ask another teacher about this and he gave me this solution :

After expanding : 2x^3 + 3x^2 + 3x + 2 = 0 -->
2x^3 + 2x^2 + x^2 + 2x + x + 2 = 0 -->
2x^2(x + 1) + x(x + 1) + 2(x + 1) = 0 -->
(x + 1)(2x^2 + x + 2) = 0
x = -1

This factorization (using the fact that "x = a" as a solution means "x - a" is a factor) requires that one know the solution -- which is what you're trying to find! Prior knowledge of the answer is required in order to know how to split the 3x^2 and 3x terms, in order to create the factorization necessary to find the answer.

In other words, now that you've been shown the "magic" method of "working backwards from the solution, which they won't be giving you on the test", can you use this "method" to answer any of the other questions? For instance, you say that the following is another of your exercises:

. . . . .x^4 + 4x^3 + 3x^2 -2x - 1 = 55

How would you apply the magic to this equation? If I whispered to you that a solution was x = 2, would you be able to apply the magic?

...which is why the instructor's "method" is not particularly helpful.

 

[Steven G]

I tried another part of my homework, and got even more complicated equation, I am not sure how is teacher expecting us to do this :

(x(x + 2))^2 - (x+1)^2 = 55
expand -->
x^4 + 4x^3 + 3x^2 -2x - 1 = 55

I quickly checked over other exercises, and they all had a similar pattern : a number is lower or higher by 1.
Looking at my first equation I had problems with :

(x + 1)(x^2 + 2) + (x+2)(x^2 + 1) = 2

We can see the same pattern : (x + 1) and (x + 2), (x^2 + 2) and (x^2 + 1).
Is there something we can do with this ?

"a number is lower or higher by 1" means think x= -1 or x=1. This will not always work as we also need the signs to cooperate.