Solving for x

A

asissa

New member
Joined
Sep 6, 2009
Messages
26
This is the problem: 2x^3 + 3x^2 + 3 = 8x. I don't think my answer is right, but I'm at a loss. Can you show me where I'm going wrong? Thanks!

Answer 1
2x^3 + 3x^2 + 3 = 8x
2x^3 + 3x^2 - 8x + 3 = 0
(2x^3 + 3x^2) - (8x + 3)
x^2(2x + 3) + 1(-8x + 3)
(2x + 3) (x^2 + 1) (-8x+ 3)
x= -3/2,1,-1,3/8
 
Let's see...

1) 2x^3 + 3x^2 + 3 = 8x

You started with a cubic equation.

2) (2x + 3) (x^2 + 1) (-8x+ 3)

Somehow, this turned into a quartic expression.

Yup, that's very wrong.

Give it another go and be careful, this time.

RBGTHGANH
 
asissa said:
2x^3 + 3x^2 + 3 = 8x
2x^3 + 3x^2 - 8x + 3 = 0
(2x^3 + 3x^2) - (8x + 3)
x^2(2x + 3) + 1(-8x + 3)
(2x + 3) (x^2 + 1) (-8x+ 3)
x= -3/2,1,-1,3/8
Click to expand...
A cubic has 3, not 4 solutions.
Are you attending math classes?
 
First of all, yes, I knew it was wrong. I was looking for help as to where exactly I went wrong (hence, I showed my work). And second, ouch, but yes, you hit the nail on the head. I do not attend a math class. I know it has been said that you can't teach an old dog new tricks, but give me credit for trying LOL.
 
2x^3 + 3x^2 + 3 = 8x
2x^3 + 3x^2 - 8x + 3 = 0

Generally, line things up in descending powers of x. You did that.

2x^3 + 3x^2 - 8x + 3 = 0
(2x^3 + 3x^2) - (8x + 3)

Equations and expressions are not the same thing. Don't switch back and forth.

When you change just a piece, try a test value to see if you messed it up.
Let's take the last two terms and x = 2
-8x + 3 ==> -8(2) + 3 = -16 + 3 = -13
-(8x + 3) ==> -(8(2) + 3) = -(16 + 3) = -19
Whoops! Something went wrong.

(2x^3 + 3x^2) - (8x + 3)
x^2(2x + 3) + 1(-8x + 3)

The idea behind factoring is to take advantage of FACTORS. You must know the difference between a FACTOR and a TERM.
The distributive property of multiplication over addition is this: a*(b+c) = ab + ac

When you change just a piece, try a test value to see if you messed it up.
Let's take the last two terms and x = 2
-(8x + 3) ==> -(8(2) + 3) = -(16 + 3) = -19
+1(-8x + 3) ==> -8(2) + 3 = -16 + 3 = -13
Whoops! Something went wrong.

x^2(2x + 3) + 1(-8x + 3)
(2x + 3) (x^2 + 1) (-8x+ 3)

The distributive property of multiplication over addition is this: a*(b+c) = ab + ac

When you change just a piece, try a test value to see if you messed it up.
Let's take the whole expression and x = 1
x^2(2x + 3) + 1(-8x + 3) ==> 1^2(2(1) + 3) + 1(-8(1) + 3) = 1(2+3) + (-8+3) = 5 + (-5) = 0
(2x + 3) (x^2 + 1) (-8x+ 3) ==> (2(1) + 3) (1^2 + 1) (-8(1)+ 3) = (2+3)(1+1)(-8+3) = 5*2*(-5) = -50
Whoops! Something went wrong.

(2x + 3) (x^2 + 1) (-8x+ 3)
x= -3/2,1,-1,3/8

Equations and expressions are not the same thing. Don't switch back and forth.
Since you ahve no equation, your leap to solutions is quite illogical. Solutions to what?
The square root of -1 is NOT 1.

Lesson: You may wish to start smaller.
 
Hello, asissa!

\(\displaystyle \text{Solve: }\:2x^3 + 3x^2 + 3 \:=\: 8x\)

\(\displaystyle 2x^3 + 3x^2 + 3 \:=\: 8x\)

\(\displaystyle 2x^3 + 3x^2 - 8x + 3 \:=\: 0\)

\(\displaystyle (2x^3 + 3x^2) - (8x + 3)\:=\:0\)

\(\displaystyle x^2(2x + 3) + 1(-8x + 3) \:=\:0\)

\(\displaystyle (2x + 3) (x^2 + 1) (-8x+ 3) \:=\:0\) . ??
If you factor like that, you missed more than "a math class".
Click to expand...

\(\displaystyle \text{We have: }\:2x^3 + 3x^2 - 8x + 3 \:=\:0\)

\(\displaystyle \text{Note that }x = 1\text{ is a root of the equation.}\)
. . \(\displaystyle \text{Hence, }(x-1)\text{ is a factor.}\)

\(\displaystyle \text{We find that: }\:2x^3 + 3x^2 - 8x + 3 \:=\:(x-1)(2x^2+5x-3)\)

. . \(\displaystyle \text{which factors to: }\:(x-1)(2x-1)(x+3)\)

\(\displaystyle \text{Therefore: }\:x \;=\;1,\:\tfrac{1}{2},\:\text{-}3\)

 
Thank you. Obviously I'm having a bad math day, to say the least. My mistakes were glaringly obvious once you both spelled it out. I appreciate your help.
 
Great link, thanks. I've been using those lessons quite a bit - they are great. As you can tell from my work, that was the direction I was going (however wrongly :D ). However, when I didn't get a like possible factor, I froze and totally lost my way. I also intuitively see where Soroban shows that 1 is a factor in the equation, and go from there. I guess at the root of my problem is when I see a cubic and it doesn't easily factor (as shown in the clip) and say the numbers are larger than in the original equation and a factor cannot be intuitively worked out, I blank. I need to go back and work on it quite a bit more obviously!
 
asissa said:
Great link, thanks. I've been using those lessons quite a bit - they are great. As you can tell from my work, that was the direction I was going (however wrongly :D ). However, when I didn't get a like possible factor, I froze and totally lost my way. I also intuitively see where Soroban shows that 1 is a factor in the equation, and go from there. I guess at the root of my problem is when I see a cubic and it doesn't easily factor (as shown in the clip) and say the numbers are larger than in the original equation and a factor cannot be intuitively worked out, I blank. I need to go back and work on it quite a bit more obviously!
Click to expand...

Don't fret too much over factorizing cubics or quatrics - by hand, those can become quite tedious. There are numerical methods - that can quickly give you answers.
 
You must log in or register to reply here.
Top