# Math question about reducing + my attempt but something is wrong

1. Jan 29, 2012

### mimi.janson

1. The problem statement, all variables and given/known data

reduce this one

f´(x)=(12x^3+12x)*(1/2 x^2+3)+(3x^4+6x^2+1)*(x)

i have to make all the () disappear and make it shorter

2. Relevant equations

the thing i have to use are the normal rules about ^n

3. The attempt at a solution
ive tried to reduce it this way but i dont know why i just feel its wrong so if someone could show me how to do step by step

f´(x)=6x^5+36x^3+6x^2+36+4x^4+7x^2+x

2. Jan 29, 2012

### DivisionByZro

Your answer is not correct. It'll be easier to help you if you show us exactly what you've tried. It's a straightforward problem, just expand the brackets with FOIL and combine like terms.

$$a*(b+c+d)+b*(a+g+h) = ab+ac+ad+ab+ag+ah = 2ab + ac + ad + ag + ah$$

That's almost all you have to do.

3. Jan 29, 2012

### mimi.janson

yes i know the rules you have posted but my problem is that i cant only say a*(b+c) since i have a*(b^n + c^n)

4. Jan 29, 2012

### DivisionByZro

I can't really post a solution, it's against forum rules. Show us what you've done, and we can see where you went wrong. Here's a better example:

$$x^{2}(3x^{4}+6x^{3}-\frac{1}{2}x^{2}) = 3x^{6}+6x^{5}-\frac{1}{2}x^{4}$$

Last edited: Jan 29, 2012
5. Jan 29, 2012

### Ray Vickson

Of course you can: just set B = b^n and C = c^n; now you have a*(B+C). You need to realize that in rules such as a*(b+c) = a*b + a*c, the a, b and c can be ANY algebraic or numerical quantities.

RGV

6. Jan 29, 2012

### mimi.janson

so if we just use numbers as an example and lets say
a*(b^n + c^n)

2*(3^4+5^6) then that mean you are telling me that usally you have to solve the ^ first so it ends up being 2*(81+15625) ? and then 162+31250 ? but it seems so strange how i can use it since i have like x^n ? i dont really know what x is anyway

f´(x)=(12x^3+12x)*(1/2 x^2+3)+(3x^4+6x^2+1)*(x)

when i look at the (3x^4+6x^2+1)*(x)
i dont know what (3x^4)*x is should i say 4x^4 or 3x^5 ? i dont know if my x goes to the 3x or to the ^4

7. Jan 29, 2012

### DivisionByZro

The rule is:

$$ax^{n}\cdot x^{m} = ax^{n+m}$$

Now you can use this to expand your brackets, making sure to preserve the order of operations (Sometimes remembered using BEDMAS : brackets, exponents, division/multiplication, addition/subtraction).

8. Jan 29, 2012

### mimi.janson

then it means my (3x^4+6x^2+1)*(x)
becomes (3x^4+6x^2+1) since there isnt any x^n but only x on the right side

9. Jan 29, 2012

### Ray Vickson

No, I never said what you claim I am saying. I say that 2*(3^4+5^6) equals 2*(3^4) + 2*(5^6). Whether or not you choose to put 3^4 = 81 and/or 5^6 = 15625 before, or after the expansion, or not at all, is purely a matter of personal choice.

I don't see you problem with having x^n; x^n is just another algebraic quantity, no better or worse than plain x itself. In your example y = x*(3x^4 + 6x^2 + 2), you just expand each term separately, so y = x*(3x^4) + x*(6x^2) + x*(2) = 3x^5 + 6x^3 + 2x. That's all there is to it.

Finally, your question "i dont know what (3x^4)*x is" puzzles me. You just keep all the x's together and all the numbers separate, so x*(3x^4) = 3 x x^4 = 3x^5. You most certainly cannot get (3x^4)*x = 4x^4 as you suggest; just plug in a value such as x = 1 or x = 2 and see if it works (t doesn't).

RGV

Last edited: Jan 29, 2012
10. Jan 29, 2012

### Ray Vickson

No, I never said what you claim I am saying. I say that 2*(3^4+5^6) equals 2*(3^4) + 2*(5^6). Whether or not you choose to put 3^4 = 81 and/or 5^6 = 15625 before, or after the expansion, or not at all, is purely a matter of personal choice.

I don't see you problem with having x^n; x^n is just another algebraic quantity, no better or worse than plain x itself. In your example y = x*(3x^4 + 6x^2 + 2), you just expand each term separately, so y = x*(3x^4) + x*(6x^2) + x*(2) = 3x^5 + 6x^3 + 2x. That's all there is to it.

Finally, your question "i dont know what (3x^4)*x is" puzzles me. You just keep all the x's together and all the numbers separate, so x*(3x^4) = 3 x x^4 = 3x^5. You most certainly cannot get (3x^4)*x = 4x^4 as you suggest; just plug in a value such as x = 1 or x = 2 and see if it works (it doesn't).

RGV

11. Jan 29, 2012

### mimi.janson

ok sorry i misunderstood it

if i have x*2x^6 = 2x^7 which means i just add it to the 6 right and if i have 5x^3*4x^2 itl make 20x^6 ? i really liked your last explanation thank you alot

12. Jan 29, 2012

### DivisionByZro

Almost there.
$$5x^{3}\cdot4x^{2} = (4\cdot5)x^{3+2} = 20x^{5}$$

Here's the difference between adding polynomials and multiplying them:

$$3x + 2x + 2x^{2} - x^{2} = (3x+2x)+(2x^{2} - x^{2}) = 5x + x^{2}$$

All the ones with equal powers (Ex: 2x and 3x have equal powers, 1 and 1: 2x=2x^1) are grouped together.
You can add and subtract monomials/polynomials with the same base (i.e. x^{2} and 2x have the same base, x) AND the same powers. So:

You can't add/subtract these two any more than this:

$$2x - 3x^{2}$$

$$3x+4x$$

Because they have the same base and the same power.

Now for multiplying:

$$3x^{2} \cdot 3x = (3)\cdot x^{2+1} = 3x^{3}$$

You can multiply two monomials as long as they have the same base ("x" in this example). You cannot further simplify/multiply these two:

$$a^{2} and x^{3}$$ because they aren't the same monomials.

Here's a final example to summarize all of this:

\begin{align*} & x(1+3x+2x^{2}-3x^{4}) + 2x^{2}(1-\frac{1}{2}x^{2}) \\ & = (x\cdot1)+(x\cdot3x)+(x\cdot2x^{2})+(x\cdot -3x^{4})+(2x^{2}\cdot1)+ (2x^{2}\cdot \frac{-1}{2}x^{2}) \\ & = x+3x^{1+1}+2x^{2+1}-3x^{1+4} + 2x^{2} - (2\cdot \frac{1}{2})(x^{2+2}) \\ & = x+ 3x^{2} + 2x^{3} - 3x^{5} + 2x^{2} - x^{4} \\ & = x + 5x^{2} + 2x^{3} -x^{4} -3x^{5} \\ \end{align*}

Does this help?

Last edited: Jan 29, 2012
13. Jan 29, 2012

### mimi.janson

Does it help ? No it doesnt help

it does help me in the most perfect way ever !....i am so much sure that your explanation is even better than my book !! i really like the exampels you gave couse you can relate more instead uf just heavy rules ...its easier to understand and remember this way so i thank you very much for your help and assure you that im going to use your explanation as my personal notes to remember and train them so i can learn it all :=) thank you a thousand times again

14. Jan 29, 2012

### DivisionByZro

You're quite welcome!
People here are always happy to help you as long as you've shown some effort. Good luck!

15. Jan 29, 2012

### mimi.janson

thank you alot :=)

16. Jan 30, 2012

### Staff: Mentor

I make this 9x3

17. Jan 31, 2012

### DivisionByZro

Oops typo. Thanks for spotting it! However, for some reason I can't go back and edit it.