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.