Math question about reducing + my attempt but something is wrong

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The discussion revolves around simplifying the expression f´(x) = (12x^3 + 12x)(1/2 x^2 + 3) + (3x^4 + 6x^2 + 1)(x). Participants emphasize the importance of expanding brackets using the distributive property and combining like terms. Clarifications are provided on how to handle terms with exponents, specifically how to multiply and add polynomials correctly. The conversation highlights the need for understanding algebraic rules, such as the product of powers and the addition of like terms, to achieve the correct simplification. Overall, the thread serves as a helpful resource for mastering polynomial manipulation techniques.
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Homework Statement



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

Homework Equations



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

The Attempt at a Solution


ive tried to reduce it this way but i don't 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
 
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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.

<br /> a*(b+c+d)+b*(a+g+h) = ab+ac+ad+ab+ag+ah = 2ab + ac + ad + ag + ah<br />

That's almost all you have to do.
 
DivisionByZro said:
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.

<br /> a*(b+c+d)+b*(a+g+h) = ab+ac+ad+ab+ag+ah = 2ab + ac + ad + ag + ah<br />

That's almost all you have to do.

yes i know the rules you have posted but my problem is that i can't only say a*(b+c) since i have a*(b^n + c^n)
 
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:

<br /> x^{2}(3x^{4}+6x^{3}-\frac{1}{2}x^{2}) = 3x^{6}+6x^{5}-\frac{1}{2}x^{4}<br /> <br />
 
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mimi.janson said:
yes i know the rules you have posted but my problem is that i can't only say a*(b+c) since i have a*(b^n + c^n)

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
 
Ray Vickson said:
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

so if we just use numbers as an example and let's 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 don't 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 don't know what (3x^4)*x is should i say 4x^4 or 3x^5 ? i don't know if my x goes to the 3x or to the ^4
 
The rule is:

<br /> ax^{n}\cdot x^{m} = ax^{n+m}<br />

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).
 
DivisionByZro said:
The rule is:

<br /> ax^{n}\cdot x^{m} = ax^{n+m}<br />

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).

then it means my (3x^4+6x^2+1)*(x)
becomes (3x^4+6x^2+1) since there isn't any x^n but only x on the right side
 
mimi.janson said:
so if we just use numbers as an example and let's 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 don't 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 don't know what (3x^4)*x is should i say 4x^4 or 3x^5 ? i don't know if my x goes to the 3x or to the ^4

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 don't 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
 
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  • #10
mimi.janson said:
so if we just use numbers as an example and let's 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 don't 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 don't know what (3x^4)*x is should i say 4x^4 or 3x^5 ? i don't know if my x goes to the 3x or to the ^4

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 don't 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
Ray Vickson said:
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 don't 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

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
mimi.janson said:
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

Almost there.
<br /> 5x^{3}\cdot4x^{2} = (4\cdot5)x^{3+2} = 20x^{5}<br />

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

Adding:

<br /> <br /> 3x + 2x + 2x^{2} - x^{2} = (3x+2x)+(2x^{2} - x^{2}) = 5x + x^{2}<br /> <br />

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}

But you can add/subtract these:

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*} <br /> &amp; x(1+3x+2x^{2}-3x^{4}) + 2x^{2}(1-\frac{1}{2}x^{2}) \\<br /> &amp; = (x\cdot1)+(x\cdot3x)+(x\cdot2x^{2})+(x\cdot -3x^{4})+(2x^{2}\cdot1)+ (2x^{2}\cdot \frac{-1}{2}x^{2}) \\<br /> &amp; = x+3x^{1+1}+2x^{2+1}-3x^{1+4} + 2x^{2} - (2\cdot \frac{1}{2})(x^{2+2}) \\<br /> &amp; = x+ 3x^{2} + 2x^{3} - 3x^{5} + 2x^{2} - x^{4} \\ <br /> &amp; = x + 5x^{2} + 2x^{3} -x^{4} -3x^{5} \\ <br /> \end{align*}

Does this help?
 
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  • #13
DivisionByZro said:
Almost there.
<br /> 5x^{3}\cdot4x^{2} = (4\cdot5)x^{3+2} = 20x^{5}<br />

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

Adding:

<br /> <br /> 3x + 2x + 2x^{2} - x^{2} = (3x+2x)+(2x^{2} - x^{2}) = 5x + x^{2}<br /> <br />

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}

But you can add/subtract these:

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:

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

Does this help?

Does it help ? No it doesn't 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 I am 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
You're quite welcome! :smile:
People here are always happy to help you as long as you've shown some effort. Good luck!
 
  • #15
DivisionByZro said:
You're quite welcome! :smile:
People here are always happy to help you as long as you've shown some effort. Good luck!

thank you a lot :=)
 
  • #16
DivisionByZro said:
Now for multiplying:
3x^{2} \cdot 3x = (3)\cdot x^{2+1} = 3x^{3}
[/color]
I make this 9x3
 
  • #17
NascentOxygen said:
[/color]
I make this 9x3

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

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