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Mathematics 19 Online
OpenStudy (alones.):

x 3(x 2 + 5x + 1)

OpenStudy (math_man21):

well she said it (or he)

OpenStudy (alones.):

Oh, umm thanks but i odn't see that as my multiple choice :3 x^6 + 5x ^4 + x 3<--- wrong -.- x ^5 + 5x ^4 + x 3 x ^5 + 5x ^3 + x 3

OpenStudy (alones.):

**don't

OpenStudy (phi):

x^3 means x*x*x x^2 means x*x if you distribute the x^3 times each term in side the parens, the first multiplication is x^3 * x^2 which is x*x*x*x*x what is that using "exponents" ?

OpenStudy (alones.):

x^5

OpenStudy (phi):

so the answer will have x^5 in it now the next term is x^3 * 5*x or if we rearrange that: 5* x*x*x * x what is that using exponents

OpenStudy (alones.):

5x^4?

OpenStudy (phi):

yes

OpenStudy (alones.):

Oh wait..so i add them like this

OpenStudy (phi):

the last term is x^3 *1 or x*x*x*1 or just x*x*x or x^3

OpenStudy (alones.):

Oh i see that would be x ^3

OpenStudy (phi):

yes, and the whole thing is x^5 + 5x^4 + x^3

OpenStudy (alones.):

so i add them all Oh at first i thought i had to subtract... thank you

OpenStudy (alones.):

thank you the steps actualy helped.

OpenStudy (phi):

There are rules, and if you get used to them, you can do these problems.

OpenStudy (alones.):

Yeah i need to memorize them

OpenStudy (phi):

better is to understand why they work for example, with numbers, if you had 2*(1+3) that means you have "two" of everything inside the parens (think of the ( ) as a package) in other words you have 2 1's written 2*1 and two threes: 2*3 2*(1+3) = 2*1 + 2*3 that is "distributing" the 2 and that rule also works for letters. x(1+3) = 1*x + 3*x

OpenStudy (alones.):

Oh yeah the only step i remeber always is distubite..

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