Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (animelover98):

Expand the following using the binomial theorem and Pascal's triangle. (x + 2)6

OpenStudy (anonymous):

1 121 1331 so on

OpenStudy (anonymous):

x^6 must be the first term

OpenStudy (anonymous):

really finish the triangle to the 6th row what do you get?

OpenStudy (animelover98):

I don't understand. The triangle is 1 121 1331?

OpenStudy (anonymous):

okay well the triangle is jjust a tool to quickly solve this problem

OpenStudy (anonymous):

i strongly suggest to google a explanation of the triangle, as this will save you hundreds of hours.

OpenStudy (animelover98):

So, I would expand the triangle to the 6th row because (x+2) is raised to the 6th?

OpenStudy (animelover98):

I understand how to get the triangle now, thank you. But how do I solve it?

OpenStudy (animelover98):

1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 5 1 1 6 15 20 6 1

OpenStudy (perl):

the sixth row will be the coefficients

OpenStudy (animelover98):

Ah ok

OpenStudy (animelover98):

So (1x+2) (6x+2) ect?

OpenStudy (anonymous):

so for example so that I don't answer the question for you ill do (x+2)^5

OpenStudy (anonymous):

actually ill do (x+y)^5

OpenStudy (anonymous):

x^5+x^4y+x^3y^2+x^2y3+xy^4+y^5

OpenStudy (anonymous):

that's the answer to my example^^^

OpenStudy (anonymous):

thats without the constants inforont of variables.

OpenStudy (anonymous):

I only showed how to do the coefficients

OpenStudy (anonymous):

x^5+5x^4y+10x^3y^2+10x^2y3+5xy^4+y^5 is the full answere

OpenStudy (animelover98):

Ok So I would do x^6+12x^5+60x^4+160x^3+240^2+192x+64?

OpenStudy (anonymous):

I think you did it right but look at your triangle its wrong 5th row should have extra 10

OpenStudy (anonymous):

this then messes up your 6th row

OpenStudy (animelover98):

Oh ok!

OpenStudy (anonymous):

the triangle method is never required to do, but just know that later in your math career its probably best for you to rememorize it and utilize it, as it can save you tons of time.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!