Question

Use Pascal's triangle to expand (x-7)^5

Answer 1

You need the 6th row of the triangle.

Answer 2

you can use the general formula for this, which is
\[n!/r!(n-r)!\]
where here n=5 (as it is the exponent of the bracket)
and r = 0 to 5, where for the first term it equals 0, 2nd term it = 1, ...
this helps you find the co-efficients of each of the terms
i.e. the whole nth row of the triangle

Answer 3

Those numbers will be the coefficients of the following terms: x^5(-7)^0 x^4(-7)^1 x^3(-7)^2 x^2(-7)^3 x^1(-7)^4 x^0(-7)^5

Answer 4

Because that row has 1,5,10,10,5,1 and those are the coeffiecients.

Answer 5

Also the x (or first term in a binomial) exponent goes down left to right and the -7 (or second term in the binomial) goes up from left to right.

Answer 6

Yes. Those are the coefficients

Answer 7

So you get 1x^5 -35x^4+490x^3-3430x^2+12005x-16807

Answer 8

can we do it step by step I cant understand how this triangle works

Answer 9

\[___x ^{5}(-7)^{0}+___x ^{4}(-7)^{1}+___x ^{3}(-7)^{2}+___x ^{2}(-7)^{3}+___x ^{1}-7^{4}+___x ^{0}(-7)^{5}\]

Answer 10

Oh well. I don't know why that didn't write it right. I'll try again.

Answer 11

\[x ^{5}(-7)^{0}+ x ^{4}(-7)^{1}+ x ^{3}(-7)^{2}+ x ^{2}(-7)^{3}+ x ^{1}(-7)^{4}+ x ^{0}(-7)^{5}\]

Answer 12

Now if you simplify you will get:\[x ^{5}-7x ^{4}-49x ^{3}-343x ^{2}+2401x ^{1}-16807\]

Answer 13

Now put it the coefficients from Pascal's Triangle.

Answer 14

\[(1)x ^{5}-(5)(7x ^{4})+(10)(49x ^{3})-(10)(343x ^{2})+(5)(2401x ^{1})-(1)(16809)\]

Answer 15

is this right? coeficients for power 5 corresponds to row 5:
1, 5, 10, 10, 5, 1

(1)x^5 + 5(x^4)(-7) + 10x^3(-7)^2 + 10x^2(-7)^3 + 5x(-7)^4 + (1)(-7)^5

x^5 - 35x^4 + 490x^3 - 3430x^2 + 12005x - 16807

Answer 16

I messed up the sign on the 49 a couple steps back.

Answer 17

can you check what I did

Answer 18

Well, those are the right coefficients but that is actually row 6 since the triangle is: