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1

Use Pascal's Δ 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
(a + b)^{6} = a^{6} + 6a^{5}b + 15a^{4}b^{2} + 20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6}
1. If you notice, that coefficients are from the bottom of Pascal 's triangle.
2. The powers of a go down, from 6 to 0;
3. The exponents of b go up, from 0 to 6;
4. The sum of the powers of each therm equal 6;
5. If you multiply all coefficients you will get 2^{6} = 64 (for each row of Pascal's Δ 2^{n})
Now replace a by 7 and b by (-4x)
7^{6} - 6 · 7^{5} · 4x + 15 · 7^{4} · (-4x)^{2} - 20 · 7^{3} · (4x)^{3} + 15 · 7^{2} · (-4x)^{4} - 6 · 7 · (4x)^{5} + (-4x)^{6}

There were two answers from yesterday that were not posted when I posted my response. Had they been present I would not have replicated their efforts. The tedious nature of Wyzant withtheir sluggish responses caused this duplicity.

You can expand binomials using Pascal's Triangle to find the exponents

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1 1 5 10 10 5 1 n=6

This can be formed by starting with 1, then row 2 is 1,1, then the two ones in row two add to make the 2 in row 3, etc.

rewrite the binomial as (-4x+7)6 for simplicity's sake

We split the binomial into 2 terms and the exponent for x decreases while the other exponent increases, then multiply by the appropriate number in Pascal's Triangle.

## Comments

There were two answers from yesterday that were not posted when I posted my response. Had they been present I would not have replicated their efforts. The tedious nature of Wyzant withtheir sluggish responses caused this duplicity.

George- I feel the same way, sometimes I will submit an answer and it comes up instantly and sometimes it takes days. Very frustrating.