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# what is (7-4x)^6

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 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
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 26 = 64 (for each row of Pascal's Δ 2n)
Now replace a by 7 and b by (-4x)
76 - 6 · 75 · 4x + 15 · 74 · (-4x)2 - 20 · 73 · (4x)3 + 15 · 72 · (-4x)4 - 6 · 7 · (4x)5 + (-4x)6

(7 – 4x)^6

= 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

= 7^6 - (6*7^5)(4x) + (15*7^4)(16x^2) - (20*7^3)(64 x^3) + (15*7^2)(256 x^4) - (6*7)(1024 x^5) + 4096

The rest is arithmetic.

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.

3/19/2013

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

3/24/2013

You can use binomial expansion.

(7-4x)^6 = ∑{i = 0, 6} C(6, i) (7)^(6-i) (-4x)^i

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.

1(-4x)6(7)0= -4096x6

5(-4x)5(7)1= -35840x5

etc...