Pascal's Triangle:
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
1 7 21 35 35 21 7 1
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For x^5 we use this row: 1 5 10 10 5 1
That means the expansion of (a+x)^5 = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5
Note that, for each term in the expansion, the exponents add up to 5. a^5, right? 5ax^4 has an exponent of 1 on the a and an exponent of 4 on the x, for a total of 5. This is one of the properties of Pascal's Triangle. If we were in the row 1 21 35 35 21 7 1, the exponents in each term would add up to 7.
Since we're interest just in the x^3 term we can look at the row (1 5 10 10 5 1) and see that, starting from the left, we start with x^0, then x^1, then x^2, and so on. So the x^3 term will be here: 1 5 10 10 5 1.
The x^3 part of (a+x)^5 should be 10?x^3, with the ? being a^something. But the exponents should total 5, so the x^3 should be with a^2, giving us 10a^2x^3. If we look at the full expansion above, that's right.
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How can we apply this to (1 - 2x)^6 without doing the full expansion?
We need the row of Pascal's Triangle with the 6: 1 6 15 20 15 6 1
The x^3 term will be: 1 6 15 20 15 6 1
The x^3 term is 20(1)^3(-2x)^3
20(1)^3(-2x)^3 = 20 * 1 * -8x^3 = -160x^3
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The question asks us to put the two x^3 parts together like this:
10a^2x^3 - 160x^3 = -120x^3
divide everything by x^3
10a^2 - 160 = -120
10a^2 = 40
a^2 = 4
a = +-2
Answer: a = 2, a = -2