# 6.8 – Pascal’s Triangle and the Binomial Theorem.

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6.8 – Pascal’s Triangle and the Binomial Theorem

The Binomial Theorem Strategy only: how do we expand these? 1.(x + 2) 2 2.(2x + 3) 2 3.(x – 3) 3 4.(a + b) 4

The Binomial Theorem Solutions 1.(x + 2) 2 = x 2 + 2(2)x + 2 2 = x 2 + 4x + 4 2.(2x + 3) 2 = (2x) 2 + 2(3)(2x) + 3 2 = 4x 2 + 12x + 9 3.(x – 3) 3 = (x – 3)(x – 3) 2 = (x – 3)(x 2 – 2(3)x + 3 2 ) = (x – 3)(x 2 – 6x + 9) = x(x 2 – 6x + 9) – 3(x 2 – 6x + 9) = x 3 – 6x 2 + 9x – 3x 2 + 18x – 27 = x 3 – 9x 2 + 27x – 27 4.(a + b) 4 = (a + b) 2 (a + b) 2 = (a 2 + 2ab + b 2 )(a 2 + 2ab + b 2 ) = a 2 (a 2 + 2ab + b 2 ) + 2ab(a 2 + 2ab + b 2 ) + b 2 (a 2 + 2ab + b 2 ) = a 4 + 2a 3 b + a 2 b 2 + 2a 3 b + 4a 2 b 2 + 2ab 3 + a 2 b 2 + 2ab 3 + b 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4

THAT is a LOT of work! Isn’t there an easier way?

Introducing: Pascal’s Triangle Take a moment to copy the first 6 rows. What patterns do you see? Row 5 Row 6

Use Pascal’s Triangle to expand (a + b) 5. The Binomial Theorem Use the row that has 5 as its second number. In its simplest form, the expansion is a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + b 5. The exponents for a begin with 5 and decrease. 1a 5 b 0 + 5a 4 b 1 + 10a 3 b 2 + 10a 2 b 3 + 5a 1 b 4 + 1a 0 b 5 The exponents for b begin with 0 and increase. Row 5

Use Pascal’s Triangle to expand (x – 3) 4. The Binomial Theorem First write the pattern for raising a binomial to the fourth power. 14641 Coefficients from Pascal’s Triangle. (a + b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 Since (x – 3) 4 = (x + (–3)) 4, substitute x for a and –3 for b. (x + (–3)) 4 = x 4 + 4x 3 (–3) + 6x 2 (–3) 2 + 4x(–3) 3 + (–3) 4 = x 4 – 12x 3 + 54x 2 – 108x + 81 The expansion of (x – 3) 4 is x 4 – 12x 3 + 54x 2 – 108x + 81.

The Binomial Theorem Use the Binomial Theorem to expand (x – y) 9. Write the pattern for raising a binomial to the ninth power. (a + b) 9 = 9 C 0 a 9 + 9 C 1 a 8 b + 9 C 2 a 7 b 2 + 9 C 3 a 6 b 3 + 9 C 4 a 5 b 4 + 9 C 5 a 4 b 5 + 9 C 6 a 3 b 6 + 9 C 7 a 2 b 7 + 9 C 8 ab 8 + 9 C 9 b 9 Substitute x for a and –y for b. Evaluate each combination. (x – y) 9 = 9 C 0 x 9 + 9 C 1 x 8 (–y) + 9 C 2 x 7 (–y) 2 + 9 C 3 x 6 (–y) 3 + 9 C 4 x 5 (–y) 4 + 9 C 5 x 4 (–y) 5 + 9 C 6 x 3 (–y) 6 + 9 C 7 x 2 (–y) 7 + 9 C 8 x(–y) 8 + 9 C 9 (–y) 9 = x 9 – 9x 8 y + 36x 7 y 2 – 84x 6 y 3 + 126x 5 y 4 – 126x 4 y 5 + 84x 3 y 6 – 36x 2 y 7 + 9xy 8 – y 9 The expansion of (x – y) 9 is x 9 – 9x 8 y + 36x 7 y 2 – 84x 6 y 3 + 126x 5 y 4 – 126x 4 y 5 + 84x 3 y 6 – 36x 2 y 7 + 9xy 8 – y 9.

Let’s Try Some Expand the following a) (x-y 5 ) 3 b)(3x-2y) 4

Let’s Try Some Expand the following (x-y 5 ) 3

Let’s Try Some Expand the following (3x-2y) 4

Let’s Try Some Expand the following (3x-2y) 4

How does this relate to probability? You can use the Binomial Theorem to solve probability problems. If an event has a probability of success p and a probability of failure q, each term in the expansion of (p + q) n represents a probability. Example: 10 C 2 * p 8 q 2 represents the probability of 8 successes in 10 tries

The Binomial Theorem Brianna makes about 90% of the shots on goal she attempts. Find the probability that Bri makes exactly 7 out of 12 consecutive goals. Since you want 7 successes (and 5 failures), use the term p 7 q 5. This term has the coefficient 12 C 5. Probability (7 out of 10) = 12 C 5 p 7 q 5 = 0.0037881114Simplify. Bri has about a 0.4% chance of making exactly 7 out of 12 consecutive goals. = (0.9) 7 (0.1) 5 The probability p of success = 90%, or 0.9. 12! 5! 7!