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Algebra - Binomial theorem (introduction)

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A binomial expression is an algebraic expression containing two terms, such as (a+b)(a + b). The Binomial Theorem provides a quick way to expand powers of these expressions without performing repeated multiplication.

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For any positive integer nn, the expansion of (a+b)n(a + b)^n contains n+1n + 1 terms.

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In the expansion of (a+b)n(a + b)^n, the powers of aa decrease from nn to 00, while the powers of bb increase from 00 to nn.

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The sum of the exponents of aa and bb in each term is always equal to nn.

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The coefficients of the terms in the expansion follow the pattern of Pascal's Triangle or can be calculated using the combinations formula (nr)\binom{n}{r}.

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Pascal's Triangle: The nn-th row of the triangle gives the coefficients for the expansion of (a+b)n(a + b)^n. For example, the row 1,3,3,11, 3, 3, 1 corresponds to n=3n=3.

📐Formulae

(a+b)n=(n0)anb0+(n1)an−1b1+(n2)an−2b2+⋯+(nn)a0bn(a + b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}a^0 b^n

(nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n - r)!}

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

💡Examples

Problem 1:

Expand (x+3)3(x + 3)^3 using the Binomial Theorem.

Solution:

(x+3)3=(30)x3(3)0+(31)x2(3)1+(32)x1(3)2+(33)x0(3)3(x + 3)^3 = \binom{3}{0}x^3(3)^0 + \binom{3}{1}x^2(3)^1 + \binom{3}{2}x^1(3)^2 + \binom{3}{3}x^0(3)^3 (x+3)3=1(x3)(1)+3(x2)(3)+3(x)(9)+1(1)(27)(x + 3)^3 = 1(x^3)(1) + 3(x^2)(3) + 3(x)(9) + 1(1)(27) (x+3)3=x3+9x2+27x+27(x + 3)^3 = x^3 + 9x^2 + 27x + 27

Explanation:

We use the coefficients from the 3rd row of Pascal's Triangle (1,3,3,1)(1, 3, 3, 1). The power of xx starts at 33 and decreases, while the power of 33 starts at 00 and increases.

Problem 2:

Expand (2x−1)4(2x - 1)^4.

Solution:

(2x−1)4=(40)(2x)4(−1)0+(41)(2x)3(−1)1+(42)(2x)2(−1)2+(43)(2x)1(−1)3+(44)(2x)0(−1)4(2x - 1)^4 = \binom{4}{0}(2x)^4(-1)^0 + \binom{4}{1}(2x)^3(-1)^1 + \binom{4}{2}(2x)^2(-1)^2 + \binom{4}{3}(2x)^1(-1)^3 + \binom{4}{4}(2x)^0(-1)^4 =1(16x4)(1)+4(8x3)(−1)+6(4x2)(1)+4(2x)(−1)+1(1)(1)= 1(16x^4)(1) + 4(8x^3)(-1) + 6(4x^2)(1) + 4(2x)(-1) + 1(1)(1) =16x4−32x3+24x2−8x+1= 16x^4 - 32x^3 + 24x^2 - 8x + 1

Explanation:

Note that the second term in the binomial is −1-1. When raising a negative number to an odd power, the term becomes negative. The coefficients used are 1,4,6,4,11, 4, 6, 4, 1.

Problem 3:

Find the coefficient of the x2x^2 term in the expansion of (x+5)4(x + 5)^4.

Solution:

The general term is given by (nr)an−rbr\binom{n}{r} a^{n-r} b^r. Here n=4n=4, a=xa=x, and b=5b=5. To find x2x^2, we need n−r=2n-r = 2, which means 4−r=2⇒r=24-r = 2 \Rightarrow r = 2. Term=(42)x4−252\text{Term} = \binom{4}{2} x^{4-2} 5^2 Term=6⋅x2⋅25=150x2\text{Term} = 6 \cdot x^2 \cdot 25 = 150x^2 The coefficient is 150150.

Explanation:

We identify the specific term required by matching the exponent of xx with the general formula's power. Then we calculate the combination and the power of the constant.