Question 3 of 20beginner🔧 ApplyLong Answer4 marks

Expand each of the expressions in Exercises 1 to 5.

(2x−3)6(2x - 3)^6
Correct Answer

To expand the expression (2x−3)6(2x - 3)^6, we use the Binomial Theorem. The theorem states that the expansion of (a+b)n(a + b)^n is given by the formula:

nC0 an+nC1 an−1b+nC2 an−2b2+⋯+nCn bn{}^nC_0 \, a^n + {}^nC_1 \, a^{n-1} b + {}^nC_2 \, a^{n-2} b^2 + \dots + {}^nC_n \, b^n

For the given expression, we identify the values as a=2xa = 2x, b=−3b = -3, and n=6n = 6. Substituting these values into the formula, we write the expansion. We determine the binomial coefficients 6C0,6C1,…,6C6{}^6C_0, {}^6C_1, \dots, {}^6C_6 as 1, 6, 15, 20, 15, 6, and 1 respectively.

The expansion proceeds as follows:

(2x−3)6=6C0(2x)6(−3)0+6C1(2x)5(−3)1+6C2(2x)4(−3)2+6C3(2x)3(−3)3+6C4(2x)2(−3)4+6C5(2x)1(−3)5+6C6(2x)0(−3)6(2x - 3)^6 = {}^6C_0 (2x)^6 (-3)^0 + {}^6C_1 (2x)^5 (-3)^1 + {}^6C_2 (2x)^4 (-3)^2 + {}^6C_3 (2x)^3 (-3)^3 + {}^6C_4 (2x)^2 (-3)^4 + {}^6C_5 (2x)^1 (-3)^5 + {}^6C_6 (2x)^0 (-3)^6

Calculating each term individually:

=1(64x6)(1)+6(32x5)(−3)+15(16x4)(9)+20(8x3)(−27)+15(4x2)(81)+6(2x)(−243)+1(1)(729)= 1(64x^6)(1) + 6(32x^5)(-3) + 15(16x^4)(9) + 20(8x^3)(-27) + 15(4x^2)(81) + 6(2x)(-243) + 1(1)(729)

Simplifying the products:

=64x6−576x5+2160x4−4320x3+4860x2−2916x+729= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729

Thus, the final expansion is 64x6−576x5+2160x4−4320x3+4860x2−2916x+72964x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729.

Exercise: EXERCISE 7.1 | Q: 3 | (Page 7)
For More Understanding

Explanation

The solution applies the Binomial Theorem formula provided in the Summary section of the context. It identifies aa as 2x2x and bb as −3-3 with n=6n = 6.

It calculates the binomial coefficients (Pascal's triangle or combination formula) and simplifies the powers of 2x2x and −3-3 to arrive at the final expansion.

Solution Steps

  1. Step 1: Identify the values for the Binomial Theorem formula (a+b)n(a+b)^n. Here, a=2xa=2x, b=−3b=-3, and n=6n=6.

  2. Step 2: Write the expansion using the formula:

    (2x−3)6=6C0(2x)6(−3)0+6C1(2x)5(−3)1+6C2(2x)4(−3)2+6C3(2x)3(−3)3+6C4(2x)2(−3)4+6C5(2x)1(−3)5+6C6(2x)0(−3)6(2x - 3)^6 = {}^6C_0(2x)^6(-3)^0 + {}^6C_1(2x)^5(-3)^1 + {}^6C_2(2x)^4(-3)^2 + {}^6C_3(2x)^3(-3)^3 + {}^6C_4(2x)^2(-3)^4 + {}^6C_5(2x)^1(-3)^5 + {}^6C_6(2x)^0(-3)^6

  3. Step 3: Calculate the binomial coefficients:

    6C0=1,  6C1=6,  6C2=15,  6C3=20,  6C4=15,  6C5=6,  6C6=1{}^6C_0=1, \; {}^6C_1=6, \; {}^6C_2=15, \; {}^6C_3=20, \; {}^6C_4=15, \; {}^6C_5=6, \; {}^6C_6=1

  4. Step 4: Substitute the coefficients and simplify the powers:

    1(64x6)+6(32x5)(−3)+15(16x4)(9)+20(8x3)(−27)+15(4x2)(81)+6(2x)(−243)+1(729)1(64x^6) + 6(32x^5)(-3) + 15(16x^4)(9) + 20(8x^3)(-27) + 15(4x^2)(81) + 6(2x)(-243) + 1(729)

  5. Step 5: Perform the final multiplication and combine terms:

    64x6−576x5+2160x4−4320x3+4860x2−2916x+72964x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729
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