We read your photo as:

To expand the expressions, we will use the binomial theorem, which states that for any non-negative integer n:

Mathematics

ScanSolve AI Answer

Answers

2 parts
1

x5+30x4y+360x3y2+2160x2y3+6480xy4+7776y5x^5 + 30x^4y + 360x^3y^2 + 2160x^2y^3 + 6480xy^4 + 7776y^5

2

243−2430y+9720y2−19440y3+19440y4−7776y5243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5

Step-by-step solution

6 steps

To expand the expressions, we will use the binomial theorem, which states that for any non-negative integer nn:

(a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

The coefficients (nk)\binom{n}{k} for n=5n=5 are found in Row 5 of Pascal's triangle: 1,5,10,10,5,11, 5, 10, 10, 5, 1.

Part 1· 3 steps

Expand (x+6y)5(x+6y)^5

Here, a=xa=x and b=6yb=6y.

  1. Apply the binomial theorem with n=5n=5.

    (x+6y)5=(50)x5(6y)0+(51)x4(6y)1+(52)x3(6y)2+(53)x2(6y)3+(54)x1(6y)4+(55)x0(6y)5(x+6y)^5 = \binom{5}{0}x^5(6y)^0 + \binom{5}{1}x^4(6y)^1 + \binom{5}{2}x^3(6y)^2 + \binom{5}{3}x^2(6y)^3 + \binom{5}{4}x^1(6y)^4 + \binom{5}{5}x^0(6y)^5

  2. Substitute the binomial coefficients and simplify each term.

    =1⋅x5⋅1+5⋅x4⋅(6y)+10⋅x3⋅(36y2)+10⋅x2⋅(216y3)+5⋅x⋅(1296y4)+1⋅1⋅(7776y5)= 1 \cdot x^5 \cdot 1 + 5 \cdot x^4 \cdot (6y) + 10 \cdot x^3 \cdot (36y^2) + 10 \cdot x^2 \cdot (216y^3) + 5 \cdot x \cdot (1296y^4) + 1 \cdot 1 \cdot (7776y^5)

  3. Perform the multiplications.

    =x5+30x4y+360x3y2+2160x2y3+6480xy4+7776y5= x^5 + 30x^4y + 360x^3y^2 + 2160x^2y^3 + 6480xy^4 + 7776y^5

    The expansion of (x+6y)5(x+6y)^5 is:

    x5+30x4y+360x3y2+2160x2y3+6480xy4+7776y5\boxed{x^5 + 30x^4y + 360x^3y^2 + 2160x^2y^3 + 6480xy^4 + 7776y^5}

Part 2

243−2430y+9720y2−19440y3+19440y4−7776y5243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5

Expand (3−6y)5(3-6y)^5

Here, a=3a=3 and b=−6yb=-6y.

  1. Apply the binomial theorem with n=5n=5.

    (3−6y)5=(50)(3)5(−6y)0+(51)(3)4(−6y)1+(52)(3)3(−6y)2+(53)(3)2(−6y)3+(54)(3)1(−6y)4+(55)(3)0(−6y)5(3-6y)^5 = \binom{5}{0}(3)^5(-6y)^0 + \binom{5}{1}(3)^4(-6y)^1 + \binom{5}{2}(3)^3(-6y)^2 + \binom{5}{3}(3)^2(-6y)^3 + \binom{5}{4}(3)^1(-6y)^4 + \binom{5}{5}(3)^0(-6y)^5

  2. Substitute the binomial coefficients and simplify each term.

    =1⋅(243)⋅1+5⋅(81)⋅(−6y)+10⋅(27)⋅(36y2)+10⋅(9)⋅(−216y3)+5⋅(3)⋅(1296y4)+1⋅1⋅(−7776y5)= 1 \cdot (243) \cdot 1 + 5 \cdot (81) \cdot (-6y) + 10 \cdot (27) \cdot (36y^2) + 10 \cdot (9) \cdot (-216y^3) + 5 \cdot (3) \cdot (1296y^4) + 1 \cdot 1 \cdot (-7776y^5)

  3. Perform the multiplications.

    =243−2430y+9720y2−19440y3+19440y4−7776y5= 243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5

    The expansion of (3−6y)5(3-6y)^5 is:

    243−2430y+9720y2−19440y3+19440y4−7776y5\boxed{243 - 2430y + 9720y^2 - 19440y^3 + 19440y^4 - 7776y^5}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QThe binomial theorem is used to expand expressions of the form: a) (a + b) b) (a + b) c) (a + b)n d)…QExpand (1−2x)6 using binomial theoremQApply the binomial theorem.QExpand (2x + 3)4 using the binomial theorem.QThe general term in the binomial expansion of (a+b)ⁿ is given by the formula:
Browse by topic:Algebra
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.