Expand Using the Binomial Theorem (2x^2y^3-1/4y)^9

${(2 x^{2} y^{3} - \frac{1}{4} y)}^{9}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{9} \frac{9!}{(9 - k)! k!} \cdot {(2 x^{2} y^{3})}^{9 - k} \cdot {(- \frac{1}{4} y)}^{k}$

Step 2

Expand the summation.

$\frac{9!}{(9 - 0)! 0!} \cdot {(2 x^{2} y^{3})}^{9 - 0} \cdot {(- \frac{1}{4} y)}^{0} + \frac{9!}{(9 - 1)! 1!} \cdot {(2 x^{2} y^{3})}^{9 - 1} \cdot {(- \frac{1}{4} y)}^{1} + \dots + \frac{9!}{(9 - 9)! 9!} \cdot {(2 x^{2} y^{3})}^{9 - 9} \cdot {(- \frac{1}{4} y)}^{9}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x^{2} y^{3})}^{9} \cdot {(- \frac{1}{4} y)}^{0} + 9 \cdot {(2 x^{2} y^{3})}^{8} \cdot {(- \frac{1}{4} y)}^{1} + \dots + 1 \cdot {(2 x^{2} y^{3})}^{0} \cdot {(- \frac{1}{4} y)}^{9}$

Step 4

Simplify the polynomial result.

$512 x^{18} y^{27} - 576 x^{16} y^{25} + 288 x^{14} y^{23} - 84 x^{12} y^{21} + \frac{63 y^{19} x^{10}}{4} - \frac{63 y^{17} x^{8}}{32} + \frac{21 y^{15} x^{6}}{128} - \frac{9 y^{13} x^{4}}{1024} + \frac{9 y^{11} x^{2}}{32768} - \frac{y^{9}}{262144}$

Expand Using the Binomial Theorem (2x^2y^3-1/4y)^9

Related Questions

Support LCMGCF.com by disabling your adblocker.