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What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C5(2)5(3)

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ANSWER

[tex]10C_5(2)^{5}( - 3)^5 [/tex]

EXPLANATION

The given binomial expansion is:

[tex] {(2x - 3y)}^{10} [/tex]

Compare this to

[tex] {(a + b)}^{n} [/tex]

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

[tex] {x}^{5} {y}^{5} [/tex]

This implies that, r=5.

The terms in the expansion can be obtained using

[tex]T_{r+1}=nC_ra^{n-r}b^r[/tex]

We substitute the given values to obtain;

[tex]T_{5+1}=10C_5(2x)^{10-5}( - 3y)^5[/tex]

[tex]T_{6}=10C_5(2x)^{5}(3y)^5[/tex]

[tex]T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5} {y}^{5} [/tex]

Hence the coefficient is;

[tex]10C_5(2)^{5}( - 3)^5 [/tex]

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The coefficient of that term is:

c = 10*(2)^5*(-3)^5

How to find the coefficient?

We have the expression:

(2x - 3y)^10

We want to get the component of the x^5*y^5-term.

To get that term, we need to take the product:

(2x)*(2x)*(2x)*(2x)*(2x)*(-3y)*(-3y)*(-3y)*(-3y)*(-3y)

= (2x)^5*(-3y)^5 = (32x^5)*(-243y^5) = -7,776*x^5*y^5

And that term will appear 10 times, so the actual coefficient is:

10*-7,776 = 10*(2)^5*(-3)^5

If you want to learn more about polynomials, you can read:

https://brainly.com/question/4142886

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