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How do you find the coefficient of #x^3# in #(2x+3)^5#?

Answer 1

Theodore Amidon

The answer is #=720#

The theorem of binomials is

#(a+b)^n=sum_(k=0)^n((n),(k))a^(n-k)b^k#

Where,

#((n),(k))=(n!)/((n-k)!(k!))#

This is what we have

#(2x+3)^5#

And we need the coefficient of #x^3#

#n=5#

#a=2x#

#b=3#

#k=2#

The coefficient is

# = ( (5),(2)) * (2)^3 *(3)^2#

#=(5!)/((3!)(2!))* 8* 9#

#=10*72#

#=720#

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Answer 2

Liam Brennan

To find the coefficient of (x^3) in ((2x + 3)^5), you can use the binomial theorem or Pascal's triangle. In this case, using the binomial theorem:

[ \text{Coefficient of } x^3 = \binom{5}{2} \cdot (2x)^2 \cdot 3^3 ]

[ = \frac{5!}{2! \cdot (5 - 2)!} \cdot (2x)^2 \cdot 3^3 ]

[ = \frac{5 \cdot 4}{2 \cdot 1} \cdot (2x)^2 \cdot 3^3 ]

[ = 10 \cdot (4x^2) \cdot 27 ]

[ = 1080x^2 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.