Question

Find the coefficient of square of x in the expansion of (2x+3)^2 (x-5)

Answer 1

Step-by-step explanation:

The general term in the expansion of (2−x+3x

2

)

6

is

r!s!t!

6!

2

r

(−x)

s

(3x

2

)

t

, where r+s+t=6

=

r!s!t!

6

2

r

×(−1)

s

×3

t

×x

s+2t

,

where r+s+t=6

For the coefficient of x

5

, we must have

s+2t=5

But r+s+t=6

∴s=5−2t and r=1+t

where, 0≤r,s,t≤6

Now, t=0⇒r=1,s=5

t=1⇒r=2,s=3

t=2⇒r=3,s=1

Thus, there are three terms containing x

5

and coefficient of x

5

=

1!5!0!

6!

×2

1

×(−1)

5

×3

0

+

2!3!1!

6!

×2

2

×(−1)

3

×3

1

+

3!1!2!

6!

×2

3

×(−1)

1

×3

2

=−12−720−4320

=−5052.