Question

The coefficient of x2 in the expansion of (x2+2x+3)2+(x2−2x+3)2 is (a)10 (b) 20 (c)−10 (d)−20

Answer 1

$\textbf{Given:}$

$(x^2+2x+3)^2+(x^2-2x+3)^2$

$\textbf{To find:}$

$\text{Coefficient of x^2 in the expansion of (x^2+2x+3)^2+(x^2-2x+3)^2 }$

$\textbf{Solution:}$

$\text{Consider,}$

$(x^2+2x+3)^2+(x^2-2x+3)^2$

$\text{Using the identity,}$

$\boxed{\bf(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca}$

$=x^4+4x^2+9+4x^3+12x+6x^2+x^4+4x^2+9-4x^3-12x+6x^2$

$=x^4+4x^2+9+6x^2+x^4+4x^2+9+6x^2$

$=2\,x^4+8\,x^2+18+12\,x^2$

$=2\,x^4+20\,x^2+18$

$\therefore\textbf{Coefficient of \bf\,x^2 is 20}$

$\textbf{Answer:}$

$\textbf{Option (b) is correct}$

Answer 2

Step-by-step explanation:

Solution:

Consider,

(x^2+2x+3)^2+(x^2-2x+3)^2(x

2

+2x+3)

2

+(x

2

−2x+3)

2

Using the identity

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca}

(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca

=x^4+4x^2+9+4x^3+12x+6x^2+x^4+4x^2+9-4x^3-12x+6x^2=x

4

+4x

2

+9+4x

3

+12x+6x

2

+x

4

+4x

2

+9−4x

3

−12x+6x

2

=x^4+4x^2+9+6x^2+x^4+4x^2+9+6x^2=x

4

+4x

2

+9+6x

2

+x

4

+4x

2

+9+6x

2

=2\,x^4+8\,x^2+18+12\,x^2=2x

4

+8x

2

+18+12x

2

=2\,x^4+20\,x^2+18=2x

4

+20x

2

+18

{Coefficient of $\bf\,x^2$ is 20}∴Coefficient of x^2 is 20