Question

Expand (2 + x)" – (2 - x)" in ascending powers of x and simplify your result.
following. ( chapter is binomial theorem plss answer correctly no unnecessary answer)​

Answer 1

Answer:

the answer is 2x

Step-by-step explanation:

(2+x)- (2-x)

=2+x-2+x

=2-2+x+x

=2x

Answer 2

Answer:

The expansion of the given expression in ascending powers of x is 160x+72x^3+2x^5

Therefore (2+x)^5-(2-x)^5=160x+72x^3+2x^5

Step-by-step explanation:

Given expression is (2+x)^5-(2-x)^5

To find the expansion of the given expression in ascending powers of x :

First find (2+x)^5

(2+x)^5=(2+x)^{2+3}

=(2+x)^2.(2+x)^3

=(2^2+2(2)(x)+x^2).(2^3+3(2)^2(x)+3(2)(x^2)+x^3)

=(4+4x+x^2).(8+12x+6x^2+x^3)

=4(8+12x+6x^2+x^3)+4x(8+12x+6x^2+x^3)+x^2(8+12x+6x^2+x^3)

=4(8)+4(12x)+4(6x^2)+4(x^3)+4x(8)+4x(12x)+4x(6x^2)+4x(x^3)+x^2(8)+x^2(12x)+x^2(6x^2)+x^2(x^3)

=32+48x+24x^2+4x^3+32x+48x^2+24x^3+4x^4+8x^2+12x^3+6x^4+x^5

=32+80x+80x^2+40x^3+10x^4+x^5

Therefore (2+x)^5=32+80x+80x^2+40x^3+10x^4+x^5

Similarly we can find (2-x)^5

(2-x)^5=(2-x)^{2+3}

=(2-x)^2.(2-x)^3

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

=(4-4x+x^2).(8-12x+6x^2-x^3)

=4(8-12x+6x^2-x^3)-4x(8-12x+6x^2-x^3)+x^2(8-12x+6x^2-x^3)

=4(8)+4(-12x)+4(6x^2)-4(-x^3)-4x(8)-4x(-12x)-4x(6x^2)-4x(-x^3)+x^2(8)+x^2(-12x)+x^2(6x^2)+x^2(-x^3)

=32-48x+24x^2+4x^3-32x+48x^2-24x^3+4x^4+8x^2-12x^3+6x^4-x^5

=32-80x+80x^2-32x^3+10x^4-x^5

Therefore (2-x)^5=32-80x+80x^2-32x^3+10x^4-x^5

Substituting the values in (2+x)^5-(2-x)^5 we get

(2+x)^5-(2-x)^5=32+80x+80x^2+40x^3+10x^4+x^5-(32-80x+80x^2-32x^3+10x^4-x^5)

=32+80x+80x^2+40x^3+10x^4+x^5-32+80x-80x^2+32x^3-10x^4+x^5

=160x+72x^3+2x^5

Therefore the expansion of the