Question

Find the third term of (x^2 + 2)^4.

Answer 1

Answer:

Step-by-step explanation:

$in~the~expansion~of~(x+a)^n\\T_{r+1}=ncr x^{n-r}a^r\\r+1=3\\r=3-1=2\\n=4\\4c2=\frac{4 \times 3}{2 \times 1} =6\\T_{2}=4(x^2)^{4-2}(2)^2\\=16(x^2)^2\\=16x^4$

Answer 2

Answer:

24*x^4

Step-by-step explanation:

We want to find the third term of the equation:

(x^2 + 2)^4

So we need to expand that.

Remember the rule:

(a + b)^2 = a^2 + 2*a*b + b^2

(a + b + c)^2 = a^2 + b^2 + c^2 + 2*a*b + 2*b*c + 2*c*a

Then we can write:

(x^2 + 2)^4 = (x^2 + 2)^2*(x^2 + 2)^2

and we have:

(x^2 + 2)^2 = (x^4 + 2*2*x^2 + 2^2) = (x^4 + 4*x^2 + 4)

Then we have:

(x^2 + 2)^4 = (x^2 + 2)^2*(x^2 + 2)^2

(x^2 + 2)^4 = (x^4 + 4*x^2 + 4)^2

And we can expand this as:

(x^4)^2 + (4*x^2)^2 + 4^2 + 2*(x^4)*(4*x^2) + 2*(4*x^2)*(4) + 2*(x^4)*(4)

now we have this expanded, so we need to simplify this:

x^8 + 16*x^4 + 16 + 8*x^6 + 32*x^2 + 8*x^4

Putting the terms in order (larger degrees go in the left) we get:

x^8 + 8*x^6 + 16*x^4 + 8*x^4 + 32*x^2 + 16

x^8 + 8*x^6 + (16 + 8)*x^4 + 32*x^2 + 16

x^8 + 8*x^6 + 24*x^4 + 32*x^2 + 16

So the third term would be the third counting from the left, which is:

24*x^4