Is there a trick to solving this problem faster?

[weak lodge]

binomial expansion

or ig binomial theorem

[zinc shore]

Just like a difference of two squares, the numerator term will have a factor of (x + 2) - x = 2 and leave behind a degree 4 polynomial.

The next term in the expansion will be

$\binom{5}{1}x^{5-1}\times 2^1$

making the leading term

$\frac{\binom{5}{1}x^4\times 2^1}{2} = \frac{5x^4 \times 2}{2} = 5x^4$

[nocturne spear]

Jay

[plush siren]

@zinc shore can you explain a bit more? Like how you ended up with the 1 times 5 and 5-1.

[zinc shore]

Each term in the expansion of $(x + 2)^5$

follows the pattern $\binom{5}{k}x^{5-k}2^k$

taking integer values of $k$ starting at $k=0$

and continuing until $k = 5$.

This is what the statement of the expansion

$(x+2)^5 = \sum_{k=0}^5 \binom{5}{k}x^{5-k}2^k$

actually means.

So, the first term is $\binom{5}{0}x^{5-0}2^0 = 1 \times x^5 \times 1 = x^5$

And, the second term is $\binom{5}{1}x^{5-1}2^1 = 5 \times x^4 \times 2 = 10x^4$

And, the third term is $\binom{5}{2}x^{5-2}2^2 = 10 \times x^3 \times 4 = 40x^3$

And so on...

The complete expansion is then:

$(x+2)^5 = \sum_{k=0}^5 \binom{5}{k}x^{5-k}2^k$

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

and hence,

$\frac{(x+2)^5 - x^5}{2} = \frac{x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32 - x^5}{2}$

$= \frac{10x^4 + 40x^3 + 80x^2 + 80x + 32}{2}$

$ = 5x^4 + 20x^3 + 40x^2 + 40x + 16$

[nocturne spear]

Jay

[plush siren]

@zinc shore Okay, thank you so much for explaining it to me clearly. I understand how you got the answer but I’m still not really familiar with this form of math so I’ll try to complete some more College Algebra practices so I can get familiar!

[zinc shore]

You could expand $(x+2)^5$ in the longer (manual) way...

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

$(x+2)^4 = \left[(x+2)^2\right]^2 =\left[x^2 + (4x+4)\right]^2$

$= \left(x^2\right)^2 + 2x^2(4x+4) + (4x+4)^2$

$= x^4 + 8x^3 + 8x^2 + (4x)^2 + 2(4x)(4) + 4^2$

$= x^4 + 8x^3 + 24x^2 + 32x + 16$

Hence, $(x+2)^5 = (x+2)\left(x^4 + 8x^3 + 24x^2 + 32x + 16\right)$

$= \left(x^5 + 8x^4 + 24x^3 + 32x^2 + 16x\right) + \left(2x^4 + 16x^3 + 48x^2 + 64x + 32\right)$

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

as predicted

[nocturne spear]

Jay

[zinc shore]

And, it then follows that:

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

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

$\frac{(x+2)^5 - x^5}{2} = \frac{10x^4 + 40x^3 + 80x^2 + 80x + 32}{2}$

$= 5x^4 + 20x^3 + 40x^2 + 40x + 16$