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Expand the product (x - 2)^2 (x + 2)^3. What is the product of the nonzero coefficients of the resulting expression, including the constant term?

 May 8, 2022
 #1
avatar+9457 
0

Whenever you have \((x + a)^n\) and the exponent is bigger than 2, you want to use binomial theorem to expand it.

Otherwise, you can use sum of square identity (a + b)^2 = a^2 + 2ab + b^2.

 

\(\quad(x - 2)^2 (x + 2)^3 \\= (x^2 - 4x + 4)(\binom{3}0x^3 + \binom{3}1 x^2 \cdot 2+ \binom{3}2 x\cdot 2^2 + \binom{3}32^3) \\= (x^2 - 4x + 4)(x^3 + 6x^2 + 12x + 8)\)

 

Afterwards, you just expand it out like this:

 

 

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

 

And then expand each clump. It is troublesome, but it will work out nicely.

 May 8, 2022
 #2
avatar+1750 
+1

We can write this as: \((x-2) \times (x+2) \times (x-2) \times (x+2) \times (x+2)\)

 

Recall the identity: \((a-b)(a+b) = a^2-b^2\)

 

This means we can rewrite the equation as: \((x^2-4) \times (x^2-4) \times (x-2)\)

 

We know that \((x^2-4)(x^2-4) = x^2 \times x^2 -4 \times x^2 - 4\times x^2 + 16 = x^4-8x^2+16\)

 

Now, we have: \((x^4-8x^2+16)(x+2)\)

 

Can you expand this?

 May 9, 2022
 #3
avatar+9457 
0

This method is nicer and is easier to do! 

MaxWong  May 9, 2022

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