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Polynomial

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47
3

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
+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
+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
+9457
0

This method is nicer and is easier to do!

MaxWong  May 9, 2022