Question

разложите на множители 16 a) ---- _ b^4 81b) a^2-x^2+4x-4

Answer 1

$\frac{16}{81}-b^4=(\frac{4}{9}-b^2)(\frac{4}{9}+b^2)$

 

a²-x²+4x-4a=(a-x)(a+x)-4(a-x)=(a-x)(a+x-4)

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Answer 2

16 + a) 16 can be factored into 4 * 4 a^2 + 81 can be factored into (a+9)(a-9)

So, the factored form of 16 + a) would be 4 * 4 * (a+9)(a-9).

b^4 + 81b) This expression can be factored by taking out the common factor b: b(b^3 + 81)

a^2 - x^2 + 4x - 4 This expression can be factored using the difference of squares formula: (a + x + 2)(a - x + 2)

So, the factored form of a^2 - x^2 + 4x - 4 would be (a + x + 2)(a - x + 2).