Question

Phân tích đa thức thành nhân tử:

k^5-k^3+k^2-1

2m^2-72+96n-32n^2

(b-3a)^2-4b^2+12ab

(a^2-3a-10)^2-4(a^2-10)^2+12a(a^2-10)

Answer 1

1.

$k^5-k^3+k^2-1=(k^5-k^3)+(k^2-1)=k^3(k^2-1)+(k^2-1)=(k^2-1)(k^3+1)$

$=(k-1)(k+1)(k+1)(k^2-k+1)=(k-1)(k+1)^2(k^2-k+1)$

2. 

$2m^2-72+96n-32n^2$

$=2(m^2-36+48n-16n^2)$

$=2[m^2-(16n^2-48n+36)]$

$=2[m^2-(4n-6)^2]=2(m-4n+6)(m+4n-6)$

 

Answer 2

3.
$(b-3a)^2-4b^2+12ab=(b-3a)^2-(4b^2-12ab)=(b-3a)^2-4b(b-3a)$

$=(b-3a)(b-3a-4b)=(b-3a)(-3a-3b)=3(3a-b)(a+b)$

4.

$(a^2-3a-10)^2-4(a^2-10)^2+12a(a^2-10)$

$=(a^2-3a-10)^2-4(a^2-10)(a^2-10-3a)$

$=(a^2-3a-10)(a^2-3a-10-4a^2+40)$

$=(a^2-3a-10)(-3a^2-3a+30)$

$=-3(a^2-3a-10)(a^2+a-10)$

$=-3(a-5)(a+2)(a^2+a-10)$