Factor completely 2a^3 +8a^2 + 10a?

Question

factor
a^3 - 125b^3

(2x)^3 + y^3

a^ 3 + 2ab^2 - 2a^2 b -4 b^3

Answer 1

a^3 - 125b^3
=a^3-(5b)^3
=(a-5b)[a^2+a(5b)+(5b)^2]
=(a-5b)(a^2+5ab+25b^2)

(2x)^3 + y^3
=(2x+y)[(2x)^2+(2x)(y)+y^2]
=(2x+y)(4x^2+2xy+y^2)

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

Answer 2

1) 2a(5 + 4a + a^2)

2) a^3 + -125b^3

3) (2x+y)(4x^2-2yx+y^2)

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

Answer 3

2a^3 + 8a^2 + 10a
= 2a(a^2 + 4a + 5)
= 2a[a^2 + 4a + 4 + 1]
= 2a(a + 2 + i)(a + 2 -- i)

a^3 -- 125b^3 = (a -- 5b)(a^2 + 5ab + 25b^2)

(2x)^3 + y^3 = (2x + y)(4x^2 -- 2xy + y^2)

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

Answer 4

first divide each term by 2 (a^3+4a^2+10a)
then factor and a out of the equation a(a^2+4a+5)
finally use the quadratic formula to factor (a^2+4a+5)

Answer 5

= 2a(a^2 + 4a + 5)
= 2a[(a+2)^2 - 2^2 + 5] (completing the square)
= 2a[(a+2)^2 +1]

Answer 6

1)
2a^3+8a^2+10a
Factor out 2a:
2a(a^2+4a+5)

2)
a^3-125b^3
(a-b)(a^2+5ab+25b^2)

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

Answer 7

wouldn't it be great if (a-5b)^3 came to a^3-125b^3...why are things like this so superficially simple..?