Algebra Problem help?

Question

1. Factor (4x+5)^3 + (4x+5)^2

2. Factor 3(x+2)(x-1)+15(x+2)^2(x-1)

3. Factor 2x^3 - 5x^2+10x-25

4. Factor (125x^6)-8

Answer 1

1. let (4x+5) be y
so sub y into your question
y^3 + y^2
= (y^2) (y+1) .... (1)
sub (4x+5) into (1)
(4x+5)^2 (4x+5+1)
= (4x+5)^2(4x+6)
= (4x+5)^2 (2)(2x+3)
= 2(2x+3)(4x+5)^2

2. Let y=(x+2) z=(x-1)
3yz + 15yyz
=3yz(1+5y) ....(1)
sub y=(x+2) z=(x-1) into (1)
3(x+2)(x-1)[1+5(x+2)]
=3(x+2)(x-1)[1+5x+10]
=3(x+2)(x-1)(5x+11)

3. 2x^3 - 5x^2+10x-25
= x^2(2x-5)+ 5(2x-5)
= (x^2+5)(2x-5)

4. (125x^6)-8
= (5x^2)^3-2^3
let a=(5x^2) b=2
a^3-b^3
=(a-b)(a^2+ab+b^2)
=(5x^2-2)[(5x^2)^2+(5x^2)(2)
+(2^2)]
=(5x^2-2)[25x^4+10x^2+4]

Answer 2

1. Factor (4x+5)^3 + (4x+5)^2
= (4x+5)^2[(4x+5) +1]
=(4x+5)^2(4x+6)

2. Factor 3(x+2)(x-1)+15(x+2)^2(x-1)
=3(x+2)(x-1)[ 1+5(x+2)]
= 3(x+2)(x-1)(5x+11)

3. Factor 2x^3 - 5x^2+10x-25
x^2(2x-5) + 5(2x-5)
= (x^2+5)(2x-5)

4. Factor (125x^6)-8
= (5x^2 -2)(25x^4 + 10x^2 +4)

Answer 3

1. (4x + 5)^3 + (4x + 5)^2 = (4x + 5)^2*(4x + 5 + 1).

2. Hint: 3*(x + 2)*(x - 1) is a common factor.

3. Hint: Factor x^2 from the first two terms, and 5 out of the last two; continue.

4. This is the difference of two cubes.

Answer 4

(4x+5)^3 + (4x+5)^2
=(4x+5)^2 * (4x+6)


3(x+2)(x-1)+15(x+2)^2(x-1)
=3(x+2)(x-1) ( 5x+11)


2x^3 - 5x^2+10x-25
=(x-2,5)(x^2+10)


(125x^6)-8
=(5x^2)^3-2^3
=(5x^2-2)(25x^4+10x^2+4)