How do I factor these?

Question

1.) 5x^2 - 3x + 17
2.) a^3 + b^3 + a^2 - b^2
3.) -6q^5 - 6000q^2
4.) 27(a+x)^6 + 8(b+y)^3

Answer 1

I assume you want a factorisation over the integers.
Compute the discriminant of 1):
b²-4ac = 9-4(85) < 0.
Since this is not a square, 1) is not factorable.

2). a³+b³ = (a+b)(a²-ab+b²)
a²-b² = (a+b)(a-b).
So a+b is a common factor.
Taking it out, we get
(a+b)(a²-ab+b²+a-b)

3). Take out the common factor:
-6q² is a common factor.
We get -6q²(q³+1000).
The second factor is the sum of 2 cubes.
So we finally get
-6q²(q+10)(q²-10q+100).

4). This is also the sum of 2 cubes.
So we get
[3(a+x)² + 2(b+y)][9(a+x)^4 - 6(a+x)²(b+y) +4(b+y)²].

Hope that helps!

Answer 2

2) a^3 + a^2 + b^3 - b^2 = a( 1^3 +1^2) -b (1^2 + 1^3)
= (a-b) (1^3 + 1^2)

Answer 3

use the formula
x= -b+(in roots)b^2-4ac\2a

here b=-3,a=5,c=17
2)(a+b)(a^2-ab+b^2)+(a-b)(a+b)
take common(a+b)
a+b{a^2-ab+b^2+a-b}
(a+b)( ")

3)-6q^2 is common
-6q^2(q^3+1000)

use the formula of (a^3+b^3) =(a+b)(a^2-ab+b^2)