how do you factor these???

Question

6c^3 - 3c^2 - 45c= 0

x^4 - 10x^2 + 9 = 0

a^4 + a^2 - 2 = 0

4m^4 + 17m^2 + 4 = 0


i would apreciate it if you show step by step on at least a few so i can get the rest....

Answer 1

6c^3 - 3c^2 - 45c= 0
3c(2c^2 -c-15)=0
3c [ 2c^2-6c+5c-15]=0
3c [ 2c(c-3) +5(c-3)]=0
3c(c-3)(2c+5)=0
so c can be 0 or 3 or -5/2

Answer 2

6c^3 - 3c^2 - 45c= 0
Determine the greatest common factor among the three terms, which is 3c. Then divide each term by 3c, you will get:
3c (2c^2 - c - 15)
We can still extract factors of (2c^2 - c - 15).
Factor out the first and last terms:
2c^2 = (2c) (c)
-15 = (3) (-5) or (-3) (5)

Combine them, taking consideration of the middle term (-c)
The middle term should be the sum of the products of the means and extremes.

Let's try this combinations:
(2c + 3) (c -5)
3c + (-15c) = -12c ---- so this is not the right combinations

Let's try other combinations:
(2c -3) (c + 5) ---- -3c + 10c = 7c ----- not likely

Let's try another combinations:
(2c+ 5) (c - 3)
-5c + (-6c) = -c --- now, we get the right combinations

Therefore the factors of (2c^2 - c - 15) are:
3 (2c +5) (c - 3)
-------------------------------------------------------------------------------

x^4 - 10x^2 + 9

There's no common factor, so let's proceed to the next step; determine the factors of the first and last terms:
x^4 = (x^2) (x^2)

Take a look at the numerical coefficient of the middle term. It's 10, therefore, you need a 9 and a 1 to make it ten, right? Let's consider the signs : the middle term has a - sign which would lead us to think that both factors of 9 should also be negative. And to check further, the last term has a positive sign, which gives us the conclusion that the signs of the factors of 9 are both negative (because negative x negative = positive.

9 = (-9)(-1)

Let's now combine the factors:
(x^2 - 9) (x^2 -1)

You can now try the other two problems.

Answer 3

6c^3 - 3c^2 - 45c=0
c(6c² - 3c - 45)= 0 => c=0 is a root
2(2c² - c - 15) = 0
2(2c +5)(c - 3) = 0 => c = -5/2 or c = 3
The rest of them are kinda similar.

Doug

Answer 4

1. 3c(2c^2 - c - 15)=0
3c(2c + 5)(c - 3)=0
c = 0, -5/2, 3
2. x^4 - 10x^2 + 9 = 0
(x^2 - 9)(x^2 - 1) = 0
[(x+3)(x-3)][(x+1)(x-1)] = 0
x=3,-3,1,-1
3. a^4 + a^2 - 2 = 0
(a^2 + 2)(a^2 - 1) = 0
(a^2 + 2)[(a+1)(a-1)]=0
a=+/- sqrt(2), 1, -1
4. 4m^4 + 17m^2 + 4 = 0
(2m^2 + 4)(2m^2 + 1) = 0
2m^2 + 4 = 0
2m^2 = -4
m^2 = -2
m = +/- sqrt(-2), +/- sqrt (-1/2)