# Math Help - Second degree equation (difficult)

1. ## Second degree equation (difficult)

C) x^3 - 8x^2 + 15x = 0

I don´t know how to solve this the formulae I usually use I can´t use it since it is x^3 and not x^2... I don´t know what the formulae is called in English.

D) 4(3-3x) (8-2x^2)= 0

Help anyone? :/

2. For C, factor out an x on the LHS. That corresponds to the solution x = 0. Then what remains is a quadratic, for which there are numerous methods of solution. For D, each factor could be zero, so solve them separately.

Make sense?

3. Originally Posted by Ackbeet
For C, factor out an x on the LHS. That corresponds to the solution x = 0. Then what remains is a quadratic, for which there are numerous methods of solution. For D, each factor could be zero, so solve them separately.

Make sense?
Yes, it does. =) Just unsure how to solve D... I tried but failed.

4. Originally Posted by Bluerabbits
Yes, it does. =) Just unsure how to solve D... I tried but failed.
What did you try? Please show all your working. I hope part of your effort involved the null factor law.

5. Originally Posted by mr fantastic
What did you try? Please show all your working. I hope part of your effort involved the null factor law.
4(3-3x)(8-2x^2)=0
I just divided 4 and then I got:

(3-3x)=0 or (8-2x^2)=0

(3-3x)=0
x=1

(8-2x^2)=0
2(4-x^2)=0
4-x^2 = 0
x=±2

6. Originally Posted by Bluerabbits
4(3-3x)(8-2x^2)=0
I just divided 4 and then I got:

(3-3x)=0 or (8-2x^2)=0

(3-3x)=0
x=1

(8-2x^2)=0
2(4-x^2)=0
4-x^2 = 0
x=±2
So why did you say you "tried but failed"?

7. Well in the beginning I was unsure how to approach the number and I didn´t get the right answer. Then I got some help and understood what I did wrong. =)