# Thread: Factoring Quadratic Equations

1. ## Factoring Quadratic Equations

Can i get some help these are the last 5 problems that im stuck with. -_-

(COMPLETED)1) (X-Y)^3 +27

(COMPLETED)2) X^2-A^2+X-A

(COMPLETED)3) 3x^(-3/2) - 12x^(1/2)

4) x^(1/2) (x-4)^(-1/2) +3x^(-1/2) (x-4)^(-3/2)

(COMPLETED)5) x^(2n+2) +x^(n+2)-56x^2

for 3 im think its (3x^-3/2)(1-4x^2) but yeah im not sure. but everything else im clueless at right now. thxs for hte help!

2. 1) $\displaystyle 27 = 3^3$, difference of two cubes.

2) If you factorize (difference of two squares) the two first terms, you get $\displaystyle (x - a)(x + a) + (x - a)$. Can you see the common factor ?

Now I have no time for the other ones, sorry. I did 40% of the job though !

3. Originally Posted by Nismo
...

3) 3x^(-3/2) - 12x^(1/2)

...

for 3 im think its (3x^-3/2)(1-4x^2) but yeah im not sure. but everything else im clueless at right now. thxs for hte help!
1. Your result is OK but unfinished:

2. When factoring out a term from a sum you have to divide each summand by the factor. In your case:

$\displaystyle 3x^{-\frac32} - 12x^{\frac12} = 3x^{-\frac32} \left(\dfrac{3x^{-\frac32}}{3x^{-\frac32}} - \dfrac{12x^{\frac12}}{3x^{-\frac32}} \right) = 3x^{-\frac32}\left(1-4x^2\right)$

3. The sum in the bracket is a difference of squares which can be factorized immediately. Therefore the result is:

$\displaystyle 3x^{-\frac32} - 12x^{\frac12} = 3x^{-\frac32}(1-2x)(1+2x)$

4. thxs earboth i just noticed that i didnt factor it all the way out -_- it was like 12ish i think when i was doing this.. -_-

also for that 1st problem i was getting

((x-y)-3)(x^2-2xy+y^2+3x-3y+9) but it looks wrong. is it?

and for 2 i actually did factor out to that

but didnt know where to go from thier. i thought it would be somthing like

(x-a)((1+1)(1-1))-1 but i dont think its that.

5. 1. $\displaystyle a^3+b^3= (a+b)(a^2-ab+b^2)$
you need to leave (x-y) intact and not multiply it out.

5. Take out a factor of x^2 and then it may look familiar.

Originally Posted by Nismo
and for 2 i actually did factor out to that

but didnt know where to go from thier. i thought it would be somthing like

(x-a)((1+1)(1-1))-1 but i dont think its that.
take out (x-a)

(x-a)(x+a)+(x-a)=(x-a)((x+a)+1)

6. Originally Posted by Krahl
1. $\displaystyle a^3+b^3= (a+b)(a^2-ab+b^2)$
you need to leave (x-y) intact and not multiply it out.

5. Take out a factor of x^2 and then it may look familiar.

take out (x-a)

(x-a)(x+a)+(x-a)=(x-a)((x+a)+1)

i see how you did it, and that it does distribute out to the x^2-y^2+x-a but im still getting marked wrong on it o_0

wouldnt it just be (x-a)(x+a+1) without the prenthesis in side?

7. not sure about #4. this is what i am up to

$\displaystyle \frac{(x - 3)(x - 1)}{\sqrt{x^2 - 4x}(x - 4)}$

8. Originally Posted by Nismo
i see how you did it, and that it does distribute out to the x^2-y^2+x-a but im still getting marked wrong on it o_0

wouldnt it just be (x-a)(x+a+1) without the prenthesis in side?
If your doing one of them online things then sometimes extra brackets do effect answers. try it without the brackets see what it says

9. naw found out my teacher did his coding wrong. and he said it was right! thxs for the help! although i still gotta finish the rest of the problems!! ha.

10. 5. Take out a factor of x^2 and then it may look familiar.

x^(2n+2) +x^(n+2)-56x^2= $\displaystyle x^2x^{2n}+x^2x^n-56x^2$
$\displaystyle =x^2(x^{2n}+x^n-56)=x^2((x^n)^2+x^n-56)$

can you factorise the stuff in the brackets?
hint y^2+y-56=(y+8)(y-7)

11. Originally Posted by Krahl
5. Take out a factor of x^2 and then it may look familiar.

x^(2n+2) +x^(n+2)-56x^2= $\displaystyle x^2x^{2n}+x^2x^n-56x^2$
$\displaystyle =x^2(x^{2n}+x^n-56)=x^2((x^n)^2+x^n-56)$

can you factorise the stuff in the brackets?
hint y^2+y-56=(y+8)(y-7)

haha i actually got that far last night with the x^2((x^n)^2+x^n-56) but i didnt know how to factor it out from thier although i knew it had to deal with 7 and 8. but yeah i got it fianlly! thxs!