1. ## factor

hello all, i was scanning my memory for some of the questions that were on my math exam, could you please confirm the following answers:

1.) is k^2 - 2k + 4 a factor of K^3 - 8? My answer: NO.

2.) For a solution to have two real and unequal solution, the discriminant should be? my answer: GREATER THAN 0

3.) Simplify (a-b)^2 - (a+b)^2. My answer: -4ab

4.) There are two decks of 52 playing cards, find the probability of obtaining a club or a king: my answer: 16/52 or 4/13.

5.) x^2 - 6x + 1 in turning point form has values of a,h and k as? my answer: a = 1, h = 3, k = 2.

6.) x^2 - 5x +2 factorised completely is? My answer: (x-5+root7/2) (x-5-root7/2)

thanks guys, i hope i did well

2. Originally Posted by andrew2322
hello all, i was scanning my memory for some of the questions that were on my math exam, could you please confirm the following answers:

1.) is k^2 - 2k + 4 a factor of K^3 - 8? My answer: NO.

2.) For a solution to have two real and unequal solution, the discriminant should be? my answer: GREATER THAN 0

3.) Simplify (a-b)^2 - (a+b)^2. My answer: -4ab

4.) There are two decks of 52 playing cards, find the probability of obtaining a club or a king: my answer: 16/52 or 4/13.

5.) x^2 - 6x + 1 in turning point form has values of a,h and k as? my answer: a = 1, h = 3, k = 2.

6.) x^2 - 5x +2 factorised completely is? My answer: (x-5+root7/2) (x-5-root7/2)

thanks guys, i hope i did well
nope

3. ## hey

how is question 5 wrong?

4. Originally Posted by andrew2322
how is question 5 wrong?
x^2 - 6x + 1 = (x - 3)^2 - 8

5. Jhevon pointed out your errors (and my work agrees), but I'll see if I can lead you on the right track:

Originally Posted by andrew2322
5.) x^2 - 6x + 1 in turning point form has values of a,h and k as? my answer: a = 1, h = 3, k = 2.
The standard form is $y = a(x - h)^2 + k$. To get the original equation in this form, we must complete the square. Ignore the constant term for now, and concentrate on the expression $x^2 - 6x$. We want to add a term to this that will give us a perfect square. If we consider expressions of the form $x^2 + bx$, then the constant term that we need to add is $\left(\frac b2\right)^2$. Why? Because a perfect square of the form $\left(m + n\right)^2 = m^2 + 2mn + n^2$, which gives $m^2 = x^2,\;2mn = bx,\text{ and so }n^2 = \left(\frac{bx}{2m}\right)^2 = \left(\frac{bx}{2x}\right)^2 = \left(\frac b2\right)^2$.

So, we add $\left(\frac b2\right)^2$ to our expression to form a perfect square, and then we can factor:

$x^2 - 6x = x^2 - 6x + \left(\frac62\right)^2 - \left(\frac62\right)^2$

$= x^2 - 6x + 9 - 9$

$= x^2 - 2\cdot3\cdot x + 3\cdot3 - 9$

$= (x - 3)^2 - 9$

So going back to our original equation, we find that

$x^2 - 6x + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8$

and $\left\{\begin{array}{rcr}
a & = & 1\\
h & = & 3\\
k & = & -8
\end{array}\right.$
.

Originally Posted by andrew2322
6.) x^2 - 5x +2 factorised completely is? My answer: (x-5+root7/2) (x-5-root7/2)
This one is, in form, close enough to the actual solution that I might suspect a typo. If not, just go over your work once more as you probably just made a silly mistake somewhere.