# Placement tests

• Jun 18th 2007, 08:55 AM
michele921
Placement tests
I have to take a placement test for my college orientation, and i have no problem with the highschool level math, but the college level looks rediculous.:confused:

if i post the link then maybe someone could explain how to do it? its really short, not a lot of problems. and the answers are given lol i just cant get to them. Thanks. :o

<3 michele

http://newton.newhaven.edu/soar/math.pdf <-- test

• Jun 18th 2007, 09:03 AM
Pterid
Are there any questions in particular you'd like to discuss? Surely you can't be stuck on all of them!
• Jun 18th 2007, 11:56 AM
michele921
Well how about maybe 2, 3 and 7?:)
• Jun 18th 2007, 12:05 PM
Pterid
Sure. :)

Question 2: We've got a function of $x$, $f(x) = x^3 - 2x^2 + x - 1$. We want $f(-x)$, so write $-x$ in place of $x$ in the expression:

$f(-x) = (-x)^3 - 2(-x)^2 + (-x) - 1 = -x^3 - 2x^2 - x - 1$

using $(-x)^2 = x; (-x)^3 = -x$.

(I need to pop out, to anyone else for the others?)
• Jun 18th 2007, 12:17 PM
michele921
wow thanks, its a lot less scary looking now! your the best! :D
• Jun 18th 2007, 01:04 PM
Jonboy
For # 3 whenever you have an expression in the form.. $\frac{a}{b}\,=\,\frac{c}{d}$..you can multiply diagonally and the following is true.. $a\,\cdot\,d\,=\,b\,\cdot\,c$

So we have:.. $\frac{3}{1}\,=\,\frac{8\,-\,r}{r}$

Cross multiply:.. $(1)(8\,-\,r)\;=\;3r$

Solve for $r$. You should get 2 for $r$.

Now for # 7 are you familiar with the factoring method for quadratics?

Whenver you have an expression in the form.. $ax^2\,+\,bx\,+\,c$.. you look for 2 numbers that multiply to get you $c$ and add up to $b$.

So our quadratic is:.. $\;x^2\,+\,x\,-\,12$

Since a is 1, we always start with this: $(x\,+\;\;)(x\,+\;\;)$

But since c is negative, we now have: $(x\,+\;\;)(x\,-\;\;)$

We know $6\,\cdot\,2=\,12$

Also $12\,\cdot\,1\,=\,12$

Plus $4\,\cdot\,3\,=\,12$

We need the two multiples to add up to 1. These are $4$ and $-3$.

So the factored form is:.. $\boxed{(x\,+\,4)(x\,-\,3)}$