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## Replies to: SAT Math Problems Thread

• Registered User Posts: 304 Member
edited September 2016
@tiffym I solved this problem quickly by imagining the graphs. Imagine y=3; it's a straight line. The graph y=ax^2 + b will always be a parabola where the vertext is located at (0, b). This means that, in order for the two equations to intersect at two locations, either the parabola has to be concave up (i.e. a>0) and b<3 OR the parabola has to be concave down (i.e. a<0) and b>3. (B) is the only answer that fits this criterion. This might not be the ideal way to view the question, but it was quick and easy that might come in handy.
• Registered User Posts: 1,289 Senior Member
@ObitoSigma I think that this is a really nice solution. Stronger students can probably follow your reasoning pretty easily. Students weaker in math can actually put each of the equations in their graphing calculator along with y=3, and see which one yields 2 points of intersection (if a calculator is allowed).
• Registered User Posts: 12 New Member
For a polynomial p x ( ), the value of p(3) is −2.
Which of the following must be true about p x ( ) ?
A) x − 5 is a factor of p x ( ).
B) x − 2 is a factor of p x ( ).
C) x + 2 is a factor of p x ( ).
D) The remainder when p x ( ) is divided
by x − 3 is −2.

Can someone simplify the answer and the way.
• Registered User Posts: 1,289 Senior Member
The answer is choice D by a direct application of the remainder theorem.

The remainder when the polynomial p(x) is divided by x - r is p(r).
• Registered User Posts: 4,747 Senior Member
@Amitre yes, this follows from the remainder theorem.

It is also worth knowing that r is a root of polynomial p(x) if and only if x-r is a factor of p(x) (that is, p(x) = (x-r)q(x) for some polynomial q(x)).

Also, use p or p(x) -- it is not p times ().
• Registered User Posts: 12 New Member
Thank you guys
• Registered User Posts: 2 New Member
Yes, the factor theorem and the remainder theorem are all useful. For last minutes studying, you might find the following videos very helpful!

• Registered User Posts: 2 New Member
Every polynomial can be written using the division algorithm for polynomials as P(x)=(x-a)Q(x)+r(x) where Q(x) is a polynomial of degree n-1 when P(x) is of degree n. Then, r(x) has to be a constant in this particular case because it has to be of lower degree than (x-a), therefore, r(x) is of degree 0, hence a constant.

So, we have:

P(x)=(x-a)Q(x)+r for a constant r.

Then, if (x-a) is a factor of P(x), then, P(a)=0 meaning, 0=P(a)=(a-a)Q(a)+r so, 0=0+r, that is, the remainder r must be zero if in fact (x-a) is a factor of P(x).

Moreover,

If (x-a) is not a factor, we have:

P(a)=(a-a)Q(a)+r which implies

P(a)=r, which is in fact the case. That is, when you divide a polynomial by x-b, then, either P(b)=0, that is, x-b is a factor, or P(b)=r which is that the value of the remainder is the value of the function at x=b.

Hope this helps!
• Registered User Posts: 146 Junior Member
Hi there,
there is this one problem that I am stuck on for some reason. Would be grateful if anyone can help.
So there is a circle with a radius of two and it has an arc. IN the figure, points A and B lie on the circle with center O. If the length of minor arc AB is less than pi/2 but grater than pi/4, what is one possible value of x
• Registered User Posts: 146 Junior Member
nvm I figured it out any number between 22.5 and 45 would work.
• Registered User Posts: 146 Junior Member
Here is another problem:
A group of people were asked if they are or are not registered organ donors. An equal number of men and women were surveyed and partial results of the data are shown. Three times as many men responded that they were not registered organ donors than men who responded that they were registered organ donors, and fifty more women responded that that they were not registered organ donors than women who responded that they were registered organ donors. If an individual is selected at random from this group, what is the probability that the person is a man who is a registered organ donor. There are total of 350 organ donors and a total of 650 not organ donors.
• Registered User Posts: 1,079 Senior Member
As is almost always the case, you can puzzle this out without using algebra (though algebra might be quicker).

We need 300 more non-donors than donors. From the women, we get 50 more. So we need 250 more from the men. Since there are three times as many non-donor men than donor men, we need a number that when tripled, gives us the 250 surplus. You may realize that double the number must be 250 or you may do trial and error to land on 125 donor men, 375 non-donor. So 500 men over all, 1000 people total and 125 who are male donors.

I'll let someone else post an algebra-based answer. Where is the problem from?
• Registered User Posts: 119 Junior Member
A thread filled with math problems :)
Are all the problems from the SAT or can there be other types of math problems? Either way, I'm definitely going to be working through them in the next few days.
• Registered User Posts: 1 New Member
edited June 2017
You are great!
• Registered User Posts: 27 Junior Member
I used 375 and found that both C and E worked. I was confused at first until I saw "the next larger integer". I guess you have to make sure you read the question right.