<p>1) Susan discovered by trial and error that one root of the equation 2x^3-9x^2+16x-12=0 is 2. What are the other roots?</p>
<p>2) Which of the following is an equivalent expression for r in terms of S and t whenever r,S, and t are all distinct and S= rt-3/r-t</p>
<p>answer: st-3/S-t</p>
<p>thanks!</p>
<ol>
<li><p>Divide polynomial by x-2, find the roots of the resulting quadratic. But a simpler way is to let p,q, and 2 be the roots, such that p+q+2 = 9/2 → p+q = 5/2, and 2pq = 6 → pq = 3 (by Vieta’s formulas). Solving for p and q shouldn’t be too difficult.</p></li>
<li><p>Your question is too ambiguous. Use parentheses wherever needed, or LaTeX if you know it. Right now it looks like S = rt - (3/r) - t.</p></li>
</ol>
<p>S= [(r)(t)-3] divided by r-t</p>
<p>Oh okay. S = (rt - 3)/(r-t) is another way to put it. (Or $S = \frac{rt - 3}{r-t}$ in LaTeX).</p>
<p>Multiply both sides by r-t:
Sr - St = rt - 3</p>
<p>Isolate r:
Sr - rt = St - 3</p>
<p>Factor r:
r(S-t) = St - 3 → r = (St - 3)/(S-t)</p>
<p>See, just simple algebra…</p>
<p>simple for you! haha…</p>
<p>I get it now and it is so easy but I went a different wy and got all messed up. I multiplied both sides by r-t like you did. But then I subtracted 3 from both sides instead of doing what you did. Therefore, I after I isolated r, my answer looked nothing like your ( the correct) answer. </p>
<p>How do I know the correct order to do things to isolate the r? DId you just know by looking at the correct answer? </p>
<p>Also, I still dont get the first one…I have no idea how to divide that thing by x-2 and have never heard of the vieta formula…I suppose if I only miss 2 questions that is pretty good for me!</p>
<p>After multiplying by r-t, just move all the terms containing r to one side. I didn’t look at the correct answer until after I solved it.</p>
<p>For the first problem, do you know polynomial or synthetic division?</p>
<p>Also, Vieta’s formulas are very simple to understand. They essentially give expressions for the sum and product of roots of a polynomial.</p>
<p>Suppose our cubic polynomial is in the form ax^3 + bx^2 + cx + d = 0. This has three roots (denote p,q,r) and can be factored to a(x-p)(x-q)(x-r) = 0. If you expand and equate the coefficients, you will see that p+q+r = -b/a, and pqr = -c/a. So in order to find the other two roots, I used Vieta’s formulas to obtain two equations in terms of p and q.</p>
<p>[Vieta’s</a> Formulas - AoPSWiki](<a href=“http://www.artofproblemsolving.com/Wiki/index.php/Vieta’s_Formulas]Vieta’s”>Art of Problem Solving)</p>
<p>Edit: Sorry I made a little mistake on post #2. pq should equal -3, not 3.</p>
<p>nope, I don’t remember it, but I will look it up i guess.</p>
<p>Alright…yeah, it always helps to know at least one of the methods. I would suggest knowing long division since synthetic division only works when you divide by a polynomial in the form x-k (I don’t even remember synthetic division that well).</p>
<p>You said p.q.r = -c/a but c is 16 a is 2 so shouldnt it be 8? As r is 2 p.q should be 4 instead of 3
Im not very sure if it is c/a or -c/a
Im sure that it is -b/a</p>
<p>Wow, idk what I’m doing.</p>
<p>a(x-p)(x-q)(x-r) = ax^3 - a(p+q+r)x^2 + … - a(pqr) = ax^3 + bx^2 + cx + d for coefficients a,b,c,d. Then</p>
<p>-a(p+q+r) = b → p+q+r = -b/a
-a(pqr) = d → pqr = -d/a</p>
<p>That should clear everything. Ignore the other stuff I had previously posted. Sorry about that…</p>
<p>So if p, q, and 2 are the roots, then</p>
<p>p+q+2 = -b/a = 9/2 → p+q = 5/2</p>
<p>2pq = -d/a = 12/2 = 6 → pq = 3 (I used the correct coefficient but somehow put in “-c”).</p>
<p>Substitute p = (5/2) - q:</p>
<p>q((5/2) - q) = 3 → -q^2 + (5/2)q - 3 = 0 → 2q^2 - 5q + 6 = 0 (multiplied by -2).</p>
<p>By quadratic formula, q = (5 ± sqrt(25 - 4(12)))/4 = (5 ± i*sqrt(23))/4, so these are the values of p and q. I checked this using WolframAlpha, everything’s correct.</p>