Difference between revisions of "2023 AMC 12B Problems/Problem 6"

Revision as of 19:39, 15 November 2023

Solution 1

$P(x)$ is a product of $(x-r_n)$ or 10 terms. When $x < 1$ , all terms are $< 0$ , but $P(x) > 0$ because there is an even number of terms. The sign keeps alternating $+,-,+,-,....,+$ . There are 11 intervals, so there are $\boxed{\textbf{6}}$ positives and 5 negatives. $\boxed{\textbf{(C) 6}}$

~ $\textbf{Techno}\textcolor{red}{doggo}$

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 ==Solution 1==
-Looking at the polynomial, we can use Vieta's formulas. Let's say the roots are r, s, and t. Then
-<cmath>r+s+t = -a</cmath>
-<cmath>rs+rt+st = b</cmath>
-<cmath>rst = 6</cmath>
-We know that r, s, and t are integers, so we can list the possibilities for distinct solutions using the third formula. We find that <math>(r,s,t)</math> can be <math>(1,2,3), (-1,-2,3), (-1,2,-3), (1,2,-3),</math> and <math>(1,-1,-6)</math>. Remember that r, s, and t have to be different. That's 5 solutions. We can also use those values to find the values of a and b to check that none of them are distinct, and we'll find that they are <math>(6,11), (0,-7), (-2,-5),  (-4,1),</math> and <math>(-6,-1)</math>. This means that there are <math>\boxed{5}</math>
-~lprado
+<math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because there is an even number of terms. The sign keeps alternating <math>+,-,+,-,....,+</math>.  There are 11 intervals, so there are <math>\boxed{\textbf{6}}</math> positives and 5 negatives. <math>\boxed{\textbf{(C) 6}}</math>
+~<math>\textbf{Techno}\textcolor{red}{doggo}</math>

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Difference between revisions of "2023 AMC 12B Problems/Problem 6"

Revision as of 19:39, 15 November 2023

Solution 1