Difference between revisions of "2008 iTest Problems/Problem 85"
Rockmanex3 (talk | contribs) (Solution to Problem 85 (credit to tenniskidperson3 and official solution) -- system with a hidden form) |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
p(x) &= (x^2 + ax + c)(x^2 + bx + d) \\ | p(x) &= (x^2 + ax + c)(x^2 + bx + d) \\ | ||
− | &= x^4 + (a+ | + | &= x^4 + (a+b)x^3 + (ab+c+d)x^2 + (ad+bc)x + cd \\ |
− | &= x^4 + 15x^3 + 78x + 160x + 96 | + | &= x^4 + 15x^3 + 78x^2 + 160x + 96 |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Factoring <math>p(x)</math> with the [[Rational Root Theorem]] results in <math>(x+4)(x+4)(x+1)(x+6)</math>. By the [[Fundamental Theorem of Algebra]], we know that <math>x+4, x+4, x+1, x+6</math> are all the linear factors of the polynomial, so the quadratic factors can only be multiplied from these linear factors. | Factoring <math>p(x)</math> with the [[Rational Root Theorem]] results in <math>(x+4)(x+4)(x+1)(x+6)</math>. By the [[Fundamental Theorem of Algebra]], we know that <math>x+4, x+4, x+1, x+6</math> are all the linear factors of the polynomial, so the quadratic factors can only be multiplied from these linear factors. |
Latest revision as of 22:18, 15 August 2020
Problem
Let be a solution to the system Find the greatest possible value of .
Solution
Note that when multiplying quadratics, terms add up similar to the equations of a system, so let Factoring with the Rational Root Theorem results in . By the Fundamental Theorem of Algebra, we know that are all the linear factors of the polynomial, so the quadratic factors can only be multiplied from these linear factors.
There are only two possible distinct groupings (not counting rearrangements) -- and . In the first case, , and in the second case, . The largest of the two options is .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 84 |
Followed by: Problem 86 | |
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