Sabendo que p(x)=q(x) onde p(x)=22x3+7x2-6 e q(x)=(3m-15)x3+7x2-6, calcule o valor de m
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Solução!
![p(x)=q(x)\\\\\\
p(x)=22 x^{3}+7 x^{2} -6\\\\\\
q(x)=(3m-15) x^{3}+7 x^{2} -6 \\\\\\\
22 x^{3}+7 x^{2} -6=(3m-15) x^{3}+7 x^{2} -6\\\\\\
22 x^{3}+7 x^{2} -7 x^{2} -6+6=(3m-15) x^{3}\\\\\\\
22 x^{3}=(3m-15) x^{3}. (\dfrac{1}{ x^{3} }) \\\\\
22= 3m-15\\\\\
-3m=-15-22\\\\\
-3m=-37\\\\\
m= \dfrac{-37}{-3}\\\\\
m= \dfrac{37}{3}](https://tex.z-dn.net/?f=p%28x%29%3Dq%28x%29%5C%5C%5C%5C%5C%5C%0Ap%28x%29%3D22+x%5E%7B3%7D%2B7+x%5E%7B2%7D+-6%5C%5C%5C%5C%5C%5C%0Aq%28x%29%3D%283m-15%29+x%5E%7B3%7D%2B7+x%5E%7B2%7D+-6++%5C%5C%5C%5C%5C%5C%5C%0A22+x%5E%7B3%7D%2B7+x%5E%7B2%7D+-6%3D%283m-15%29+x%5E%7B3%7D%2B7+x%5E%7B2%7D+-6%5C%5C%5C%5C%5C%5C%0A22+x%5E%7B3%7D%2B7+x%5E%7B2%7D+-7+x%5E%7B2%7D+-6%2B6%3D%283m-15%29+x%5E%7B3%7D%5C%5C%5C%5C%5C%5C%5C%0A22+x%5E%7B3%7D%3D%283m-15%29+x%5E%7B3%7D.+%28%5Cdfrac%7B1%7D%7B+x%5E%7B3%7D+%7D%29++%5C%5C%5C%5C%5C%0A22%3D+3m-15%5C%5C%5C%5C%5C%0A-3m%3D-15-22%5C%5C%5C%5C%5C%0A-3m%3D-37%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7B-37%7D%7B-3%7D%5C%5C%5C%5C%5C%0Am%3D+%5Cdfrac%7B37%7D%7B3%7D++++++ "p(x)=q(x)\\\\\\
p(x)=22 x^{3}+7 x^{2} -6\\\\\\
q(x)=(3m-15) x^{3}+7 x^{2} -6 \\\\\\\
22 x^{3}+7 x^{2} -6=(3m-15) x^{3}+7 x^{2} -6\\\\\\
22 x^{3}+7 x^{2} -7 x^{2} -6+6=(3m-15) x^{3}\\\\\\\
22 x^{3}=(3m-15) x^{3}. (\dfrac{1}{ x^{3} }) \\\\\
22= 3m-15\\\\\
-3m=-15-22\\\\\
-3m=-37\\\\\
m= \dfrac{-37}{-3}\\\\\
m= \dfrac{37}{3}")
Boa noite!
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Solução!
Boa noite!
Bons estudos!
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