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Akimi4 [234]
2 years ago
9

Can someone help me with number 7 the circled one?

Mathematics
1 answer:
Ivenika [448]2 years ago
5 0
Your answer:
<span>Equation:
x(x+1) = 2(x+1)+77
-----

x^2+x = 2x+79
x^2-x-79 = 0
------
This quadratic does not have an integer solution.</span>
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KATRIN_1 [288]

Answer:

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Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}\\\\-3x^5+11x^4+33x^3-26x^2-36x-6\\\\\mathrm{and\:the\:divisor\:}x^3+6x^2-3x-5:\\\\\mathrm{Quotient}=\frac{-3x^5}{x^3}=-3x^2\\\\\mathrm{Multiply\:}x^3+6x^2-3x-5\mathrm{\:by\:}-3x^2\:\:\rightarrow\:\:-3x^5-18x^4+9x^3+15x^2\\\\\mathrm{Subtract\:}-3x^5-18x^4+9x^3+15x^2\mathrm{\:from\:}-3x^5+11x^4+33x^3-26x^2-36x-6\mathrm{\:to\:get\:new\:remainder}.\\\\\mathrm{Remainder}=29x^4+24x^3-41x^2-36x-6

=-3x^2+\frac{29x^4+24x^3-41x^2-36x-6}{x^3+6x^2-3x-5}

Repeat the steps and you will reach a point where no further division is possible.

<u />

<u />=-3x^2+29x-150+\frac{946x^2-341x-756}{x^3+6x^2-3x-5}<u />

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