An Algebra Problem by Chakravarthy B.

Algebra Level 3

x 3 + x 2 + 2 x + 36 x − 5 = 3 x 3 + 6 x + 183 3 x − 15 \frac{x^3+x^2+2x+36}{x-5}=\frac{3x^3+6x+183}{3x-15}

Find the number of non-negative integer solutions to the equation above.

5 We can't say 2 1 0

Chakravarthy B by chakravarthy b , 17, India

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1 solution

Jordan Cahn
Mar 8, 2019

Note that x 3 + x 2 + 2 x + 36 x − 5 = 3 x 3 + 6 x + 183 3 x − 15 x 3 + x 2 + 2 x + 36 x − 5 = x 3 + 2 x + 61 x − 5 x 3 + x 2 + 2 x + 36 = x 3 + 2 x + 61 Assuming x ≠ 5 x 2 = 25 x = ± 5 \begin{aligned} \frac{x^3+x^2+2x+36}{x-5} &= \frac{3x^3 + 6x + 183}{3x-15} \\ \frac{x^3+x^2+2x+36}{x-5} &= \frac{x^3 + 2x + 61}{x-5} \\ x^3 + x^2 + 2x + 36 &= x^3 + 2x + 61 && \color{#3D99F6}\text{Assuming }x\neq 5 \\ x^2 &= 25 \\ x &= \pm 5 \end{aligned} But we assumed that x ≠ 5 x\neq 5 . Thus x = − 5 x=-5 is the only solution and there are 0 \boxed{0} non-negative solutions.

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