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Algebra Level 5

If the polynomial p ( x ) = x 8 + 98 x 4 + 1 p(x) = x^{8} + 98x^{4} + 1 can be expressed by two factors with integer coefficients as p ( x ) = ( a x 4 + b x 3 + c x 2 + d x + e ) ( f x 4 + g x 3 + h x 2 + i x + j ) p(x) = (ax^{4} + bx^{3} + cx^{2} + dx + e) (fx^{4} + gx^{3} + hx^{2} + ix + j) where a , b , c , d , e , f , g , h , i , j a,b,c,d,e,f,g,h,i,j are integers.

Then find the value of ∣ a + b + c + d + e + f + g + h + i + j ∣ |a+b+c+d+e+f+g+h+i+j| .


The answer is 20.

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Mohit Gupta by Mohit Gupta , 20, India

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1 solution

x 8 + 98 x 4 + 1 = 0 x 4 + 1 x 4 = − 98 x 4 + 2 + 1 x 4 = − 96 x 2 + 1 x 2 = ± 4 6 i x 2 − 2 + 1 x 2 = − 2 ± 4 6 i x − 1 x = ± ( 2 ± 6 i ) ( x − 1 x ± 2 ) 2 = − 6 ( x − 1 x ) 2 ± 4 ( x − 1 x ) + 10 = 0 x 2 − 2 + 1 x 2 ± 4 x ∓ 4 x + 10 = 0 x 4 ± 4 x 3 + 8 x 2 ∓ 4 x + 1 = 0 x^8+98x^4+1=0 \\ x^4+\frac{1}{x^4}=-98 \\ x^4+2+\frac{1}{x^4}=-96 \\ x^2+\frac{1}{x^2}=\pm 4\sqrt{6}i \\ x^2-2+\frac{1}{x^2}=-2\pm 4\sqrt{6}i\\ x-\frac{1}{x}=\pm (2\pm \sqrt{6}i) \\ \left(x-\frac{1}{x}\pm 2\right)^2=-6 \\ \left(x-\frac{1}{x}\right)^2\pm 4\left(x-\frac{1}{x}\right)+10=0 \\ x^2-2+\frac{1}{x^2}\pm 4x \mp \frac{4}{x}+10=0 \\ x^4\pm 4x^3+8x^2 \mp 4x+1=0

So, we must have p ( x ) = ( x 4 + 4 x 3 + 8 x 2 − 4 x + 1 ) ( x 4 − 4 x 3 + 8 x 2 + 4 x + 1 ) p(x)=(x^4+4x^3+8x^2-4x+1)(x^4-4x^3+8x^2+4x+1) .

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Did the same way (+1).

Aditya Sky - 5 years, 1 month ago

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