If the equation x^5-3x^4-5x^3+27x^2-32x+12=0 | vedclass

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(C) Let $f(x) = x^5-3x^4-5x^3+27x^2-32x+12$. To find the roots,we test for small integer values. $f(1) = 1-3-5+27-32+12 = 0$,so $(x-1)$ is a factor. $

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$3$

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(C) Let $f(x) = x^5-3x^4-5x^3+27x^2-32x+12$. To find the roots,we test for small integer values. $f(1) = 1-3-5+27-32+12 = 0$,so $(x-1)$ is a factor. $f'(x) = 5x^4-12x^3-15x^2+54x-32$. $f'(1) = 5-12-15+54-32 = 0$,so $(x-1)^2$ is a factor. $f''(x) = 20x^3-36x^2-30x+54$. $f''(1) = 20-36-30+54 = 8 
eq 0$. Thus,$x=1$ is a root of multiplicity $2$. Dividing $f(x)$ by $(x-1)^2 = x^2-2x+1$,we get $x^3-x^2-6x+12$. Testing $x=2$: $8-4-12+12 = 4 
eq 0$. Testing $x=-3$: $-27-9+18+12 = -6 
eq 0$. Testing $x=2$ in $f'(x)$ gives $f'(2) = 5(16)-12(8)-15(4)+54(2)-32 = 80-96-60+108-32 = 0$. So $x=2$ is a root of $f(x)$ and $f'(x)$,meaning $(x-2)^2$ is a factor. Dividing $f(x)$ by $(x-1)^2(x-2)^2 = (x^2-2x+1)(x^2-4x+4) = x^4-6x^3+13x^2-12x+4$,we get $(x+3)$. The roots are $1, 1, 2, 2, -3$. The non-repeated root is $-3$. The prime number that divides $-3$ is $3$.

If the equation $x^5-3x^4-5x^3+27x^2-32x+12=0$ has repeated roots,then the prime number that divides the non-repeated root of this equation is

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