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As a result, we have the following solution: $$ int rac1x(x+1)(x+2)(x+3),, dx = rac16ln|x| - rac12ln|x+1| + rac12ln|x+2| - rac16ln|x+3| + rac12ln|x+1| + rac12ln|x+1| + rac12ln|x+1| + rac12ln C $$Now we get the following integral for $m = 4$: $$ int rac1x(x+1)(x+2)(x+3)(x+4),, dx $$eginalign* rac1x(x+1)(x+2)(x+3)(x+4) &= racAx + racBx+1 + racCx+2 + racDx+3 + racEx+4

(X+1)(X+2)(X+3)(X+6)-3X^2 Solution

96,48,32,24,16,12,8,6,4,3,2,1 All rational roots of a polynomial have the form racpq, where p divides the constant term 96 and q divides the leading coefficient 1. List all potential candidates racpq. x=-6 Find one such root by testing out all the integer values, beginning with the least absolute value and working your way up. If no integer roots are discovered, experiment with fractions.

Factorize (X+1)(X+2)(X+3)(X+6)-3X^2

That is incorrect. You can't do that because, unlike multiplying two reals, where the only way to obtain $0$ is if one of them is $0$, receiving a $6$ as a result of a product does not need one of the components to be $6$. Your reasoning would have been incorrect in the reals, and it would have been incorrect modulo $9$ even if you had gotten the congruence $x(x-1)equiv 0pmod9$. The reason for this is because while working modulo $9$, it is conceivable for a product to be $0$ when neither component is $0$: for example, $(3)(3)equiv 0pmod9$. So, while working modulo $9$, you can't even use this kind of input when the product is $0$, much alone when it isn't.

According to John Hughes and Gerry Myerson, the polynomial $x5-1$ divides into linear components over the field $F=BbbZ 11$. Because the multiplicative group $K*$ is cyclic of order ten (we can easily prove that $2$ is a generator), all non-zero squares in $K$ are $x5-1$ zeros. This results in the linear factors $$ x-1,x-4,x-9,x-16=x-5,

If (X+1)(X+2)(X+3)(X+6)=3X^2

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