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Erfan1
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How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?
Erfan said:How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?
You can do as M R suggested or you can put $v=u+1/u$ in $v^2+5v+4=0$. You should get $u^4 + 5u^3 + 6u^2 + 5u + 1 = 0$.Erfan said:How do I reduce u^4 + 5u^3 + 6u^2 + 5u + 1 = 0 to v^2 + 5v + 4 = 0 by using v = u + 1/u ?
The roots of a polynomial equation are the values of the variable that make the equation equal to zero. In other words, they are the solutions to the equation.
One way to find the roots of a polynomial equation is by using substitution. This involves replacing the variable in the equation with a value and solving for the resulting expression to equal zero. This process is repeated for each root of the equation.
Finding the roots of a polynomial equation can help in understanding the behavior of the equation and its graph. It also allows for solving real-world problems that involve finding the values of a variable.
No, not all polynomial equations can be solved using substitution. Some equations may have irrational or complex roots that cannot be found using substitution alone. In these cases, other methods such as factoring or using the quadratic formula may be necessary.
The fundamental theorem of algebra states that a polynomial equation of degree n will have n roots, including complex roots. This means that the number of roots is equal to the degree of the polynomial equation.