SOLUTION: Find an equation for each of the sets of given solutions: ({{{sqrt(2) }}}, {{{ -sqrt(2) }}}, 4) and (-2, 2, -i, i). I know that in the first part of quadratic, {{{ x^3 }}} because

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Find an equation for each of the sets of given solutions: ({{{sqrt(2) }}}, {{{ -sqrt(2) }}}, 4) and (-2, 2, -i, i). I know that in the first part of quadratic, {{{ x^3 }}} because      Log On


   

Question 997920: Find an equation for each of the sets of given solutions: (sqrt%282%29+, +-sqrt%282%29+, 4) and (-2, 2, -i, i). I know that in the first part of quadratic, +x%5E3+ because there are 3 solutions, but then how would you figure out the rest of the equation?
Found 2 solutions by josgarithmetic, solver91311:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
You are hoping for something y=ax%5E3%2Bbx%5E2%2Bcx%2Bd, you were given the three solutions (which will be "roots" for when y=0), and you can form the equation ( I should say, "function") in FACTORED form.

Start with a format, %28x-r%29%28x-s%29%28x-t%29=0;
The three roots are given.
%28x-sqrt%282%29%29%28x-%28-sqrt%282%29%29%29%28x-4%29=0
and you may like to simplify and finish in general form.
%28x-sqrt%282%29%29%28x%2Bsqrt%282%29%29%28x-4%29=0
highlight_green%28%28x%5E2-2%29%28x-4%29=0%29----if you want to finish at factors which are irreducible linear or quadratics;
x%5E3-2x-4x%5E2%2B8=0
highlight%28x%5E3-4x%5E2-2x%2B8=0%29-----general form.

Your other set of solution values for a polynomial equation contains four members, and the degree of the equation will be four, and you would use the same kind of method or process just done for the first set of equation solutions you had.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If is a zero of a polynomial, then is a factor of the polynomial. For your first one:



Do the indicated multiplications and collect like terms. Hint: The product of two conjugates is the difference of two squares.

John

My calculator said it, I believe it, that settles it