Finding an nth degree polynomial given the zeros and one value (MAC 1105 Sec 3.4) Show
by Gabrielle Noyes 9 years ago
I teach high school mathematics at...
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I have no idea how I would figure this out. The problem states: Find an nth-degree polynomial function with real coefficients satisfying the given conditions. by the way, this is the answer: My question is, how do I get this? I'm stumped.
An nth-degree polynomial has exactly n roots (considering multiplicity). The roots of a polynomial are exactly the same as the zeros of the corresponding polynomial function. So, they say "zeros" and I'm calling them roots. Since n = 3, you need 3 roots. They gave you two of them: 2 and 5i. Strict Complex roots always come in conjugate pairs, so right away we know that the third root is -5i. Each root forms a factor (x - root) of the polynomial. The factorized polynomial function takes this form: f(x) = a (x - root1) (x - root2) (x - root3) where a is the leading coefficient Substitute the known values for f(x), x, root1, root2, root3. Solve the resulting equation for the leading coefficient a. If you want to expand the factorized polynomial, you can multiply it all out, at the very end. You should end up with the posted answer. If I wrote anything that you don't understand, please ask specific questions. Otherwise, show whatever work that you can, to get more assistance. Cheers ~ Mark
Once I set it up this is what I get. Am I setting it up wrong or am I multiplying it wrong? This is my setup: This is the answer I need.
A=-1. so multiply your version by -1 and change the signs. \(\displaystyle f(x)=a(x-2)(x+5i)(x-5i)\) For f(1)=26: \(\displaystyle a(1-2)\underbrace{(1-5i)(1+5i)}_{\text{this is 26}}=26\) \(\displaystyle a(-1)(26)=26\) \(\displaystyle a=-1\) \(\displaystyle -(x-2)(x+5i)(x-5i)=-x^{3}+2x^{2}-25x+50\)
dove99x said: Am I setting it up wrong or am I multiplying it wrong? Your multiplications are fine, but you did not substitute in all of the known values. f(x) = a (x - root1) (x - root2) (x - root3) Substitute the known values for f(x), x, root1, root2, root3.
Solve the resulting equation for the leading coefficient a. You can't find the value of a, without making these substitutions. Perhaps, you didn't realize that you need to know the leading coefficient, before you're able to multiply out the polynomial which defines function f. Otherwise, you seem to know what you're doing. I mean, your multiplications shown are all correct.
Cheers How do you find the nth degree of a polynomial equation?Check-out interactive examples on nth degree polynomial function with real coefficients.
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General Form of Nth Degree Polynomial.. What is a polynomial function with real coefficients?A polynomial having only real numbers as coefficients. A polynomial with real coefficients is a product of irreducible polynomials of first and second degrees.
How do you find the degree of a polynomial on a calculator?To determine the degree of the polynomial, add up the exponents of each term, and select the highest sum if the expression is having two variables. The degree of the polynomial can be determined as the largest exponent of the variable if the expression is having one variable.
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