Set each factor, in turn, equal to zero and solve for x.
Algebra
You need to multiply three terms, one for each zero. To have only two zeros, the polynomial would need to have a "double zero" (or more generally, a "multiple zero), that is, a repeated factor. In this case, the zeros can be one of the following: -1, -1, 1, with the corresponding factors: (x+1)(x+1)(x-1) or: -1, 1, 1, with the corresponding factors: (x+1)(x-1)(x-1) If you like, you can multiply these factors out to get the polynomial in standard form.
irreducible polynomial prime...i know its the same as irreducible but on mymathlab you would select prime
a2-b2 = (a-b)(a+b)
She was afraid it would be constant. (Constance) She was afraid it would be a related function.
If a polynomial has factors x-6 and x-3 it will equal 0 if either factor equals 0 since the other factor then would be multiplied by 0. ie. 0 * (x-6)=0 and 0 * (x-3)=0. so x=3 or 6
No. It would not be a polynomial function then.
If a number has equal factors, it is a perfect square and the equal factors would be square roots.
to get the factors of a number you figure out what numbers multiplied together would equal the number youre trying to get factors for
the mean is the average of the values of a population. the mean of 2, 8, 11 would be 7 while the median (equal numbers of values above and below) would be 8
Algebra
I am assuming this is: .2x4 - 5x2 - 7x, which would be a Quartic Polynomial.
You need to multiply three terms, one for each zero. To have only two zeros, the polynomial would need to have a "double zero" (or more generally, a "multiple zero), that is, a repeated factor. In this case, the zeros can be one of the following: -1, -1, 1, with the corresponding factors: (x+1)(x+1)(x-1) or: -1, 1, 1, with the corresponding factors: (x+1)(x-1)(x-1) If you like, you can multiply these factors out to get the polynomial in standard form.
Yes, over the real set of numbers. For example, the graph of y=x2+1 is a regular parabola with a vertex that is one unit above the origin. Because the vertex is the lowest point on the graph, and 1>0, there is no way for it to touch the x-axis.NOTE: But if we're considering imaginary numbers, the values "i" and "-i" would be the zeroes. I'm pretty sure that all polynomial functions have a number of zeroes equal to their degree if we include imaginary numbers.
(a+b)(a-b)
The square root of a polynomial is another polynomial that, when multiplied by itself, yields the original polynomial. Not all polynomials have a square root that is also a polynomial; for example, the polynomial (x^2 + 1) does not have a polynomial square root in the real number system. However, some polynomials, like (x^2 - 4), have polynomial square roots, which in this case would be (x - 2) and (x + 2). Finding the square root of a polynomial can involve techniques such as factoring or using the quadratic formula for quadratic polynomials.
These are values which make the denominator equal zero, therefore the system described by the transfer function would be unstable near these values.