Set each factor, in turn, equal to zero and solve for x.
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.
Algebra
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
If a number has equal factors, it is a perfect square and the equal factors would be square roots.
No. It would not be a polynomial function then.
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
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.
Algebra
I am assuming this is: .2x4 - 5x2 - 7x, which would be a Quartic Polynomial.
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.
I assume x2 + 5x - 36 is the polynomial in the question. First, look for two factors of 36 that have a difference of 5, which would be 9 and 4. It would factor into (x + 9)(x - 4). To double check, multiplying them together results in x2 - 4x + 9x - 36 = x2 + 5x - 36. If the polynomial is 5x2 +7x +2
These are values which make the denominator equal zero, therefore the system described by the transfer function would be unstable near these values.
(a+b)(a-b)