how many real roots does the polynomial 2x^5+8x-7 have
Try graphing it but you will find just one because the function looks like x^5 alone for any x except very close to the origin (close to x = 1). For small x , x<1 it looks like 8x-7 or x = 7/8 approximately for the zero
So does this polynomial, when graphed, intersect the x-axis or it just comes close?
It just intersects at the point (0, -7), right?
To determine the number of real roots of a polynomial, we need to analyze its behavior. In this case, the polynomial is 2x^5 + 8x - 7.
There is a theorem called Descartes' Rule of Signs which helps determine the number of positive and negative roots of a polynomial. However, it does not provide the exact number of real roots, only the maximum possible number.
To find the maximum number of positive real roots, we need to count the number of sign changes in the coefficients when we substitute -x into the expression. In this case, the coefficients are 2, 8, and -7.
Let's substitute -x into the polynomial:
2(-x)^5 + 8(-x) - 7 = -2x^5 - 8x - 7
Looking at the signs of the coefficients (2, -8, -7), we notice one sign change from positive to negative. So, there is at most 1 positive real root.
To find the maximum number of negative real roots, we substitute x into the polynomial:
2x^5 + 8x - 7
Looking at the signs of the coefficients (2, 8, -7), we notice two sign changes. However, we need to verify if there are any duplicate roots. We do this by evaluating the polynomial at 0:
2(0)^5 + 8(0) - 7 = -7
Since the value obtained is negative (a sign change), we can conclude that there are at most 2 negative real roots.
Therefore, the polynomial 2x^5 + 8x - 7 has at most 1 positive real root and at most 2 negative real roots.