given the following polynomial find the zero and multiply of each
were the graph touches the x axis
number of turning points
fx =(x-4)(x^2-25)
0 = (x-4)(x+5)(x-5)
Set each equal to zero and solve for x. That will give you three different zeros. Each occurs only once.
f(x) = (x-4)(x-5)(x+5)
3 roots, each of multiplicity 1
The graph crosses the axis at each root, so it has to turn around 2 times.
4
-5
5
To find the zeros of a polynomial, we set the polynomial equal to zero and solve for the value(s) of 'x' that make the equation true. Zeros of a polynomial are also known as the x-intercepts or roots of the polynomial.
1. Set the polynomial equal to zero:
(x - 4)(x^2 - 25) = 0
2. Set each factor equal to zero and solve for 'x':
x - 4 = 0
x = 4 (This is the zero/multiplicity 1)
x^2 - 25 = 0
(x - 5)(x + 5) = 0
x - 5 = 0 --> x = 5 (This is the zero/multiplicity 1)
x + 5 = 0 --> x = -5 (This is the zero/multiplicity 1)
Therefore, the zeros of the polynomial are x = 4, x = 5, and x = -5. The multiplicities of the zeros are all 1, which indicates that each zero touches the x-axis once (the graph intersects the x-axis at these points).
To find the number of turning points of a polynomial, we look at the degree of the polynomial. The degree of a polynomial is determined by the highest power of 'x' in the expression.
In this case, the degree of the polynomial is 3 (x^2 is the largest power of 'x').
The number of turning points in a polynomial is always one less than the degree. Therefore, in this case, there are 2 turning points in the graph of the polynomial fx = (x - 4)(x^2 - 25).