If a polynomial p(x), has zeros at z1, z2, z3, ...
then p(x) is a multiple of (x - z1)*(x - z2)*(x - z3)...
To get the exact form of p(x) you also need to know the order of each root. If zk has order n then the relevant factor in p(x) is (x - zk)n
A quadratic polynomial must have zeros, though they may be complex numbers.A quadratic polynomial with no real zeros is one whose discriminant b2-4ac is negative. Such a polynomial has no special name.
To find the zeros of the polynomial from the given graph, identify the points where the graph intersects the x-axis. These intersection points represent the values of x for which the polynomial equals zero. If the graph crosses the x-axis at specific points, those x-values are the zeros of the polynomial. If the graph merely touches the x-axis without crossing, those points indicate repeated zeros.
If the cubic polynomial you are given does not have an obvious factorization, then you must use synthetic division. I'm sure wikipedia can tell you all about that.
To find the quadratic polynomial whose zeros are 2 and -3, we can use the fact that a polynomial can be expressed in factored form as ( f(x) = a(x - r_1)(x - r_2) ), where ( r_1 ) and ( r_2 ) are the zeros. Here, substituting ( r_1 = 2 ) and ( r_2 = -3 ), we have ( f(x) = a(x - 2)(x + 3) ). Expanding this, we get ( f(x) = a(x^2 + x - 6) ). For simplicity, we can choose ( a = 1 ), giving us the polynomial ( f(x) = x^2 + x - 6 ).
Yes, a polynomial can have no rational zeros while still having real zeros. This occurs, for example, in the case of a polynomial like (x^2 - 2), which has real zeros ((\sqrt{2}) and (-\sqrt{2})) but no rational zeros. According to the Rational Root Theorem, any rational root must be a factor of the constant term, and if none exist among the possible candidates, the polynomial can still have irrational real roots.
when the equation is equal to zero. . .:)
A quadratic polynomial must have zeros, though they may be complex numbers.A quadratic polynomial with no real zeros is one whose discriminant b2-4ac is negative. Such a polynomial has no special name.
Polynomial fuction in standard form with the given zeros
To find the zeros of the polynomial from the given graph, identify the points where the graph intersects the x-axis. These intersection points represent the values of x for which the polynomial equals zero. If the graph crosses the x-axis at specific points, those x-values are the zeros of the polynomial. If the graph merely touches the x-axis without crossing, those points indicate repeated zeros.
by synthetic division and quadratic equation
Multiply x3 - 2x2 - 13x - 10
If the cubic polynomial you are given does not have an obvious factorization, then you must use synthetic division. I'm sure wikipedia can tell you all about that.
The zeros of a polynomial represent the points at which the graph crosses (or touches) the x-axis.
To find the quadratic polynomial whose zeros are 2 and -3, we can use the fact that a polynomial can be expressed in factored form as ( f(x) = a(x - r_1)(x - r_2) ), where ( r_1 ) and ( r_2 ) are the zeros. Here, substituting ( r_1 = 2 ) and ( r_2 = -3 ), we have ( f(x) = a(x - 2)(x + 3) ). Expanding this, we get ( f(x) = a(x^2 + x - 6) ). For simplicity, we can choose ( a = 1 ), giving us the polynomial ( f(x) = x^2 + x - 6 ).
Yes, a polynomial can have no rational zeros while still having real zeros. This occurs, for example, in the case of a polynomial like (x^2 - 2), which has real zeros ((\sqrt{2}) and (-\sqrt{2})) but no rational zeros. According to the Rational Root Theorem, any rational root must be a factor of the constant term, and if none exist among the possible candidates, the polynomial can still have irrational real roots.
The polynomial is (x + 1)*(x + 1)*(x - 1) = x3 + x2 - x - 1
Since no polynomial was given, no answer will be given.