Since there are two zeros, we have:
y = (x - (-2))(x - 7)
y = (x + 2)(x - 7)
the zeros of a function is/are the values of the variables in the function that makes/make the function zero. for example: In f(x) = x2 -7x + 10, the zeros of the function are 2 and 5 because these will make the function zero.
It is x^3 - x^2 - 4x + 4 = 0
A zero of a polynomial function - or of any function, for that matter - is a value of the independent variable (often called "x") for which the function evaluates to zero. In other words, a solution to the equation P(x) = 0. For example, if your polynomial is x2 - x, the corresponding equation is x2 - x = 0. Solutions to this equation - and thus, zeros to the polynomial - are x = 0, and x = 1.
A polynomial of degree 2.
x3 - 2x2 - 25x + 50 = 0
the zeros of a function is/are the values of the variables in the function that makes/make the function zero. for example: In f(x) = x2 -7x + 10, the zeros of the function are 2 and 5 because these will make the function zero.
The function ( f(x) = x^2 - 6x + 8 ) is a polynomial function because it is a quadratic expression. To find the zeros, we can factor it as ( (x - 2)(x - 4) ), which gives us the zeros ( x = 2 ) and ( x = 4 ). Thus, the zeros of the function are 2 and 4.
Any multiple of X^2+X/2-1/2
It is x^3 - x^2 - 4x + 4 = 0
To find the zeros of the polynomial function ( f(x) = x^3 - 2x^2 - 8x ), we first factor the expression. We can factor out ( x ) from the polynomial, giving us ( f(x) = x(x^2 - 2x - 8) ). Next, we can factor the quadratic ( x^2 - 2x - 8 ) into ( (x - 4)(x + 2) ), leading to ( f(x) = x(x - 4)(x + 2) ). Therefore, the zeros of the function are ( x = 0 ), ( x = 4 ), and ( x = -2 ).
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.
Try the quadratic formula. X = -b ± (sqrt(b^2-4ac)/2a)
A zero of a polynomial function - or of any function, for that matter - is a value of the independent variable (often called "x") for which the function evaluates to zero. In other words, a solution to the equation P(x) = 0. For example, if your polynomial is x2 - x, the corresponding equation is x2 - x = 0. Solutions to this equation - and thus, zeros to the polynomial - are x = 0, and x = 1.
A polynomial function of least degree with rational coefficients and a leading coefficient of 1 that has the zeros -7 and -4 can be constructed using the fact that if ( r ) is a zero, then ( (x - r) ) is a factor. Therefore, the polynomial can be expressed as ( f(x) = (x + 7)(x + 4) ). Expanding this, we get ( f(x) = x^2 + 11x + 28 ). Thus, the polynomial function is ( f(x) = x^2 + 11x + 28 ).
3y2-5xyz yay i figured it out!!!!
A polynomial of degree 2.
A polynomial of degree ( n ) can have at most ( n ) distinct zeros (roots) in the complex number system, according to the Fundamental Theorem of Algebra. These zeros may be real or complex, and they can also be repeated, meaning a polynomial can have fewer than ( n ) distinct zeros if some are counted multiple times (multiplicity). For example, a polynomial of degree 3 could have 3 distinct zeros, 2 distinct zeros (one with multiplicity 2), or 1 distinct zero (with multiplicity 3).