Hello,
Third degree and zeroes are -3, -1 and 2 so we can write the function as below, a being a real
[tex]a(x+3)(x+1)(x-2)[/tex]
and we know that f(1)=7
so
a * 4 * 2 *(-1) = -8a=7
a = -7/8
Then
[tex]\dfrac{-7(x+3)(x+1)(x-2)}{8}[/tex]
thanks
Using the Factor Theorem, the polynomial function is given by:
[tex]p(x) = -\frac{7}{8}(x^3 + 2x^2 - 5x - 6)[/tex]
The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, ..., x_n[/tex] is given by:
[tex]p(x) = a(x - x_1)(x - x_2)...(x - x_n)[/tex]
In which a is the leading coefficient.
In this problem, the zeros are -3, -1 and 2, hence [tex]x_1 = -3, x_2 = -1, x_3 = 2[/tex].
The function is given by:
[tex]p(x) = a(x - x_1)(x - x_2)...(x - x_n)[/tex]
[tex]p(x) = a(x + 3)(x + 1)(x - 2)[/tex]
[tex]p(x) = a(x^2 + 4x + 3)(x - 2)[/tex]
[tex]p(x) = a(x^3 + 2x^2 - 5x - 6)[/tex]
It passes through the point (1,7), which means that when [tex]x = 1, p = 7[/tex], and this us used to find a.
[tex]7 = a(1 + 2 - 5 - 6)[/tex]
[tex]-8a = 7[/tex]
[tex]a = -\frac{7}{8}[/tex]
Hence, the polynomial is:
[tex]p(x) = -\frac{7}{8}(x^3 + 2x^2 - 5x - 6)[/tex]
A similar problem is given at https://brainly.com/question/25312366
