If the solution of p(x)=0 are -4 and 11, which function could be p?
If the solutions to the polynomial equation p(x) = 0 are -4 and 11, this means that x = -4 and x = 11 are roots of the polynomial p(x). Therefore, p(x) must have factors corresponding to these roots. The factors related to the roots -4 and 11 would be (x + 4) and (x - 11), respectively.
Thus, a polynomial function p(x) that has -4 and 11 as solutions would be given by:
p(x) = a(x + 4)(x - 11)
where "a" is a non-zero constant. This constant determines the leading coefficient of the polynomial and can be any real number. If "a" is chosen to be 1, the polynomial is monic (the leading coefficient is 1). Therefore, the simplest form of the polynomial with roots -4 and 11 is:
p(x) = (x + 4)(x - 11)
Expanding this product gives:
p(x) = x^2 - 11x + 4x - 44
p(x) = x^2 - 7x - 44
So, p(x) = x^2 - 7x - 44 could be a possible polynomial with the solutions x = -4 and x = 11.
However, if you choose a different non-zero value for "a," you would get different but equivalent polynomials, all of which would have the same solutions of -4 and 11. For example, if a = 2, the polynomial would be:
p(x) = 2(x + 4)(x - 11)
p(x) = 2(x^2 - 7x - 44)
p(x) = 2x^2 - 14x - 88
Both polynomials have the same roots but different coefficients.
So what's the answer?
The answer is that there are infinitely many polynomials p(x) that could have -4 and 11 as solutions for the equation p(x) = 0, as long as the polynomial includes the factors (x + 4) and (x - 11). Each polynomial will differ by a constant factor "a" (which can be any non-zero number), as shown in the general form:
p(x) = a(x + 4)(x - 11)
Here are a few examples of specific polynomials that have -4 and 11 as solutions:
1. For a = 1:
p(x) = (x + 4)(x - 11)
p(x) = x^2 - 11x + 4x - 44
p(x) = x^2 - 7x - 44
2. For a = 2:
p(x) = 2(x + 4)(x - 11)
p(x) = 2(x^2 - 11x + 4x - 44)
p(x) = 2x^2 - 14x - 88
3. For a = -3:
p(x) = -3(x + 4)(x - 11)
p(x) = -3(x^2 - 11x + 4x - 44)
p(x) = -3x^2 + 33x - 12x + 132
p(x) = -3x^2 + 21x + 132
All of these polynomials (and any others created using different non-zero values of "a") share the same roots of -4 and 11. There isn't a single unique polynomial with these roots; instead, there is a family of polynomials defined by that factor "a."
To find a function that could be p (x) with solutions -4 and 11, we need to find a polynomial that has these roots.
A polynomial with these roots can be written in factored form as follows:
p(x) = a(x - (-4))(x - 11)
Where (x - (-4)) represents one factor which gives the solution -4, and (x - 11) represents another factor which gives the solution 11.
Expanding the equation, we get:
p(x) = a(x + 4)(x - 11)
Now, the specific value for 'a' will depend on the context or any additional information given. If no other information is available, 'a' can be any non-zero real number.
So, a possible function that could be p(x) is:
p(x) = a(x + 4)(x - 11)