Respuesta :
Answer:
Option B, D )the function [tex]x^{2} -5x+4[/tex] has real zeros at x = 1 and x =4.
Step-by-step explanation:
Given : Real zero x = 1, x = 4.
To find : Which functions have real zeros at 1 and 4.
Solution : We have real zeros x =1 , x=4.
Factor theorem states that (x-r) is a factor of the polynomial function f(x) if and only if r is a root of the function f(x).
Since, we know that the root of the function i.e f(x) are -8 and 5 then the function has the following factor:
(x-1) = 0 and (x-4) =0
Zero product property states that if ab = 0 if and only if a =0 and b =0.
By zero product property,
(x-1)(x-4) = 0
Now, distribute each terms of the first polynomial to every term of the second polynomial we get;
Now, when you multiply two terms together you must multiply the coefficient (numbers) and add the exponent.
x(x-4) -1(x-4) = 0
[tex]x^{2} -4x-x+4[/tex]= 0
Combine like terms;
[tex]x^{2} -5x+4[/tex] = 0
Since B and we can see D
[tex]-2x^{2} +10x-8[/tex]= 0
Taking common -2
-2([tex]x^{2} -5x+4[/tex]) = 0
On dividing by -2 both side
[tex]x^{2} -5x+4[/tex]
Therefore, Option B, D )the function [tex]x^{2} -5x+4[/tex] has real zeros at x = 1 and x =4.
You can use the fact that if a polynomial has a root at a, then that polynomial will have factor (x-a) assuming that polynomial is of single variable x.
The function having real zeros at 1 and 4 is given by
Option B: [tex]f(x) = x^2 - 5x +4[/tex]
What is a root of a single variable polynomial function p(x) ?
Root of a polynomial function p(x) is a value of x such that putting that value in place of x makes the output 0.
Thus, if x = k is a root of the polynomial p(x), then
p(k) = 0
It is also a fact that if x = k is a root of the polynomial p(x), then we have (x-k) as its factor.
Thus,
p(x) = g(x) (x - k)
where g(x) is another polynomial.
Using the above fact to find the functions having zeros at 1 and 4:
Zeros, also called roots, are those values for which the function outputs 0.
Let the quadratic(2 degree) polynomial be f(x),
Then by above rule, we have the the factors of f(x) as (x-1) and (x-4) since 1 and 4 are roots of f(x) for which f(x) = 0
Now, that means,
f(x) = g(x)(x-4)(x-1)
Since the given options are quadratic, thus g(x) must be a constant as (x-4)(x-1) is already forming a quadratic polynomial.
Let g(x) = "a" (a constant)
Then, we have:
[tex]f(x) = a(x-4)(x-1) = a(x^2 - 4x- x + 4) = a(x^2 - 5x +4)[/tex]
Seeing it in the options, we see that option B fits with value of a = 1
Thus, we have:
The function having real zeros at 1 and 4 is given by
Option B: [tex]f(x) = x^2 - 5x +4[/tex]
Learn more about polynomials here:
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