Question

Which functions have real zeros at 1 and 4? Check all that apply. A.f(x) = x2 + x + 4 B.f(x) = x2 – 5x + 4 C. f(x) = x2 + 3x – 4 D.f(x) = –2x2 + 10x – 8 E.f(x) = –4x2 – 16x – 1

Answer 1

Answer:

b and d

Step-by-step explanation:

Answer 2

Answer:

Option B, D )the function $x^{2} -5x+4$ 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

$x^{2} -4x-x+4$= 0

Combine like terms;

$x^{2} -5x+4$ = 0

Since B and we can see D

$-2x^{2} +10x-8$= 0

Taking common -2

-2($x^{2} -5x+4$) = 0

On dividing by -2 both side

$x^{2} -5x+4$

Therefore,  Option B, D )the function $x^{2} -5x+4$ has real zeros at x = 1 and x =4.