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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

User Jupenur
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Final answer:

The functions that have real zeros at 1 and 4 are f(x) = x^2 - 5x + 4 and f(x) = -2x^2 + 10x - 8.

Step-by-step explanation:

To find the functions that have real zeros at 1 and 4, we need to substitute these values into each function and check if the result is 0. Let's check each function:

A. f(x) = x2 + x + 4: f(1) = 12 + 1 + 4 = 1 + 1 + 4 = 6 (not zero), f(4) = 42 + 4 + 4 = 16 + 4 + 4 = 24 (not zero).

B. f(x) = x2 – 5x + 4: f(1) = 12 – 5(1) + 4 = 1 - 5 + 4 = 0 (zero), f(4) = 42 – 5(4) + 4 = 16 - 20 + 4 = 0 (zero).

C. f(x) = x2 + 3x – 4: f(1) = 12 + 3(1) – 4 = 1 + 3 – 4 = 0 (zero), f(4) = 42 + 3(4) – 4 = 16 + 12 – 4 = 24 (not zero).

D. f(x) = –2x2 + 10x – 8: f(1) = –2(1)2 + 10(1) – 8 = –2 + 10 – 8 = 0 (zero), f(4) = –2(4)2 + 10(4) – 8 = –32 + 40 – 8 = 0 (zero).

E. f(x) = –4x2 – 16x – 1: f(1) = –4(1)2 – 16(1) – 1 = –4 – 16 – 1 = -21 (not zero), f(4) = –4(4)2 – 16(4) – 1 = –64 – 64 – 1 = -129 (not zero).

Based on these calculations, the functions that have real zeros at 1 and 4 are B. f(x) = x2 – 5x + 4 and D. f(x) = –2x2 + 10x – 8.

User Zooba
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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.

User Stroi
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