Question

which function has real zeros at x = –8 and x = 5? g(x) = x2 3x – 40 g(x) = x2 3x 40 g(x) = x2 14x – 40 g(x) = x2 14x 40

Answer 1

Answer:

Option A is correct.

$x^2+3x-40$

Step-by-step explanation:

Given the real zeroes at x = -8 and x = 5.

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+8) = 0 and (x-5) =0

Zero product property states that if ab = 0 if and only if a =0 and b =0.

By zero product property,

(x+8)(x-5) = 0

Now, distribute each terms of the first polynomial to every term of the second polynomial we get;

$x(x-5) +8(x-5)$

Now, when you multiply two terms together you must multiply the coefficient (numbers) and add the exponent.

$x^2-5x+8x-40$

Combine like terms;

$x^2+3x-40$

therefore, the function $x^2+3x-40$ has real zeros at x = -8 and x =5

Answer 2

Answer:

g(x) = $x^{2} + 3x-40$ for real zero x= -8, x =5.

Step-by-step explanation:

Given : real zeros x =-8 , x =5 .

To find : Function.

Solution : We have given that:

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+8) = 0 and (x-5) =0

Zero product property states that if ab = 0 if and only if a =0 and b =0.

By zero product property,

(x+8)(x-5) = 0

Now, distribute each terms of the first polynomial to every term of the second polynomial we get;

x (x-5 ) +8(x-5)

$x^{2}-5x +8x-40$.

$x^{2} + 3x-40$.

Therefore, g(x) = $x^{2} + 3x-40$ for real zero x= -8, x =5.