Question

Factor the polynomial x^4-16. How many real zeros does the function g(x)=x^4-16 have?

Answer 1

Answer:

(x-2)(x+2)(x^2 + 4)

Step-by-step explanation:

This can be done by difference of squares:

(x^2 - 4) (x^2 + 4)

And now (x^2 - 4) can be further factored:

(x-2)(x+2)(x^2 + 4) is your answer

Clearly, There are only 2 zeroes that we can get from this, 2 and -2

Hope this helps!

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Answer 2

Answer:

$\large\boxed{x^4-16=(x^2+4)(x+2)(x-2)}$

$\large\boxed{g(x)=x^4-16\ \text{has two real zeros}\ x=-2\ \text{and}\ x=2}$

Step-by-step explanation:

$x^4-16=x^4-2^4=x^{(2)(2)}-2^{(2)(2)}\\\\\text{use}\ (a^n)^m=a^{(n)(m)}\\\\=(x^2)^2-(2^2)^2\\\\\text{use}\ a^2-b^2=(a+b)(a-b)\\\\=(x^2+2^2)(x^2-2^2)=(x^2+4)(x+2)(x-2)$

The zeros of g(x):

$g(x)=x^2-16=(x^2+4)(x+2)(x-2)\\\\g(x)=0\iff(x^2+4)(x+2)(x-2)=0\iff x^2+4=0\ \vee\ x+2=0\ \vee\ x-2=0\\\\x^2+4=0\qquad\text{subtract 4 from both sides}\\x^2=-4\qquad\bold{FALSE}\\\\x+2=0\qquad\text{subtract 2 from both sides}\\\boxed{x=-2}\\\\x-2=0\qquad\text{add 2 to both sides}\\\boxed{x=2}$