Yes, we can conclude that q(x) is ever equal to 0. This is because of the Intermediate Value Theorem, which states that if a function is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in the interval (a, b) such that f(c) = k.

In this case, the function q(x) is a polynomial, and polynomials are continuous everywhere. The values given are q(–12)=–3 and q(–8)=3. So, we have a closed interval [-12, -8] and 0 is a number between -3 and 3. Therefore, there must be some number c in the interval (-12, -8) such that q(c) = 0.

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Yes, we can conclude that q(x) is ever equal to 0. This is because of the Intermediate Value Theorem, which states that if a function is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in the interval (a, b) such that f(c) = k.

In this case, the function q(x) is a polynomial, and polynomials are continuous everywhere. The values given are q(–12)=–3 and q(–8)=3. So, we have a closed interval [-12, -8] and 0 is a number between -3 and 3. Therefore, there must be some number c in the interval (-12, -8) such that q(c) = 0.

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Yes, we can conclude that q(x) is ever equal to 0. This is because of the Intermediate Value Theorem, which states that if a function is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in the interval (a, b) such that f(c) = k.

In this case, the function q(x) is a polynomial, and polynomials are continuous everywhere. The values given are q(–12)=–3 and q(–8)=3. So, we have a closed interval [-12, -8] and 0 is a number between -3 and 3. Therefore, there must be some number c in the interval (-12, -8) such that q(c) = 0.

The function q(x) is a polynomial. If q(–12)=–3 and q(–8)=3, can you conclude that q(x) is ever equal to 0?

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