Understanding the Factor Theorem

Oct 13, 2025 by pynkocean.com 125 views

Understanding the Problem

When dealing with polynomials, it's essential to understand the properties and characteristics of these mathematical expressions. In this case, we're given a quadratic polynomial in the form of $$x^2+8x-6$$, and we need to determine which statement is true about this polynomial.

Factors of a Polynomial

A factor of a polynomial is an expression that divides the polynomial without leaving a remainder. In other words, if we have a polynomial $$p(x)$$ and an expression $$q(x)$$, then $$q(x)$$ is a factor of $$p(x)$$ if $$p(x) = q(x) \cdot r(x)$$ for some polynomial $$r(x)$$. In this case, we're looking for factors of the polynomial $$x^2+8x-6$$.

Option A: $$(x-2)$$ is a factor of the polynomial

To determine if $$(x-2)$$ is a factor of the polynomial, we need to check if $$x^2+8x-6$$ can be divided by $$(x-2)$$ without leaving a remainder. We can do this by using polynomial long division or synthetic division. If we divide $$x^2+8x-6$$ by $$(x-2)$$, we get a quotient of $$x+6$$ and a remainder of $$0$$. This means that $$(x-2)$$ is indeed a factor of the polynomial.

Option B: $$(x+2)$$ is a factor of the polynomial

To determine if $$(x+2)$$ is a factor of the polynomial, we need to check if $$x^2+8x-6$$ can be divided by $$(x+2)$$ without leaving a remainder. We can do this by using polynomial long division or synthetic division. If we divide $$x^2+8x-6$$ by $$(x+2)$$, we get a quotient of $$x+4$$ and a remainder of $$-10$$. This means that $$(x+2)$$ is not a factor of the polynomial.

Option C: $$(x-4)$$ is a factor of the polynomial

To determine if $$(x-4)$$ is a factor of the polynomial, we need to check if $$x^2+8x-6$$ can be divided by $$(x-4)$$ without leaving a remainder. We can do this by using polynomial long division or synthetic division. If we divide $$x^2+8x-6$$ by $$(x-4)$$, we get a quotient of $$x+10$$ and a remainder of $$-46$$. This means that $$(x-4)$$ is not a factor of the polynomial.

Option D: The polynomial is prime

A prime polynomial is a polynomial that has no factors other than itself and the number 1. In other words, a polynomial is prime if it cannot be divided by any other polynomial without leaving a remainder. However, we've already established that $$(x-2)$$ is a factor of the polynomial, which means that the polynomial is not prime.

Conclusion

Based on our analysis, we can conclude that the correct answer is:

A. $$(x-2)$$ is a factor of the polynomial.

This is because we've shown that $$(x-2)$$ is indeed a factor of the polynomial $$x^2+8x-6$$.

Understanding the Factor Theorem

The factor theorem states that if $$(x-a)$$ is a factor of a polynomial $$p(x)$$, then $$p(a) = 0$$. In other words, if we substitute $$a$$ into the polynomial and get a result of 0, then $$(x-a)$$ is a factor of the polynomial. We can use this theorem to verify our result.

Verifying the Factor Theorem

To verify the factor theorem, we need to substitute $$a=2$$ into the polynomial $$x^2+8x-6$$. If we do this, we get:

(2)^2+8(2)-6 = 4+16-6 = 14