Understanding the Factor Theorem in Algebra

The question asks us to identify which of the given expressions has \((mx + n)\) as a factor. To solve this, we can use the Factor Theorem.

Applying the Factor Theorem

The Factor Theorem states that if \((ax + b)\) is a factor of a polynomial \(P(x)\), then \(P\left(-\frac{b}{a}\right) = 0\). In our case, the potential factor is \((mx + n)\). Comparing this to \((ax + b)\), we have \(a = m\) and \(b = n\).

According to the Factor Theorem, if \((mx + n)\) is a factor, then substituting the root of \((mx + n) = 0\) into the polynomial should result in zero.

First, let's find the value of \(x\) for which \((mx + n) = 0\):

• \(mx + n = 0\)
• \(mx = -n\)
• \(x = -\frac{n}{m}\) (assuming \(m \neq 0\))

Now, we will substitute \(x = -\frac{n}{m}\) into each of the given options and check which expression evaluates to 0.

Testing the Options

Let's evaluate each expression with \(x = -\frac{n}{m}\).

Option 1: \(m^2x^2 + 2nx + n^2\)

Substitute \(x = -\frac{n}{m}\):

\(m^2\left(-\frac{n}{m}\right)^2 + 2n\left(-\frac{n}{m}\right) + n^2\)

\(= m^2\left(\frac{n^2}{m^2}\right) - \frac{2n^2}{m} + n^2\)

\(= n^2 - \frac{2n^2}{m} + n^2\)

\(= 2n^2 - \frac{2n^2}{m}\)

This expression is not necessarily 0.

Option 2: \(m^2x^2 + 2mnx + n^2\)

Substitute \(x = -\frac{n}{m}\):

\(m^2\left(-\frac{n}{m}\right)^2 + 2mn\left(-\frac{n}{m}\right) + n^2\)

\(= m^2\left(\frac{n^2}{m^2}\right) - 2n^2 + n^2\)

\(= n^2 - 2n^2 + n^2\)

\(= 0\)

Since substituting \(x = -\frac{n}{m}\) into this expression results in 0, \((mx + n)\) is a factor of this expression.

Verification by Factoring

Let's examine the expression from Option 2: \(m^2x^2 + 2mnx + n^2\). We can try to factor this expression directly.

Notice the form of the expression:

• The first term, \(m^2x^2\), is the square of \(mx\), i.e., \((mx)^2\).
• The last term, \(n^2\), is the square of \(n\), i.e., \((n)^2\).
• The middle term, \(2mnx\), is \(2\) times the product of \(mx\) and \(n\), i.e., \(2(mx)(n)\).

This matches the pattern of a perfect square trinomial: \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = mx\) and \(b = n\).

So, \(m^2x^2 + 2mnx + n^2\) can be factored as \((mx + n)^2\).

\((mx + n)^2 = (mx + n)(mx + n)\)

This confirms that \((mx + n)\) is indeed a factor of \(m^2x^2 + 2mnx + n^2\).

Conclusion

Using either the Factor Theorem or direct factoring, we find that \((mx + n)\) is a factor only of the expression \(m^2x^2 + 2mnx + n^2\).

Revision Table: Key Concepts

Additional Information: Factoring and Roots

Understanding factoring and roots is crucial in algebra. When we say \((mx + n)\) is a factor of a polynomial, it means that \((mx + n)\) divides the polynomial evenly, leaving no remainder. The roots (or zeros) of a polynomial are the values of \(x\) for which the polynomial equals zero. The Factor Theorem connects factors and roots: if \((x - k)\) is a factor, then \(k\) is a root, and vice-versa.

For a linear factor \((ax + b)\), the root is \(x = -b/a\). If substituting this root into a polynomial \(P(x)\) gives \(P(-b/a) = 0\), then \((ax + b)\) is a factor of \(P(x)\).

In this specific problem, the expression \(m^2x^2 + 2mnx + n^2\) is a quadratic expression. Quadratic expressions can often be factored into two linear factors. Recognizing it as a perfect square \((mx+n)^2\) directly shows its factors are \((mx+n)\) and \((mx+n)\).

This problem demonstrates two ways to check for factors: using the Factor Theorem by testing the root, and by attempting to factor the expression directly.