Problem 45 - Elementary Algebra

Question

For the following problems, factor the polynomials. $$ (9 a-b) w-(9 a-b) x $$

Step-by-Step Solution

Answer

If so, find the factors. Answer: No, the polynomial \(m^2 + m + 1\) cannot be factored using real numbers since it has complex roots.

Step 1: Check for roots using the quadratic formula

We'll use the quadratic formula to determine the type of roots the given polynomial has. The quadratic formula is given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where \(a\), \(b\), and \(c\) are the coefficients of the polynomial \(ax^2 + bx + c\). For the given polynomial, \(m^2 + m + 1\), we have \(a = 1\), \(b = 1\), and \(c = 1\). Let's substitute these values into the quadratic formula: $$m = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(1)}}{2(1)}$$.

Step 2: Calculate the discriminant

The discriminant (\(\Delta\)) is the part of the quadratic formula that determines the type of roots the polynomial has. It's given by \(\Delta = b^2 - 4ac\), where \(a\), \(b\), and \(c\) are the coefficients of the polynomial \(ax^2 + bx + c\). In this case, we have: $$\Delta = (1)^2 - 4(1)(1) = 1 - 4 = -3$$.

Step 3: Determine the roots and check for factorability

Since the discriminant \(\Delta = -3\), it is negative. When the discriminant is negative, the quadratic polynomial has complex roots and cannot be factored using real numbers.

Step 4: Conclusion

The given polynomial, \(m^2 + m + 1\), cannot be factored using real numbers since it has complex roots. Therefore, the factored form of the polynomial remains the same: $$m^2 + m + 1$$

Key Concepts

Quadratic Formula

When you encounter a quadratic equation of the form \(ax^2 + bx + c = 0\), the quadratic formula is a powerful tool to find its roots. This formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). To use it, you simply insert the coefficients of your equation into the formula.

For instance, with the polynomial \(m^2 + m + 1\), the coefficients are \(a=1\), \(b=1\), and \(c=1\). Plugging these into the quadratic formula gives us the potential roots of the polynomial. However, the nature of these roots hinges on the value of the discriminant within the formula, which leads us into our next key topic.

Discriminant

The discriminant can be found in the quadratic formula under the square root: \(b^2 - 4ac\). It informs us about the nature of the roots the quadratic equation will have. Here is what the different values of the discriminant tell us:

• If \(\Delta > 0\), the equation has two distinct real roots.
• If \(\Delta = 0\), the equation has one real root.
• If \(\Delta < 0\), the equation has complex roots, which cannot be represented by real numbers.

In our exercise with the polynomial \(m^2 + m + 1\), we calculate the discriminant to be \(-3\). Since it's negative, we know without further calculation that the roots will not be real numbers, indicating that the polynomial cannot be factored into real-number factors.

Complex Roots

When the discriminant is negative, as we saw with \(m^2 + m + 1\), the quadratic equation will yield complex roots. These roots take the form of \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. The presence of \(i\), which is the square root of \(-1\), indicates that we're now dealing with numbers outside the realm of real numbers.

Even though complex roots may seem less intuitive, they still provide meaningful solutions in the context of certain mathematical problems and applications. Complex roots come as conjugate pairs, meaning if \(a + bi\) is a root, then \(a - bi\) is also a root. In the context of our exercise, recognizing that a polynomial has complex roots tells us that we won't be able to factor it using real numbers, which is an essential insight for solving such problems.