Problem 73 - A Graphical Approach to College Algebra

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=6 x^{3}+25 x^{2}+3 x-4 ; \quad k=-4$$

Step-by-Step Solution

Answer

The factors of \(P(x)\) are \((x + 4)(3x - 1)(x + 1)\).

Step 1: Understand the Given Problem

We are given a polynomial \(P(x) = 6x^3 + 25x^2 + 3x - 4\) and a root \(k = -4\). Our task is to factor this polynomial into linear factors using the known root.

Step 2: Use Synthetic Division

Since \(k = -4\) is a root of \(P(x)\), \(x + 4\) is a factor. We'll use synthetic division to divide \(P(x)\) by \(x + 4\) to find the quotient, which is a quadratic polynomial.

Step 3: Perform Synthetic Division

Write the coefficients of \(P(x)\) as \([6, 25, 3, -4]\). Perform synthetic division using \(-4\):1. Bring down the 6. 2. Multiply by \(-4\) to get \(-24\). Add to 25 to get 1.3. Multiply by \(-4\) to get \(-4\). Add to 3 to get \(-1\).4. Multiply by \(-4\) to get 4. Add to \(-4\) to get 0.The result is \(6x^2 + x - 1\) with no remainder.

Step 4: Factor the Quadratic

Now, we need to factor \(6x^2 + x - 1\) into linear factors. We can use the quadratic formula to find the roots of \(6x^2 + x - 1 = 0\).

Step 5: Use Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = 1\), and \(c = -1\).1. Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4 \cdot 6 \cdot (-1) = 25\).2. Compute the square root: \(\sqrt{25} = 5\).3. Find the roots: \(x = \frac{-1 \pm 5}{12}\).The roots are \(x = \frac{4}{6} = \frac{2}{3}\) and \(x = \frac{-6}{6} = -1\).

Step 6: Write the Polynomial in Factored Form

The linear factors of \(P(x)\) are \((x + 4)\), \((6x - 2)\), and \((x + 1)\). Hence, the fully factored form of \(P(x)\) is:\[P(x) = (x + 4)(6x - 2)(x + 1)\].

Step 7: Simplify the Factors

The factor \(6x - 2\) can be simplified to \(3x - 1\). Therefore, the fully simplified factored form of \(P(x)\) is:\[P(x) = (x + 4)(3x - 1)(x + 1)\].

Key Concepts

Synthetic Division

Synthetic division is a streamlined way to divide polynomials when you know one of its roots. Here, we used it knowing that the polynomial root is given as \(k = -4\). Since \(k = -4\) is a root, the factor \(x + 4\) should correctly divide the polynomial.For synthetic division, follow these steps:

• Set up by writing down the coefficients of the polynomial, which for the given problem are 6, 25, 3, and -4.
• Use the root \(-4\), the opposite of \(4\) from the factor \(x + 4\).

Start division by bringing down the leading coefficient, which is 6. Multiply 6 by \(-4\) and add this result to the next coefficient (25) to secure the new number in the quotient. This step creates a new sequence of coefficients from which the quadratic factor can be extracted once the remainder is zero, indicating a complete division.In our exercise, after performing synthetic division, the polynomial divides into \(6x^2 + x - 1\) with no remainder.

Quadratic Formula

The next step in factoring involves using the quadratic formula to find the roots of the remaining quadratic, \(6x^2 + x - 1\), derived from synthetic division. The quadratic formula is crucial when factoring quadratics that do not factor neatly using common numbers or factoring techniques.Recall the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, identify \(a = 6\), \(b = 1\), and \(c = -1\). Plug these into the formula, beginning by calculating the discriminant \(b^2 - 4ac\), which helps determine the nature of the roots:

• The discriminant is \(25\), indicating two distinct real roots as it's positive.
• Compute the square root, \(\sqrt{25} = 5\).
• Find the roots: \(x = \frac{-1 \pm 5}{12}\). After completing the calculations, you get the roots \(x = \frac{2}{3}\) and \(x = -1\).

These roots help convert the quadratic into its linear factors.

Linear Factors

The last phase involves writing the polynomial in its linear factor form using the roots and division results. Initially, with a known root of \(k = -4\), we already have a factor \(x + 4\).After achieving the quadratic roots \(\frac{2}{3}\) and \(-1\) from the quadratic formula, these are converted into factors:

• For the root \(\frac{2}{3}\), the associated factor is \(3x - 2\), derived from setting \(3x - 2 = 0\).
• For the root \(-1\), the factor is \(x + 1\).

Initially, from synthetic division, the expression was \((x + 4)(6x - 2)(x + 1)\). Simplifying the middle factor gives \((x + 4)(3x - 1)(x + 1)\), which is the simplest form of the factored polynomial. This completes the full factorization of the polynomial into linear factors, making solving polynomial equations or analyzing properties of the function more accessible.