Problem 74 - A Graphical Approach to Precalculus with Limits

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=8 x^{3}+50 x^{2}+47 x-15 ; \quad k=-5$$

Step-by-Step Solution

Answer

Factorization is \((x + 5)(x - \frac{1}{2})(x + 1.5)\).

Step 1: Understand the problem

We need to factor the cubic polynomial \(P(x) = 8x^3 + 50x^2 + 47x - 15\) into linear factors, knowing that \(x = -5\) is one of its zeros. This means \(x + 5\) is a factor of \(P(x)\).

Step 2: Use Synthetic Division

First, divide \(P(x)\) by \(x + 5\) using synthetic division. This will help us find the quotient, which is a quadratic polynomial that, together with \(x + 5\), forms the complete factorization of \(P(x)\).

Step 3: Set up Synthetic Division

Write the coefficients of \(P(x)\), which are 8, 50, 47, and -15. Using synthetic division, write -5 on the left and draw a line. Follow the synthetic division process, adding and multiplying as appropriate.

Step 4: Perform Synthetic Division

Start the synthetic division. Bring down 8, then multiply it by -5 to get -40. Add -40 to 50 to get 10. Multiply 10 by -5 to get -50, then add -50 to 47 to get -3. Multiply -3 by -5 to get 15, and finally, add 15 to -15 to get 0. This confirms \(x + 5\) is a factor of \(P(x)\), as the remainder is 0. The result is a new polynomial: \(8x^2 + 10x - 3\).

Step 5: Factor the Quadratic Polynomial

Next, factor the quadratic polynomial \(8x^2 + 10x - 3\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = 10\), and \(c = -3\). Calculate the discriminant \(b^2 - 4ac = 100 + 96 = 196\).

Step 6: Solve for the Roots Using the Quadratic Formula

Solve \(x = \frac{-10 \pm \sqrt{196}}{16}\). \(\sqrt{196} = 14\), so the roots are \(x = \frac{-10 + 14}{16} = \frac{1}{2}\) and \(x = \frac{-10 - 14}{16} = -1.5\). Thus, the polynomial factors are \(x - \frac{1}{2}\) and \(x + 1.5\).

Step 7: Write Final Factorization

Combine all the linear factors to express \(P(x)\) as \(P(x) = (x + 5)(x - \frac{1}{2})(x + 1.5)\). Since these adjustments maintain equivalence, this is the factorization of the original polynomial.

Key Concepts

Synthetic Division

Synthetic Division is a simplified method used to divide polynomials. It is especially handy when dividing by a linear binomial of the form \(x - c\). In this method, you only work with the coefficients of the polynomial, making the process less cumbersome than long division. Here's how it works:

• Write down the coefficients of the polynomial you want to divide, in this case: 8, 50, 47, and -15.
• On the left side, write the zero you're using, here it's -5 because \(x + 5\) is a factor, so the opposite sign is used.
• Draw a horizontal line below the coefficients and another parallel line to the side of the zero.

The process involves bringing down the leading coefficient, multiplying it by the zero, adding the result to the next coefficient, and repeating. When it's complete and the remainder is zero, it confirms that the divisor is a factor.

Quadratic Formula

The Quadratic Formula is a reliable method for finding the roots of a quadratic equation of the form \(ax^2 + bx + c = 0\). This formula is your go-to tool when factoring a quadratic is not straightforward. The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's how it applies to our example:

• Identify coefficients: \(a = 8\), \(b = 10\), \(c = -3\).
• Calculate the discriminant: \(b^2 - 4ac = 100 + 96 = 196\).
• Take the square root: \(\sqrt{196} = 14\).
• Use the formula to find roots: \(x = \frac{-10 \pm 14}{16}\).

The result gives us the roots \(x = \frac{1}{2}\) and \(x = -1.5\), which further factor the quadratic portion of our original polynomial.

Cubic Polynomial

A Cubic Polynomial is a polynomial of degree three, typically in the form \(ax^3 + bx^2 + cx + d\). The highest exponent, which is three, dictates the shape of the graph and can have up to three real roots. In our example, the cubic polynomial given is:\(8x^3 + 50x^2 + 47x - 15\).Cubic polynomials can be challenging to tackle, especially when factoring. Finding one root (in our case, \(x = -5\)) simplifies the process significantly, because you can use it to perform synthetic division and reduce the polynomial to a quadratic one, which is much easier to manage.

Zeros of Polynomial

Zeros of a polynomial are the values of \(x\) that make the polynomial equal to zero. In simpler terms, they are the solutions to the polynomial equation \(P(x) = 0\). These zeros are crucial because they correspond to the x-intercepts of the polynomial graph. In our exercise, we are initially given one zero \(k = -5\). This information directly helps us identify one of the factors of the polynomial as \(x + 5\). Once one zero is known, the remaining zeros (if any) can be found by factoring the reduced polynomial. After using synthetic division with \(x + 5\) and applying the quadratic formula, we determined the complete set of zeros:\[-5, \frac{1}{2}, \text{and } -1.5.\]Understanding zeros is vital for polynomial factorization, as they provide a complete picture of how the polynomial behaves and are essential for creating a fully factored form.