Problem 74 - A Graphical Approach to College Algebra

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

The factorization of \(P(x)\) is \((x + 5)(4x - 1)(2x + 3)\).

Step 1: Verify the Zero

Substitute \(k = -5\) into \(P(x)\) to verify that it is indeed a zero of the polynomial. \[ P(-5) = 8(-5)^3 + 50(-5)^2 + 47(-5) - 15 \]Evaluate each term: \[ 8(-125) + 50(25) + 47(-5) - 15 = -1000 + 1250 - 235 - 15 = 0 \]Since \(P(-5) = 0\), \(k = -5\) is confirmed as a zero of \(P(x)\).

Step 2: Perform Synthetic Division

Since \(x + 5\) is a factor of \(P(x)\), we will use synthetic division to divide \(P(x)\) by \(x + 5\). Write down the coefficients: \[ 8, 50, 47, -15 \]Perform the division using \(-5\):1. Bring down 8.2. Multiply 8 by \(-5\), add to 50, yield 10.3. Multiply 10 by \(-5\), add to 47, yield \(-3\).4. Multiply \(-3\) by \(-5\), add to \(-15\), yield 0.The quotient is \(8x^2 + 10x - 3\) with a remainder of 0.

Step 3: Factor the Quotient

Now factor the quadratic \(8x^2 + 10x - 3\). We look for two numbers whose product is \(-24\) (\(8 \times -3\)) and sum is \(10\). These numbers are 12 and \(-2\).Rewrite the middle term and factor by grouping:\[ 8x^2 + 12x - 2x - 3 \]Group and factor:\[ 4x(2x + 3) - 1(2x + 3) \]Factor out \((2x + 3)\):\[ (4x - 1)(2x + 3) \]

Step 4: Write the Complete Factorization

Combine all factors, including the factor from synthetic division. Since \(x + 5\) divided the polynomial without a remainder, the complete factorization of \(P(x)\) is:\[ P(x) = (x + 5)(4x - 1)(2x + 3) \]

Key Concepts

Synthetic Division

Synthetic division is a shorthand method of polynomial division, particularly when dividing by a linear factor. It's a simpler alternative to long division for polynomials and is often used to find the quotient and remainder quickly. To use synthetic division, you need:

• A polynomial that you want to divide, such as \( P(x) = 8x^3 + 50x^2 + 47x - 15 \).
• A known zero or root of the polynomial, given by a linear factor in the form \( x - k \) or \( x + k \). In our exercise, we use \( x + 5 \).
• Since the zero \( k \) is \(-5\), you use this number to perform the synthetic division.

Start by writing down the coefficients of the polynomial: \( 8, 50, 47, -15 \).
Then perform synthetic division by following these steps:

• Write the zero, \(-5\), on the outside.
• Bring down the first coefficient, 8, to start. This is your first number.
• Multiply this number by \(-5\) and add it to the next coefficient. Continue this process through all the coefficients.

The synthetic division yields a quotient of \(8x^2 + 10x - 3\) and confirms \( k = -5 \) as a correct zero since the remainder is 0.

Quadratic Factorization

After using synthetic division, we simplify \(P(x)\) to a quadratic polynomial \(8x^2 + 10x - 3\). Quadratic factorization is the process of breaking this expression into two simpler binomial expressions. Here's how to do it:

• First, find a pair of numbers that multiply to give \(8 \times -3 = -24\).
• These numbers also need to add up to the middle coefficient, which is 10. The numbers 12 and -2 work, as \(12 \times -2 = -24\) and \(12 + (-2) = 10\).

Split the middle term using these numbers: \(8x^2 + 12x - 2x - 3\).
This allows us to group terms for easier factoring:

• Group as \((8x^2 + 12x) + (-2x - 3)\).
• Factor each group. The first group becomes \(4x(2x + 3)\) and the second group becomes \(-1(2x + 3)\).

Finally, factor out the common binomial factor \((2x + 3)\), giving the factorization \((4x - 1)(2x + 3)\).

Zero of a Polynomial

Finding the zero of a polynomial is crucial as it points to values of \(x\) for which the polynomial equals zero. These values provide insight into the roots or solutions of the polynomial equation. In our exercise, we start with a cubic polynomial and are given that \( -5 \) is a zero.

• To verify this zero, substitute \(k = -5\) into the polynomial \(P(x)\).
• Calculate each term: \( P(-5) = 8(-5)^3 + 50(-5)^2 + 47(-5) - 15 \).
• If this computation results in 0, then \( -5 \) is a correct zero for \(P(x)\).

In our case, after substitution, we verified: \(-1000 + 1250 - 235 - 15 = 0\). This confirms \(k = -5\) as a zero of the polynomial.
Subsequently, knowing this zero, we can divide the polynomial by \(x + 5\), simplifying the process of factorization as we pursue further factorization using the resulting quadratic quotient.