Problem 72 - A Graphical Approach to College Algebra

Question

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=2 x^{3}-3 x^{2}-5 x+6 ; \quad k=1$$

Step-by-Step Solution

Answer

The linear factors of \(P(x)\) are \((x - 1)(2x - 3)(x + 2)\).

Step 1: Verify k is a Zero of P(x)

Since we are given that \(k = 1\) is a zero of \(P(x)\), we should first confirm that substituting \(x = 1\) into \(P(x)\) results in zero. Calculate \(P(1)\):\[P(1) = 2(1)^3 - 3(1)^2 - 5(1) + 6 = 2 - 3 - 5 + 6 = 0\]Since \(P(1) = 0\), \(k = 1\) is indeed a zero of \(P(x)\).

Step 2: Use Synthetic Division

Perform synthetic division to divide \(P(x)\) by \(x - k = x - 1\):Use the coefficients of \(P(x): 2, -3, -5, 6\).1 | 2 -3 -5 6 | 2 -1 -6 ---------------- 2 -1 -6 0The division gives a quotient of \(2x^2 - x - 6\) and a remainder of 0, confirming \(x - 1\) is a factor.

Step 3: Factor the Quotient Further

Now, factor the quadratic equation \(2x^2 - x - 6\). To factor \(2x^2 - x - 6\), look for two numbers that multiply to \(-12\) (\(2 \times -6\)) and add to \(-1\): The numbers are \(-3\) and \(4\).Rewrite the middle term: \(2x^2 - 3x + 2x - 6\).Factor by grouping:\[(2x^2 - 3x) + (2x - 6) = x(2x - 3) + 2(2x - 3)\]Now factor out the common factor, \(2x - 3\):\[(2x - 3)(x + 2)\]

Step 4: Write Final Factored Form

Combine your result from synthetic division with the factors of the quotient:So, \(P(x)\) can be factored as:\[P(x) = (x - 1)(2x - 3)(x + 2)\]

Key Concepts

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear factor of the form \(x - k\). It is particularly useful for easily determining whether \(k\) is a root of the polynomial.
To perform synthetic division, follow these steps:

• Write down the coefficients of the polynomial in order.
• Place \(k\) on the outside, as if making a division array, but only the divisor is represented (like \(x - k\)).
• Bring down the first coefficient as is.
• Multiply \(k\) by this number and place the result under the next coefficient.
• Add the column of numbers and write the result below. Repeat this process for each coefficient.

After completing these steps, the last number you obtained is the remainder. If it is zero, \(x - k\) is indeed a factor of the polynomial. The other numbers provide the coefficients of the reduced polynomial.

Zero of a Polynomial

The term "zero of a polynomial" refers to any value \(k\) that satisfies the equation \(P(k) = 0\). In other words, it is the value of \(x\) at which the polynomial equals zero.
Factoring a polynomial often involves finding its zeros. When you know a zero of the polynomial, you can divide the polynomial by \(x - k\) (where \(k\) is the zero) to simplify the polynomial into lower degree factors.
Verifying that a specific \(k\) is a zero is straightforward: simply substitute \(k\) into the polynomial and perform the arithmetic. If the result is zero, \(k\) is indeed a zero. This confirmation step was seen in the original problem where inserting \(k = 1\) into the polynomial did result in zero.

Quadratic Factorization

Quadratic factorization refers to the process of breaking down a quadratic expression (like \(ax^2 + bx + c\)) into a product of two simpler expressions, which usually take the form \((px + q)(rx + s)\). To factor a quadratic formula correctly, follow these steps:

• Multiply the leading coefficient \(a\) and the constant term \(c\) to find the product \(ac\).
• Find two numbers that multiply to give \(ac\) and add up to the middle coefficient \(b\).
• Use these two numbers to split the middle term, then factor by grouping.

This technique was used in the solution to factor \(2x^2 - x - 6\) into \((2x - 3)(x + 2)\), making use of number pairs that can reconstruct the expression correctly. The process of quadratic factorization simplifies solving for roots or finding further zeros when the quadratic stands as part of a higher degree polynomial.

Polynomial Roots

Polynomial roots are values for \(x\) where the polynomial equals zero. They are essentially the solutions to the polynomial equation \(P(x) = 0\).
Finding the roots of a polynomial often involves two main strategies: factoring and the quadratic formula. Factoring, as seen in the problem, allows one to break down the polynomial into simpler linear factors. Each factor of the form \(x - r\) suggests a root at \(x = r\).
The roots provide significant insight into the behavior of the polynomial. For example, graphs of polynomials will intersect the x-axis at points corresponding to the roots. In the provided problem, the polynomial was successfully factored into \((x - 1)(2x - 3)(x + 2)\), revealing the roots at \(x = 1\), \(x = \frac{3}{2}\), and \(x = -2\), confirming their utility in finding x-intercepts and analyzing polynomial functions.