Chapter 4: Problem 40
Find all rational zeros of the polynomial, and write the polynomial in factored form. $$ P(x)=12 x^{3}-20 x^{2}+x+3 $$
Short Answer
Expert verified
The rational zeros are \(\frac{1}{2}\), and \(P(x)\) in factored form is \((2x - 1)(3x + 1)(2x - 3)\).
Step by step solution
01
Apply the Rational Root Theorem
The Rational Root Theorem states that any rational zero of a polynomial, given by \(P(x) = a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0\), has the form \(\frac{p}{q}\) where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\). For the polynomial \(P(x) = 12x^3 - 20x^2 + x + 3\), the possible rational zeros are the factors of \(3\) divided by the factors of \(12\). These include \(\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}\).
02
Test Possible Rational Zeros
We now need to test these possible rational zeros by substituting them into \(P(x)\) and checking if \(P(x) = 0\). Start with \(x = 1\):\(P(1) = 12(1)^3 - 20(1)^2 + 1 + 3 = -4\), which is not zero.Now try \(x = -1\):\(P(-1) = 12(-1)^3 - 20(-1)^2 + (-1) + 3 = -30\), which is not zero.Now try \(x = 3\):\(P(3) = 12(3)^3 - 20(3)^2 + 3 + 3 = 96\), which is not zero.After testing all values, we find that \(P(\frac{1}{2}) = 0\) is a zero.
03
Perform Synthetic Division
Now that we have a zero \(x = \frac{1}{2}\), we use synthetic division to divide \(P(x)\) by \(x - \frac{1}{2}\). Set up the synthetic division:\[\begin{array}{r|rrrr}\frac{1}{2} & 12 & -20 & 1 & 3 \ & & 6 & -7 & -3 \\hline & 12 & -14 & -6 & 0 \\end{array}\]The quotient polynomial is \(12x^2 - 14x - 6\). The remainder is zero, which confirms \(\frac{1}{2}\) as a root. The polynomial can now be written as: \(P(x) = (2x - 1)(12x^2 - 14x - 6)\).
04
Factor the Quadratic Polynomial
We now need to factor the quadratic polynomial \(12x^2 - 14x - 6\). We can try to factor by grouping or use the quadratic formula. Let's try factoring by grouping:Find two numbers that multiply to \(12 \times (-6) = -72\) and add to \(-14\). These are \(-18\) and \(4\). Rewrite the polynomial:\(12x^2 - 18x + 4x - 6\).Group and factor:\(6x(2x - 3) + 2(2x - 3)\).The common factor is \((2x - 3)\), so:\((6x + 2)(2x - 3) = 2(3x + 1)(2x - 3)\).Thus, the factored form of \(P(x)\) is \((2x - 1)\cdot (2) \cdot (3x + 1) \cdot (2x - 3)\).
05
Verify Factored Form
Verify that the product \((2x - 1)(6x + 2)(2x - 3)\) matches the original polynomial:Multiply \((2x - 1)\) and \((6x + 2)\):\( (2x - 1)(6x + 2) = 12x^2 + 4x - 6x - 2 = 12x^2 - 2x - 2 \).Now multiply by \( (2x - 3)\):\((12x^2 - 2x - 2)(2x - 3) = 24x^3 - 36x^2 - 4x + 6x + 4x - 6 \).Simplify to get the original polynomial:\(24x^3 - 40x^2 + x + 3\).Thus, the factored form is confirmed correct:\((2x - 1)(3x + 1)(2x - 3)\), each multiplied by 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Factoring
Polynomial factoring is simply rewriting a polynomial as a product of simpler polynomials, much like breaking a number down into its prime factors. This process can reveal important properties of the polynomial and simplify solving equations.
- For a given polynomial, a possible factorization involves expressing the polynomial as a product of linear or lower-degree polynomials.
- For instance, given a polynomial like \(P(x) = 12x^3 - 20x^2 + x + 3\), once we determine it can be divided by \(x - \frac{1}{2}\), we know it can be factored to that extent.
- The process involves systematically trying potential factors or using techniques like synthetic division and factoring by grouping.
Synthetic Division
Synthetic division is a simplified method for dividing a polynomial by a linear factor of the form \(x - c\). It's faster and more efficient than traditional polynomial long division. This process is especially useful when checking potential roots derived from the Rational Root Theorem.
- To perform synthetic division, list the coefficients of the polynomial in a row as your starting point.
- Bring down the first coefficient, multiply by \(c\) (the root you're testing), and add to the next coefficient, continuing across the row.
- If the division is exact (remainder zero), the value \(c\) is a root of the polynomial, and the resulting coefficients form the quotient polynomial.
Quadratic Polynomial
A quadratic polynomial is a second-degree polynomial with the general form \(ax^2 + bx + c\). Solving quadratic polynomials is straightforward through several methods: factoring, using the quadratic formula, or completing the square.
- In the context of factoring a larger polynomial, the remaining quotient after an initial division can often be a quadratic polynomial, like the \(12x^2 - 14x - 6\) found in our exercise.
- You can factor quadratics by grouping or by identifying two numbers that multiply to \(a \cdot c\) and add to \(b\).
- If straightforward factoring isn't possible, the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) always provides the solution.